{"id":98,"date":"2019-08-20T17:02:26","date_gmt":"2019-08-20T21:02:26","guid":{"rendered":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/chapter\/exponential-functions\/"},"modified":"2022-06-01T10:39:28","modified_gmt":"2022-06-01T14:39:28","slug":"exponential-functions","status":"publish","type":"chapter","link":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/chapter\/exponential-functions\/","title":{"raw":"Exponential Functions","rendered":"Exponential Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\nIn this section, you will:\n<ul>\n \t<li>Evaluate exponential functions.<\/li>\n \t<li>Find the equation of an exponential function.<\/li>\n \t<li>Use compound interest formulas.<\/li>\n \t<li>Evaluate exponential functions with base e.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137640062\">India is the second most populous country in the world with a population of about[latex]\\,1.25\\,[\/latex]billion people in 2013. The population is growing at a rate of about[latex]\\,1.2%\\,[\/latex]each year[footnote]<a href=\"http:\/\/www.worldometers.info\/world-population\/\">http:\/\/www.worldometers.info\/world-population\/<\/a>. Accessed February 24, 2014.[\/footnote] . If this rate continues, the population of India will exceed China\u2019s population by the year[latex]\\,2031.[\/latex]When populations grow rapidly, we often say that the growth is \u201cexponential,\u201d meaning that something is growing very rapidly. To a mathematician, however, the term <em>exponential growth <\/em>has a very specific meaning. In this section, we will take a look at <em>exponential functions<\/em>, which model this kind of rapid growth.<\/p>\n\n<div id=\"fs-id1165135159940\" class=\"bc-section section\">\n<h3>Identifying Exponential Functions<\/h3>\n<p id=\"fs-id1165137446191\">When exploring linear growth, we observed a constant rate of change\u2014a constant number by which the output increased for each unit increase in input. For example, in the equation[latex]\\,f\\left(x\\right)=3x+4,[\/latex]the slope tells us the output increases by 3 each time the input increases by 1. The scenario in the India population example is different because we have a <em>percent<\/em> change per unit time (rather than a constant change) in the number of people.<\/p>\n\n<div id=\"fs-id1165137477143\" class=\"bc-section section\">\n<h4>Defining an Exponential Function<\/h4>\n<p id=\"fs-id1165137553324\">A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, 2.5% of the population was vegan, adhering to a diet that does not include any animal products\u2014no meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up 10% of the U.S. population in 2015, 40% in 2019, and 80% in 2021.<\/p>\n<p id=\"fs-id1165134069131\">What exactly does it mean to <em>grow exponentially<\/em>? What does the word <em>double <\/em>have in common with <em>percent increase<\/em>? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media.<\/p>\n\n<ul id=\"fs-id1165134042783\">\n \t<li><strong>Percent change <\/strong>refers to a <em>change<\/em> based on a <em>percent<\/em> of the original amount.<\/li>\n \t<li><strong>Exponential growth <\/strong>refers to an <em>increase<\/em> based on a constant multiplicative rate of change over equal increments of time, that is, a <em>percent<\/em> increase of the original amount over time.<\/li>\n \t<li>Exponential decay refers to a <em>decrease<\/em> based on a constant multiplicative rate of change over equal increments of time, that is, a <em>percent<\/em> decrease of the original amount over time.<\/li>\n<\/ul>\n<p id=\"fs-id1165137760753\">For us to gain a clear understanding of <span class=\"no-emphasis\">exponential growth<\/span>, let us contrast exponential growth with <span class=\"no-emphasis\">linear growth<\/span>. We will construct two functions. The first function is exponential. We will start with an input of 0, and increase each input by 1. We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of 0, and increase each input by 1. We will add 2 to the corresponding consecutive outputs. See <a class=\"autogenerated-content\" href=\"#Table_04_01_01\">(Figure)<\/a>.<\/p>\n\n<table id=\"Table_04_01_01\" summary=\"Eight rows and three columns. The first column is labeled, \u201cx\u201d, which goes from 0 to 6; the second column is labeled, \u201cf(x)=2^x\u201d; and the third column is labeled, \u201cg(x) = 2x\u201d. The following values are for the function f: (0, 1), (1, 2), (2, 4), (3, 8), (4, 16), (5, 32), and (6, 64). The following values are for the function g: (0, 0), (1, 2), (2, 4), (3, 6), (4, 8), (5, 10), and (6, 12).\">\n<thead>\n<tr>\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/th>\n<th>[latex]g\\left(x\\right)=2x[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>1<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>2<\/td>\n<td>2<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>4<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>8<\/td>\n<td>6<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>16<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>32<\/td>\n<td>10<\/td>\n<\/tr>\n<tr>\n<td>6<\/td>\n<td>64<\/td>\n<td>12<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165135190676\">From <a class=\"autogenerated-content\" href=\"#Table_04_01_01\">(Figure)<\/a> we can infer that for these two functions, exponential growth dwarfs linear growth.<\/p>\n\n<ul id=\"fs-id1165137725808\">\n \t<li><strong>Exponential growth <\/strong>refers to the original value from the range increases by the <em>same percentage<\/em> over equal increments found in the domain.<\/li>\n \t<li><strong>Linear growth<\/strong> refers to the original value from the range increases by the <em>same amount<\/em> over equal increments found in the domain.<\/li>\n<\/ul>\n<p id=\"fs-id1165137561507\">Apparently, the difference between \u201cthe same percentage\u201d and \u201cthe same amount\u201d is quite significant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding 2 to the output whenever the input was increased by one.<\/p>\n<p id=\"fs-id1165135445949\">The general form of the <span class=\"no-emphasis\">exponential function<\/span> is[latex]\\,f\\left(x\\right)=a{b}^{x},\\,[\/latex]where[latex]\\,a\\,[\/latex]is any nonzero number,[latex]\\,b\\,[\/latex]is a positive real number not equal to 1.<\/p>\n\n<ul id=\"fs-id1165137635065\">\n \t<li>If[latex]\\,b&gt;1,[\/latex]the function grows at a rate proportional to its size.<\/li>\n \t<li>If[latex]\\,0&lt;b&lt;1,[\/latex] the function decays at a rate proportional to its size.<\/li>\n<\/ul>\n<p id=\"fs-id1165137465225\">Let\u2019s look at the function[latex]\\,f\\left(x\\right)={2}^{x}\\,[\/latex]from our example. We will create a table (<a class=\"autogenerated-content\" href=\"#Table_04_01_02\">(Figure)<\/a>) to determine the corresponding outputs over an interval in the domain from[latex]\\,-3\\,[\/latex]to[latex]\\,3.[\/latex]<\/p>\n\n<table summary=\"Two rows and eight columns. The first row is labeled, \u201cx\u201d, and the second row is labeled, \u201cf(x)=2^x\u201d. Reading the columns as ordered pairs, we have the following values: (-3, 2^(-3)=1\/8), (-2, 2^(-2)=1\/4), (-1, 2^(-1)=1\/2), (0, 2^(0)=1), (1, 2^(1)=2), (2, 2^(2)=4), and (3, 2^(3)=8).\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>[latex]-3[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\n<td>[latex]{2}^{-3}=\\frac{1}{8}[\/latex]<\/td>\n<td>[latex]{2}^{-2}=\\frac{1}{4}[\/latex]<\/td>\n<td>[latex]{2}^{-1}=\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]{2}^{0}=1[\/latex]<\/td>\n<td>[latex]{2}^{1}=2[\/latex]<\/td>\n<td>[latex]{2}^{2}=4[\/latex]<\/td>\n<td>[latex]{2}^{3}=8[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137640874\">Let us examine the graph of[latex]\\,f\\,[\/latex]by plotting the ordered pairs we observe on the table in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_006\">(Figure)<\/a>, and then make a few observations.<\/p>\n\n<div id=\"CNX_Precalc_Figure_04_01_006\" class=\"medium aligncenter\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140059\/CNX_Precalc_Figure_04_01_006.jpg\" alt=\"Graph of Companies A and B\u2019s functions, which values are found in the previous table.\" width=\"731\" height=\"479\"> <strong>Figure 1.<\/strong>[\/caption]\n\n<\/div>\n<p id=\"fs-id1165137408862\">Let\u2019s define the behavior of the graph of the exponential function[latex]\\,f\\left(x\\right)={2}^{x}\\,[\/latex]and highlight some its key characteristics.<\/p>\n\n<ul id=\"fs-id1165137566018\">\n \t<li>the domain is[latex]\\,\\left(-\\infty ,\\infty \\right),[\/latex]<\/li>\n \t<li>the range is[latex]\\,\\left(0,\\infty \\right),[\/latex]<\/li>\n \t<li>as[latex]\\,x\\to \\infty ,f\\left(x\\right)\\to \\infty ,[\/latex]<\/li>\n \t<li>as [latex]\\,x\\to -\\infty ,f\\left(x\\right)\\to 0,[\/latex]<\/li>\n \t<li>[latex]\\,f\\left(x\\right)\\,[\/latex]is always increasing,<\/li>\n \t<li>the graph of[latex]\\,f\\left(x\\right)\\,[\/latex]will never touch the <em>x<\/em>-axis because base two raised to any exponent never has the result of zero.<\/li>\n \t<li>[latex]\\,y=0\\,[\/latex]is the horizontal asymptote.<\/li>\n \t<li>the <em>y<\/em>-intercept is 1.<\/li>\n<\/ul>\n<div id=\"fs-id1165137442472\" class=\"textbox key-takeaways\">\n<h3>Exponential Function<\/h3>\n<p id=\"fs-id1165137911387\">For any real number[latex]\\,x,[\/latex]an exponential function is a function with the form<\/p>\n\n<div id=\"Equation_4_1_1\">[latex]f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\n<p id=\"eip-751\">where<\/p>\n\n<ul id=\"fs-id1165137401680\">\n \t<li>[latex]\\,a\\,[\/latex]is a non-zero real number called the initial value and<\/li>\n \t<li>[latex]\\,b\\,[\/latex]is any positive real number such that[latex]\\,b\\ne 1.[\/latex]<\/li>\n \t<li>The domain of[latex]\\,f\\,[\/latex]is all real numbers.<\/li>\n \t<li>The range of[latex]\\,f\\,[\/latex]is all positive real numbers if[latex]\\,a&gt;0.[\/latex]<\/li>\n \t<li>The range of[latex]\\,f\\,[\/latex]is all negative real numbers if[latex]\\,a&lt;0.[\/latex]<\/li>\n \t<li>The <em>y<\/em>-intercept is[latex]\\,\\left(0,a\\right),[\/latex]and the horizontal asymptote is[latex]\\,y=0.[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165137387381\">\n<div id=\"fs-id1165137659178\">\n<h3>Identifying Exponential Functions<\/h3>\n<p id=\"fs-id1165137601478\">Which of the following equations are <em>not<\/em> exponential functions?<\/p>\n\n<ul id=\"fs-id1165135176602\">\n \t<li>[latex]f\\left(x\\right)={4}^{3\\left(x-2\\right)}[\/latex]<\/li>\n \t<li>[latex]g\\left(x\\right)={x}^{3}[\/latex]<\/li>\n \t<li>[latex]h\\left(x\\right)={\\left(\\frac{1}{3}\\right)}^{x}[\/latex]<\/li>\n \t<li>[latex]j\\left(x\\right)={\\left(-2\\right)}^{x}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165134108513\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134108513\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134108513\"]\n<p id=\"fs-id1165137698136\">By definition, an exponential function has a constant as a base and an independent variable as an exponent. Thus,[latex]\\,g\\left(x\\right)={x}^{3}\\,[\/latex]does not represent an exponential function because the base is an independent variable. In fact,[latex]\\,g\\left(x\\right)={x}^{3}\\,[\/latex]is a power function.<\/p>\n<p id=\"fs-id1165137612252\">Recall that the base <em>b <\/em>of an exponential function is always a positive constant, and[latex]\\,b\\ne 1.\\,[\/latex]Thus,[latex]\\,j\\left(x\\right)={\\left(-2\\right)}^{x}\\,[\/latex]does not represent an exponential function because the base,[latex]\\,-2,[\/latex] is less than[latex]\\,0.[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137436342\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1165137862673\">\n<p id=\"fs-id1165137424249\">Which of the following equations represent exponential functions?<\/p>\n\n<ul id=\"fs-id1165135161022\">\n \t<li>[latex]f\\left(x\\right)=2{x}^{2}-3x+1[\/latex]<\/li>\n \t<li>[latex]g\\left(x\\right)={0.875}^{x}[\/latex]<\/li>\n \t<li>[latex]h\\left(x\\right)=1.75x+2[\/latex]<\/li>\n \t<li>[latex]j\\left(x\\right)={1095.6}^{-2x}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165137597685\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137597685\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137597685\"]\n<p id=\"fs-id1165137714559\">[latex]g\\left(x\\right)={0.875}^{x}\\,[\/latex]and[latex]j\\left(x\\right)={1095.6}^{-2x}\\,[\/latex]represent exponential functions.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137640042\" class=\"bc-section section\">\n<h3>Evaluating Exponential Functions<\/h3>\n<p id=\"fs-id1165137784783\">Recall that the base of an exponential function must be a positive real number other than[latex]\\,1.[\/latex]Why do we limit the base [latex]b\\,[\/latex]to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:<\/p>\n\n<ul id=\"fs-id1165137754880\">\n \t<li>Let[latex]\\,b=-9\\,[\/latex]and[latex]\\,x=\\frac{1}{2}.\\,[\/latex]Then[latex]\\,f\\left(x\\right)=f\\left(\\frac{1}{2}\\right)={\\left(-9\\right)}^{\\frac{1}{2}}=\\sqrt{-9},[\/latex]which is not a real number.<\/li>\n<\/ul>\n<p id=\"fs-id1165137563360\">Why do we limit the base to positive values other than [latex]1?[\/latex]Because base [latex]1\\,[\/latex]results in the constant function. Observe what happens if the base is [latex]1:[\/latex]<\/p>\n\n<ul id=\"fs-id1165137400268\">\n \t<li>Let[latex]\\,b=1.\\,[\/latex]Then[latex]\\,f\\left(x\\right)={1}^{x}=1\\,[\/latex]for any value of[latex]\\,x.[\/latex]<\/li>\n<\/ul>\n<p id=\"fs-id1165137459694\">To evaluate an exponential function with the form[latex]\\,f\\left(x\\right)={b}^{x},[\/latex]we simply substitute [latex]x\\,[\/latex]with the given value, and calculate the resulting power. For example:<\/p>\n<p id=\"fs-id1165135403544\">Let [latex]\\,f\\left(x\\right)={2}^{x}.\\,[\/latex]What is [latex]f\\left(3\\right)?[\/latex]<\/p>\n\n<div id=\"eip-id1165137643186\" class=\"unnumbered\">[latex]\\begin{array}{lll}f\\left(x\\right)\\hfill &amp; ={2}^{x}\\hfill &amp; \\hfill \\\\ f\\left(3\\right)\\hfill &amp; ={2}^{3}\\text{ }\\hfill &amp; \\text{Substitute }x=3.\\hfill \\\\ \\hfill &amp; =8\\text{ }\\hfill &amp; \\text{Evaluate the power}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137849020\">To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:<\/p>\n<p id=\"fs-id1165137849024\">Let[latex]\\,f\\left(x\\right)=30{\\left(2\\right)}^{x}.\\,[\/latex]What is[latex]\\,f\\left(3\\right)?[\/latex]<\/p>\n\n<div id=\"eip-id1165134086025\" class=\"unnumbered\">[latex]\\begin{array}{lll}f\\left(x\\right)\\hfill &amp; =30{\\left(2\\right)}^{x}\\hfill &amp; \\hfill \\\\ f\\left(3\\right)\\hfill &amp; =30{\\left(2\\right)}^{3}\\hfill &amp; \\text{Substitute }x=3.\\hfill \\\\ \\hfill &amp; =30\\left(8\\right)\\text{ }\\hfill &amp; \\text{Simplify the power first}\\text{.}\\hfill \\\\ \\hfill &amp; =240\\hfill &amp; \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137841073\">Note that if the order of operations were not followed, the result would be incorrect:<\/p>\n\n<div id=\"eip-id1165135320147\" class=\"unnumbered\">[latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165137455430\">\n<div id=\"fs-id1165137455432\">\n<h3>Evaluating Exponential Functions<\/h3>\n<p id=\"fs-id1165137767841\">Let [latex]\\,f\\left(x\\right)=5{\\left(3\\right)}^{x+1}.\\,[\/latex]Evaluate[latex]\\,f\\left(2\\right)\\,[\/latex]without using a calculator.<\/p>\n\n<\/div>\n<div id=\"fs-id1165135429364\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135429364\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135429364\"]\n<p id=\"fs-id1165137598173\">Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\n\n<div id=\"eip-id1165135208555\" class=\"unnumbered\">[latex]\\begin{array}{lll}f\\left(x\\right)\\hfill &amp; =5{\\left(3\\right)}^{x+1}\\hfill &amp; \\hfill \\\\ f\\left(2\\right)\\hfill &amp; =5{\\left(3\\right)}^{2+1}\\hfill &amp; \\text{Substitute }x=2.\\hfill \\\\ \\hfill &amp; =5{\\left(3\\right)}^{3}\\hfill &amp; \\text{Add the exponents}.\\hfill \\\\ \\hfill &amp; =5\\left(27\\right)\\hfill &amp; \\text{Simplify the power}\\text{.}\\hfill \\\\ \\hfill &amp; =135\\hfill &amp; \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137843864\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1165137762786\">\n<p id=\"fs-id1165137762788\">Let[latex]f\\left(x\\right)=8{\\left(1.2\\right)}^{x-5}.\\,[\/latex]Evaluate[latex]\\,f\\left(3\\right)\\,[\/latex]using a calculator. Round to four decimal places.<\/p>\n\n<\/div>\n<div id=\"fs-id1165137637369\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137637369\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137637369\"]\n[latex]5.5556[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137439018\" class=\"bc-section section\">\n<h4>Defining Exponential Growth<\/h4>\n<p id=\"fs-id1165137748523\">Because the output of exponential functions increases very rapidly, the term \u201cexponential growth\u201d is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.<\/p>\n\n<div id=\"fs-id1165137564690\" class=\"textbox key-takeaways\">\n<h3>Exponential Growth<\/h3>\n<p id=\"fs-id1165137834019\">A function that models exponential growth grows by a rate proportional to the amount present. For any real number[latex]\\,x\\,[\/latex]and any positive real numbers[latex]\\,a \\,[\/latex]and[latex]\\,b\\,[\/latex]such that[latex]\\,b\\ne 1,[\/latex]an exponential growth function has the form<\/p>\n\n<div id=\"fs-id1165137851784\" class=\"unnumbered aligncenter\">[latex]\\text{ }f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\nwhere\n<ul id=\"fs-id1165137863819\">\n \t<li>[latex]a\\,[\/latex]is the initial or starting value of the function.<\/li>\n \t<li>[latex]b\\,[\/latex]is the growth factor or growth multiplier per unit[latex]\\,x[\/latex].<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137644244\">In more general terms, we have an <em>exponential function<\/em>, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let\u2019s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function[latex]\\,A\\left(x\\right)=100+50x.\\,[\/latex]Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function [latex]\\,B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}.[\/latex]<\/p>\n<p id=\"fs-id1165135512493\">A few years of growth for these companies are illustrated in <a class=\"autogenerated-content\" href=\"#Table_04_01_05\">(Figure)<\/a>.<\/p>\n\n<table summary=\"Six rows and three columns. The first column is labeled, \u201cYear, x\u201d, which goes from 0 to 3; the second column is labeled, \u201cStores, Company A\u201d, which has a function of A(x) = 100+50x; and the third column is labeled, \u201cStores, Company B\u201d, which has a function of B(x)=100(1+0.5)^x. The following values are for Company A\u2019s function: (0, 100), (1, 150), (2, 200), and (3, 250). The following values are for the function Company B\u2019s function: (0, 100), (1, 150), (2, 225), and (3, 337.5).\">\n<thead>\n<tr>\n<th>Year, [latex]x[\/latex]<\/th>\n<th>Stores, Company A<\/th>\n<th>Stores, Company B<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]100+50\\left(0\\right)=100[\/latex]<\/td>\n<td>[latex]100{\\left(1+0.5\\right)}^{0}=100[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]100+50\\left(1\\right)=150[\/latex]<\/td>\n<td>[latex]100{\\left(1+0.5\\right)}^{1}=150[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]100+50\\left(2\\right)=200[\/latex]<\/td>\n<td>[latex]100{\\left(1+0.5\\right)}^{2}=225[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]100+50\\left(3\\right)=250[\/latex]<\/td>\n<td>[latex]100{\\left(1+0.5\\right)}^{3}=337.5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]x[\/latex]<\/td>\n<td>[latex]A\\left(x\\right)=100+50x[\/latex]<\/td>\n<td>[latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137653733\">The graphs comparing the number of stores for each company over a five-year period are shown in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_001\">(Figure)<\/a><strong>.<\/strong> We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.<\/p>\n\n<div id=\"CNX_Precalc_Figure_04_01_001\" class=\"small aligncenter\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140101\/CNX_Precalc_Figure_04_01_001.jpg\" alt=\"Graph of Companies A and B\u2019s functions, which values are found in the previous table.\" width=\"487\" height=\"845\"> <strong>Figure 2. <\/strong>The graph shows the numbers of stores Companies A and B opened over a five-year period.[\/caption]\n\n<\/div>\n<p id=\"fs-id1165135209682\">Notice that the domain for both functions is[latex]\\,\\left[0,\\infty \\right),[\/latex]and the range for both functions is[latex]\\,\\left[100,\\infty \\right).\\,[\/latex]After year 1, Company B always has more stores than Company A.<\/p>\n<p id=\"fs-id1165137836429\">Now we will turn our attention to the function representing the number of stores for Company B,[latex]\\,B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}.\\,[\/latex]In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and[latex]\\,1+0.5=1.5\\,[\/latex]represents the growth factor. Generalizing further, we can write this function as[latex]\\,B\\left(x\\right)=100{\\left(1.5\\right)}^{x},[\/latex]where 100 is the initial value,[latex]\\,1.5\\,[\/latex]is called the <em>base<\/em>, and[latex]\\,x\\,[\/latex]is called the <em>exponent<\/em>.<\/p>\n\n<div class=\"textbox examples\">\n<div id=\"fs-id1165137535640\">\n<div id=\"fs-id1165137535642\">\n<h3>Evaluating a Real-World Exponential Model<\/h3>\n<p id=\"fs-id1165135541867\">At the beginning of this section, we learned that the population of India was about[latex]\\,1.25\\,[\/latex]billion in the year 2013, with an annual growth rate of about[latex]\\,1.2%.\\,[\/latex]This situation is represented by the growth function[latex]\\,P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t},[\/latex] where [latex]\\,t\\,[\/latex] is the number of years since[latex]\\,2013.\\,[\/latex]To the nearest thousandth, what will the population of India be in[latex]\\,\\text{2031?}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137786632\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137786632\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137786632\"]\n<p id=\"fs-id1165137786635\">To estimate the population in 2031, we evaluate the models for[latex]\\,t=18,[\/latex] because 2031 is[latex]\\,18[\/latex] years after 2013. Rounding to the nearest thousandth,<\/p>\n\n<div id=\"eip-id1165135657117\" class=\"unnumbered\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/div>\n<p id=\"fs-id1165135394343\">There will be about 1.549 billion people in India in the year 2031.[\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135536569\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_04_01_03\">\n<div id=\"fs-id1165137635312\">\n<p id=\"fs-id1165137635314\">The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about[latex]\\,0.6%.\\,[\/latex]This situation is represented by the growth function[latex]\\,P\\left(t\\right)=1.39{\\left(1.006\\right)}^{t},[\/latex] where [latex]\\,t\\,[\/latex] is the number of years since[latex]\\,2013.[\/latex]To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in <a class=\"autogenerated-content\" href=\"#Example_04_01_03\">(Figure)<\/a>?<\/p>\n\n<\/div>\n<div id=\"fs-id1165134200184\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134200184\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134200184\"]\n<p id=\"fs-id1165134200186\">About[latex]\\,1.548\\,[\/latex]billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135526980\" class=\"bc-section section\">\n<h3>Finding Equations of Exponential Functions<\/h3>\n<p id=\"fs-id1165135526985\">In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants[latex]\\,a\\,[\/latex]and[latex]\\,b,[\/latex]and evaluate the function.<\/p>\n\n<div id=\"fs-id1165135369632\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165135180102\"><strong>Given two data points, write an exponential model.<\/strong><\/p>\n\n<ol id=\"fs-id1165135180107\" type=\"1\">\n \t<li>If one of the data points has the form[latex]\\,\\left(0,a\\right),[\/latex] then[latex]\\,a\\,[\/latex]is the initial value. Using[latex]\\,a,[\/latex] substitute the second point into the equation[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x},[\/latex] and solve for[latex]\\,b.[\/latex]<\/li>\n \t<li>If neither of the data points have the form[latex]\\,\\left(0,a\\right),[\/latex] substitute both points into two equations with the form[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x}.\\,[\/latex]Solve the resulting system of two equations in two unknowns to find[latex]\\,a\\,[\/latex]and[latex]\\,b.[\/latex]<\/li>\n \t<li>Using the[latex]\\,a\\,[\/latex]and[latex]\\,b\\,[\/latex]found in the steps above, write the exponential function in the form[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x}.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165137580876\">\n<div id=\"fs-id1165137580878\">\n<h3>Writing an Exponential Model When the Initial Value Is Known<\/h3>\n<p id=\"fs-id1165137667588\">In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function[latex]\\,N\\left(t\\right)\\,[\/latex]representing the population[latex]\\,\\left(N\\right)\\,[\/latex]of deer over time[latex]\\,t.[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135188416\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135188416\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135188416\"]\n<p id=\"fs-id1165135188418\">We let our independent variable[latex]\\,t\\,[\/latex]be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function,[latex]\\,a=80.\\,[\/latex]We can now substitute the second point into the equation[latex]\\,N\\left(t\\right)=80{b}^{t}\\,[\/latex]to find[latex]\\,b:[\/latex]<\/p>\n\n<div id=\"eip-id1165135432669\" class=\"unnumbered\">[latex]\\begin{array}{lll}N\\left(t\\right)\\hfill &amp; =80{b}^{t}\\hfill &amp; \\hfill \\\\ \\,\\,180\\hfill &amp; =80{b}^{6}\\hfill &amp; \\text{Substitute using point }\\left(6, 180\\right).\\hfill \\\\ \\,\\,\\,\\,\\,\\frac{9}{4}\\hfill &amp; ={b}^{6}\\hfill &amp; \\text{Divide and write in lowest terms}.\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,b\\hfill &amp; ={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}}\\hfill &amp; \\text{Isolate }b\\text{ using properties of exponents}.\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,b\\hfill &amp; \\approx 1.1447\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; \\text{Round to 4 decimal places}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135193230\"><strong>NOTE:<\/strong><em>Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.<\/em><\/p>\n<p id=\"fs-id1165137705073\">The exponential model for the population of deer is[latex]\\,N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}.\\,[\/latex](Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)<\/p>\nWe can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_002\">(Figure)<\/a> passes through the initial points given in the problem,[latex]\\,\\left(0,\\text{ 8}0\\right)\\,[\/latex]and[latex]\\,\\left(\\text{6},\\text{ 18}0\\right).\\,[\/latex]We can also see that the domain for the function is[latex]\\,\\left[0,\\infty \\right),[\/latex]and the range for the function is[latex]\\,\\left[80,\\infty \\right).[\/latex]\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140104\/CNX_Precalc_Figure_04_01_002.jpg\" alt=\"Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).\" width=\"487\" height=\"700\"> <strong>Figure 3.<\/strong> Graph showing the population of deer over time,[latex]\\,N\\left(t\\right)=80{\\left(1.1447\\right)}^{t},[\/latex][latex]t\\,[\/latex]years after 2006[\/caption]\n<p id=\"fs-id1165137724117\">[\/hidden-answer]<\/p>\n\n<div id=\"CNX_Precalc_Figure_04_01_002\" class=\"small aligncenter\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137761908\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1165135496544\">\n<p id=\"fs-id1165135496547\">A wolf population is growing exponentially. In 2011,[latex]\\,129\\,[\/latex]wolves were counted. By[latex]\\,\\text{2013,}\\,[\/latex]the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population[latex]\\,N\\,[\/latex]of wolves over time[latex]\\,t.[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137737094\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137737094\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137737094\"]\n<p id=\"fs-id1165137737097\">[latex]\\left(0,129\\right)\\,[\/latex]and[latex]\\,\\left(2,236\\right);\\,\\,\\,N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165135411399\">\n<div id=\"fs-id1165135411402\">\n<h3>Writing an Exponential Model When the Initial Value is Not Known<\/h3>\n<p id=\"fs-id1165135411407\">Find an exponential function that passes through the points[latex]\\,\\left(-2,6\\right)\\,[\/latex]and[latex]\\,\\left(2,1\\right).[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135646190\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135646190\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135646190\"]\n<p id=\"fs-id1165135646192\">Because we don\u2019t have the initial value, we substitute both points into an equation of the form[latex]\\,f\\left(x\\right)=a{b}^{x},[\/latex] and then solve the system for[latex]\\,a\\,[\/latex]and[latex]\\,b.[\/latex]<\/p>\n\n<ul id=\"fs-id1165134044679\">\n \t<li>Substituting[latex]\\,\\left(-2,6\\right)\\,[\/latex]gives[latex]\\,6=a{b}^{-2}\\,[\/latex]<\/li>\n \t<li>Substituting[latex]\\,\\left(2,1\\right)\\,[\/latex]gives[latex]\\,1=a{b}^{2}\\,[\/latex]<\/li>\n<\/ul>\n<p id=\"fs-id1165135361777\">Use the first equation to solve for[latex]\\,a\\,[\/latex]in terms of[latex]\\,b:[\/latex]<\/p>\n<span id=\"eip-id1619372\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140111\/CNX_Precalc_Figure_04_01_007.jpg\" alt=\"..\"><\/span>\n<p id=\"fs-id1165135180085\">Substitute[latex]\\,a\\,[\/latex]in the second equation, and solve for[latex]\\,b:[\/latex]<\/p>\n<span id=\"eip-id2591037\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140118\/CNX_Precalc_Figure_04_01_008.jpg\" alt=\"..\"><\/span>\n<p id=\"fs-id1165137761627\">Use the value of[latex]\\,b\\,[\/latex]in the first equation to solve for the value of[latex]\\,a:[\/latex]<span id=\"eip-id2811781\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140126\/CNX_Precalc_Figure_04_01_009.jpg\" alt=\"..\"><\/span><\/p>\n<p id=\"fs-id1165135168188\">Thus, the equation is[latex]\\,f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}.[\/latex]<\/p>\nWe can graph our model to check our work. Notice that the graph in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_003\">(Figure)<\/a> passes through the initial points given in the problem,[latex]\\,\\left(-2,\\text{ 6}\\right)\\,[\/latex]and[latex]\\,\\left(2,\\text{ 1}\\right).\\,[\/latex]The graph is an example of an <span class=\"no-emphasis\">exponential decay<\/span> function.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140137\/CNX_Precalc_Figure_04_01_003.jpg\" alt=\"Graph of the exponential function, f(x)=2.4492(0.6389)^x, with labeled points at (-2, 6) and (2, 1).\" width=\"487\" height=\"445\"> <strong>Figure 4. <\/strong>The graph of[latex]\\,f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}\\,[\/latex]models exponential decay.[\/caption]\n<p id=\"fs-id1165135482016\">[\/hidden-answer]<\/p>\n\n<div id=\"CNX_Precalc_Figure_04_01_003\" class=\"small aligncenter\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"ti_11_04_01\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"fs-id1165135169306\">\n<div id=\"fs-id1165135169308\">\n<p id=\"fs-id1165135169310\">Given the two points[latex]\\,\\left(1,3\\right)\\,[\/latex]and[latex]\\,\\left(2,4.5\\right),[\/latex]find the equation of the exponential function that passes through these two points.<\/p>\n\n<\/div>\n<div id=\"fs-id1165137732255\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137732255\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137732255\"]\n<p id=\"fs-id1165137732257\">[latex]f\\left(x\\right)=2{\\left(1.5\\right)}^{x}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137737999\" class=\"precalculus qa textbox shaded\">\n<p id=\"fs-id1165137738004\"><strong>Do two points always determine a unique exponential function?<\/strong><\/p>\n<p id=\"fs-id1165137663981\"><em>Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in[latex]\\,x,[\/latex] which in many real world cases involves time.<\/em><\/p>\n\n<\/div>\n<div id=\"fs-id1165137771573\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137771578\"><strong>Given the graph of an exponential function, write its equation.<\/strong><\/p>\n\n<ol id=\"fs-id1165137771583\" type=\"1\">\n \t<li>First, identify two points on the graph. Choose the <em>y<\/em>-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error.<\/li>\n \t<li>If one of the data points is the <em>y-<\/em>intercept[latex]\\,\\left(0,a\\right)[\/latex], then[latex]\\,a\\,[\/latex]is the initial value. Using[latex]\\,a,[\/latex] substitute the second point into the equation[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x},[\/latex] and solve for[latex]\\,b.[\/latex]<\/li>\n \t<li>If neither of the data points have the form[latex]\\,\\left(0,a\\right),[\/latex] substitute both points into two equations with the form[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x}.\\,[\/latex]Solve the resulting system of two equations in two unknowns to find[latex]\\,a\\,[\/latex]and[latex]\\,b.[\/latex]<\/li>\n \t<li>Write the exponential function,[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x}.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165137806415\">\n<div id=\"fs-id1165137806417\">\n<h3>Writing an Exponential Function Given Its Graph<\/h3>\n<p id=\"fs-id1165135524507\">Find an equation for the exponential function graphed in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_004\">(Figure)<\/a>.<\/p>\n\n<div id=\"CNX_Precalc_Figure_04_01_004\" class=\"medium aligncenter\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140140\/CNX_Precalc_Figure_04_01_004.jpg\" alt=\"Graph of an increasing exponential function with notable points at (0, 3) and (2, 12).\" width=\"731\" height=\"369\"> <strong>Figure 5.<\/strong>[\/caption]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135415647\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135415647\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135415647\"]\n<p id=\"fs-id1165135415649\">We can choose the <em>y<\/em>-intercept of the graph,[latex]\\,\\left(0,3\\right),[\/latex] as our first point. This gives us the initial value,[latex]\\,a=3.\\,[\/latex]Next, choose a point on the curve some distance away from[latex]\\,\\left(0,3\\right)\\,[\/latex]that has integer coordinates. One such point is[latex]\\,\\left(2,12\\right).[\/latex]<\/p>\n\n<div id=\"eip-id1165134216216\" class=\"unnumbered\">[latex]\\begin{array}{ll}\\text{ }y=a{b}^{x}\\hfill &amp; \\text{Write the general form of an exponential equation}.\\hfill \\\\ \\text{ }y=3{b}^{x}\\hfill &amp; \\text{Substitute the initial value 3 for }a.\\hfill \\\\ 12=3{b}^{2}\\hfill &amp; \\text{Substitute in 12 for }y\\text{ and 2 for }x.\\hfill \\\\ \\text{ }4={b}^{2}\\hfill &amp; \\text{Divide by 3}.\\hfill \\\\ \\text{ }b=\u00b12\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; \\text{Take the square root}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137769996\">Because we restrict ourselves to positive values of[latex]\\,b,[\/latex] we will use[latex]\\,b=2.\\,[\/latex]Substitute[latex]\\,a\\,[\/latex]and[latex]\\,b\\,[\/latex]into the standard form to yield the equation[latex]\\,f\\left(x\\right)=3{\\left(2\\right)}^{x}.[\/latex][\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135169326\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1165135431682\">\n<p id=\"fs-id1165135431684\">Find an equation for the exponential function graphed in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_005\">(Figure)<\/a>.<\/p>\n\n<div id=\"CNX_Precalc_Figure_04_01_005\" class=\"small aligncenter\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140142\/CNX_Precalc_Figure_04_01_005.jpg\" alt=\"Graph of an increasing function with a labeled point at (0, sqrt(2)).\" width=\"487\" height=\"294\"> <strong>Figure 6.<\/strong>[\/caption]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191210\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135191210\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135191210\"]\n<p id=\"fs-id1165135191213\">[latex]f\\left(x\\right)=\\sqrt{2}{\\left(\\sqrt{2}\\right)}^{x}.\\,[\/latex]Answers may vary due to round-off error. The answer should be very close to[latex]\\,1.4142{\\left(1.4142\\right)}^{x}.[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137936929\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137936934\"><strong>Given two points on the curve of an exponential function, use a graphing calculator to find the equation.<\/strong><\/p>\n\n<ol id=\"fs-id1165135369319\" type=\"1\">\n \t<li>Press <strong>[STAT].<\/strong><\/li>\n \t<li>Clear any existing entries in columns <strong>L1<\/strong> or <strong>L2.<\/strong><\/li>\n \t<li>In <strong>L1<\/strong>, enter the <em>x<\/em>-coordinates given.<\/li>\n \t<li>In <strong>L2<\/strong>, enter the corresponding <em>y<\/em>-coordinates.<\/li>\n \t<li>Press <strong>[STAT] <\/strong>again. Cursor right to <strong>CALC<\/strong>, scroll down to <strong>ExpReg (Exponential Regression)<\/strong>, and press <strong>[ENTER].<\/strong><\/li>\n \t<li>The screen displays the values of <em>a<\/em> and <em>b<\/em> in the exponential equation[latex]\\,y=a\\cdot {b}^{x}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div>\n<div>\n<h3>Using a Graphing Calculator to Find an Exponential Function<\/h3>\n<p id=\"fs-id1165137748687\">Use a graphing calculator to find the exponential equation that includes the points[latex]\\,\\left(2,24.8\\right)\\,[\/latex]and[latex]\\,\\left(5,198.4\\right).[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137738185\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137738185\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137738185\"]\n<p id=\"fs-id1165137843706\">Follow the guidelines above. First press <strong>[STAT]<\/strong>, <strong>[EDIT]<\/strong>, <strong>[1: Edit\u2026], <\/strong>and clear the lists <strong>L1<\/strong> and <strong>L2<\/strong>. Next, in the <strong>L1<\/strong> column, enter the <em>x<\/em>-coordinates, 2 and 5. Do the same in the <strong>L2<\/strong> column for the <em>y<\/em>-coordinates, 24.8 and 198.4.<\/p>\n<p id=\"fs-id1165137639756\">Now press <strong>[STAT]<\/strong>, <strong>[CALC]<\/strong>, <strong>[0: ExpReg] <\/strong>and press <strong>[ENTER]<\/strong>. The values[latex]\\,a=6.2\\,[\/latex]and[latex]\\,b=2\\,[\/latex]will be displayed. The exponential equation is[latex]\\,y=6.2\\cdot {2}^{x}.[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137771710\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_07_01_07\">\n<div id=\"fs-id1165137656116\">\n<p id=\"fs-id1165137656119\">Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6, 481.07).<\/p>\n\n<\/div>\n<div id=\"fs-id1165137656124\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137656124\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137656124\"]\n<p id=\"fs-id1165137651684\">[latex]y\\approx 12\\cdot {1.85}^{x}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137871708\" class=\"bc-section section\">\n<h3>Applying the Compound-Interest Formula<\/h3>\n<p id=\"fs-id1165137447026\">Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest. The term <em>compounding<\/em> refers to interest earned not only on the original value, but on the accumulated value of the account.<\/p>\n<p id=\"fs-id1165137447037\">The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an investment account. The term <em>nominal<\/em> is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being <em>greater<\/em> than the nominal rate! This is a powerful tool for investing.<\/p>\n<p id=\"fs-id1165135160118\">We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time[latex]\\,t,[\/latex] principal[latex]\\,P,[\/latex] APR[latex]\\,r,[\/latex] and number of compounding periods in a year[latex]\\,n:[\/latex]<\/p>\n\n<div id=\"eip-986\" class=\"unnumbered aligncenter\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/div>\n<p id=\"fs-id1165137935717\">For example, observe <a class=\"autogenerated-content\" href=\"#Table_04_01_03\">(Figure)<\/a>, which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases.<\/p>\n\n<table summary=\"Six rows and two columns. The first column is labeled, \u201cFrequency\u201d, and the second column is labeled, \u201cValue after 1 Year\u201d. Reading the rows from left to right, we have that Annually is valued at 100, Semiannually at 102.50, Quarterly at 103.81, Monthly at 104.71, and Daily at 105.16.\">\n<thead>\n<tr>\n<th>Frequency<\/th>\n<th>Value after 1 year<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Annually<\/td>\n<td>$1100<\/td>\n<\/tr>\n<tr>\n<td>Semiannually<\/td>\n<td>$1102.50<\/td>\n<\/tr>\n<tr>\n<td>Quarterly<\/td>\n<td>$1103.81<\/td>\n<\/tr>\n<tr>\n<td>Monthly<\/td>\n<td>$1104.71<\/td>\n<\/tr>\n<tr>\n<td>Daily<\/td>\n<td>$1105.16<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1165137793679\" class=\"textbox key-takeaways\">\n<h3>The Compound Interest Formula<\/h3>\n<p id=\"fs-id1165135184167\"><strong>Compound interest<\/strong> can be calculated using the formula<\/p>\n\n<div id=\"fs-id1165135184172\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/div>\n<p id=\"eip-237\">where<\/p>\n\n<ul id=\"fs-id1165137448453\">\n \t<li>[latex]A\\left(t\\right)\\,[\/latex]is the account value,<\/li>\n \t<li>[latex]t\\,[\/latex]is measured in years,<\/li>\n \t<li>[latex]P\\,[\/latex]is the starting amount of the account, often called the principal, or more generally present value,<\/li>\n \t<li>[latex]r\\,[\/latex]is the annual percentage rate (APR) expressed as a decimal, and<\/li>\n \t<li>[latex]n\\,[\/latex]is the number of compounding periods in one year.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<div>\n<div id=\"fs-id1165137812820\">\n<h3>Calculating Compound Interest<\/h3>\n<p id=\"fs-id1165137812825\">If we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years?<\/p>\n\n<\/div>\n<div id=\"fs-id1165137812830\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137812830\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137812830\"]\n<p id=\"fs-id1165137812832\">Because we are starting with $3,000,[latex]\\,P=3000.\\,[\/latex]Our interest rate is 3%, so[latex]\\,r\\text{ }=\\text{ }0.03.\\,[\/latex]Because we are compounding quarterly, we are compounding 4 times per year, so[latex]\\,n=4.\\,[\/latex]We want to know the value of the account in 10 years, so we are looking for[latex]\\,A\\left(10\\right),[\/latex]the value when[latex]\\,t\\text{ }=\\text{ }10.[\/latex]<\/p>\n\n<div id=\"eip-id1402796\" class=\"unnumbered\">[latex]\\begin{array}{lll}A\\left(t\\right)\\hfill &amp; =P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\hfill &amp; \\text{Use the compound interest formula}.\\hfill \\\\ A\\left(10\\right)\\hfill &amp; =3000{\\left(1+\\frac{0.03}{4}\\right)}^{4\\cdot 10}\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; \\text{Substitute using given values}.\\hfill \\\\ \\hfill &amp; \\approx \\text{\\$}4045.05\\hfill &amp; \\text{Round to two decimal places}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137694040\">The account will be worth about $4,045.05 in 10 years.[\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137694046\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1165137694055\">\n<p id=\"fs-id1165135180428\">An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What will the investment be worth in 30 years?<\/p>\n[reveal-answer q=\"297130\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"297130\"]about $3,644,675.88\n\n[\/hidden-answer]\n\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165135180446\">\n<div>\n<h3>Using the Compound Interest Formula to Solve for the Principal<\/h3>\n<p id=\"fs-id1165135175327\">A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child\u2019s future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to $40,000 over 18 years. She believes the account will earn 6% compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now?<\/p>\n\n<\/div>\n<div id=\"fs-id1165135175338\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135175338\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135175338\"]\n<p id=\"fs-id1165137664627\">The nominal interest rate is 6%, so[latex]\\,r=0.06.\\,[\/latex]Interest is compounded twice a year, so[latex]\\,k=2.[\/latex]<\/p>\nWe want to find the initial investment,[latex]\\,P,[\/latex] needed so that the value of the account will be worth $40,000 in[latex]\\,18\\,[\/latex]years. Substitute the given values into the compound interest formula, and solve for[latex]\\,P.[\/latex]\n<div id=\"eip-id1165131884554\" class=\"unnumbered\">[latex]\\begin{array}{lll}\\,\\,\\,\\,\\,\\,\\,A\\left(t\\right)\\hfill &amp; =P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\hfill &amp; \\text{Use the compound interest formula}.\\hfill \\\\ 40,000\\hfill &amp; =P{\\left(1+\\frac{0.06}{2}\\right)}^{2\\left(18\\right)}\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; \\text{Substitute using given values }A\\text{, }r, n\\text{, and }t.\\hfill \\\\ 40,000\\hfill &amp; =P{\\left(1.03\\right)}^{36}\\hfill &amp; \\text{Simplify}.\\hfill \\\\ \\frac{40,000}{{\\left(1.03\\right)}^{36}}\\hfill &amp; =P\\hfill &amp; \\text{Isolate }P.\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,P\\hfill &amp; \\approx \\text{\\$}13,801\\hfill &amp; \\text{Divide and round to the nearest dollar}.\\hfill \\end{array}[\/latex<\/div>\n<p id=\"fs-id1165137937589\">Lily will need to invest $13,801 to have $40,000 in 18 years.][\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_04_01_9\">\n<div id=\"fs-id1165135176727\">\n<p id=\"fs-id1165135176729\">Refer to <a class=\"autogenerated-content\" href=\"#Example_04_01_09\">(Figure)<\/a>. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?<\/p>\n\n<\/div>\n<div id=\"fs-id1165135176736\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135176736\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135176736\"]\n<p id=\"fs-id1165135176738\">$13,693<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137724961\" class=\"bc-section section\">\n<h3>Evaluating Functions with Base <em>e<\/em><\/h3>\n<p id=\"fs-id1165135684369\">As we saw earlier, the amount earned on an account increases as the compounding frequency increases. <a class=\"autogenerated-content\" href=\"#Table_04_01_04\">(Figure)<\/a> shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.<\/p>\n<p id=\"fs-id1165135684377\">Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies, listed in <a class=\"autogenerated-content\" href=\"#Table_04_01_04\">(Figure)<\/a>.<\/p>\n\n<table summary=\"Nine rows and three columns. The first column is labeled, \u201cFrequency\u201d, the second column is labeled, \u201cA(t)=(1+1\/n)^x\u201d, and the third column is labeled, \u201cValue\u201d. Reading the rows from left to right, we have that Annually has the input value of (1+1\/1)^1 which equals to \ud83d\udcb22, and Semiannually has the input value of (1+1\/2)^2 which equals to \ud83d\udcb22.25, Quarterly has the input value of (1+1\/4)^4 which equals to \ud83d\udcb22.441406, Monthly has the input value of (1+1\/12)^12 which equals to \ud83d\udcb22.613035, Daily has the input value of (1+1\/365)^365 which equals to \ud83d\udcb22.714567, Hourly has the input value of (1+1\/8766)^8766 which equals to \ud83d\udcb22.718127, One per minute has the input value of (1+1\/525960)^525960 which equals to \ud83d\udcb22.718279, and Once per second has the input value of (1+1\/31557600)^31557600 which equals to \ud83d\udcb22.718282.\"><caption>&nbsp;<\/caption>\n<thead>\n<tr>\n<th>Frequency<\/th>\n<th>[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\n<th>Value<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Annually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\n<td>$2<\/td>\n<\/tr>\n<tr>\n<td>Semiannually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\n<td>$2.25<\/td>\n<\/tr>\n<tr>\n<td>Quarterly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\n<td>$2.441406<\/td>\n<\/tr>\n<tr>\n<td>Monthly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\n<td>$2.613035<\/td>\n<\/tr>\n<tr>\n<td>Daily<\/td>\n<td>[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\n<td>$2.714567<\/td>\n<\/tr>\n<tr>\n<td>Hourly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{\\text{8760}}\\right)}^{\\text{8760}}[\/latex]<\/td>\n<td>$2.718127<\/td>\n<\/tr>\n<tr>\n<td>Once per minute<\/td>\n<td>[latex]{\\left(1+\\frac{1}{\\text{525600}}\\right)}^{\\text{525600}}[\/latex]<\/td>\n<td>$2.718279<\/td>\n<\/tr>\n<tr>\n<td>Once per second<\/td>\n<td>[latex]{\\left(1+\\frac{1}{31536000}\\right)}^{31536000}[\/latex]<\/td>\n<td>$2.718282<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137828146\">These values appear to be approaching a limit as[latex]\\,n\\,[\/latex]increases without bound. In fact, as[latex]\\,n\\,[\/latex]gets larger and larger, the expression[latex]\\,{\\left(1+\\frac{1}{n}\\right)}^{n}\\,[\/latex]approaches a number used so frequently in mathematics that it has its own name: the letter[latex]\\,e.\\,[\/latex]This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.<\/p>\n\n<div id=\"fs-id1165135511324\" class=\"textbox key-takeaways\">\n<h3>The Number <em>e<\/em><\/h3>\n<p id=\"fs-id1165135511335\">The letter <em>e<\/em> represents the irrational number<\/p>\n\n<div id=\"eip-id1165135378658\" class=\"unnumbered\">[latex]{\\left(1+\\frac{1}{n}\\right)}^{n},\\text{as}\\,n\\,\\text{increases without bound}[\/latex]<\/div>\n<p id=\"fs-id1165135369344\">The letter <em>e <\/em>is used as a base for many real-world exponential models. To work with base <em>e<\/em>, we use the approximation,[latex]\\,e\\approx 2.718282.\\,[\/latex]The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.<\/p>\n\n<\/div>\n<div id=\"Example_04_01_10\" class=\"textbox examples\">\n<div id=\"fs-id1165135344893\">\n<div id=\"fs-id1165135344895\">\n<h3>Using a Calculator to Find Powers of <em>e<\/em><\/h3>\n<p id=\"fs-id1165135545973\">Calculate[latex]\\,{e}^{3.14}.\\,[\/latex]Round to five decimal places.<\/p>\n\n<\/div>\n<div id=\"fs-id1165135192743\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135192743\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135192743\"]\n<p id=\"fs-id1165135192745\">On a calculator, press the button labeled[latex]\\,\\left[{e}^{x}\\right].\\,[\/latex]The window shows[latex]\\,\\left[e^(\\text{ }\\right].\\,[\/latex]Type[latex]\\,3.14\\,[\/latex]and then close parenthesis,[latex]\\,\\left[)\\right].\\,[\/latex]Press [ENTER]. Rounding to[latex]\\,5\\,[\/latex]decimal places,[latex]\\,{e}^{3.14}\\approx 23.10387.\\,[\/latex]Caution: Many scientific calculators have an \u201cExp\u201d button, which is used to enter numbers in scientific notation. It is not used to find powers of[latex]\\,e.[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137447188\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_04_01_10\">\n<div id=\"fs-id1165135348435\">\n<p id=\"fs-id1165135348437\">Use a calculator to find[latex]\\,{e}^{-0.5}.\\,[\/latex]Round to five decimal places.<\/p>\n\n<\/div>\n<div id=\"fs-id1165135176286\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135176286\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135176286\"]\n<p id=\"fs-id1165135176288\">[latex]{e}^{-0.5}\\approx 0.60653[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137827923\" class=\"bc-section section\">\n<h3>Investigating Continuous Growth<\/h3>\n<p id=\"fs-id1165137827929\">So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, <em>e <\/em>is used as the base for exponential functions. Exponential models that use[latex]\\,e\\,[\/latex]as the base are called <em>continuous growth or decay models<\/em>. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.<\/p>\n\n<div id=\"fs-id1165137664673\" class=\"textbox key-takeaways\">\n<h3>The Continuous Growth\/Decay Formula<\/h3>\n<p id=\"fs-id1165135453868\">For all real numbers[latex]\\,t,[\/latex]and all positive numbers[latex]\\,a\\,[\/latex]and[latex]\\,r,[\/latex]continuous growth or decay is represented by the formula<\/p>\n\n<div id=\"fs-id1165135536370\">[latex]A\\left(t\\right)=a{e}^{rt}[\/latex]<\/div>\n<p id=\"eip-101\">where<\/p>\n\n<ul id=\"fs-id1165135152052\">\n \t<li>[latex]a\\,[\/latex]is the initial value,<\/li>\n \t<li>[latex]r\\,[\/latex]is the continuous growth rate per unit time,<\/li>\n \t<li>and[latex]\\,t\\,[\/latex]is the elapsed time.<\/li>\n<\/ul>\n<p id=\"fs-id1165135560686\">If[latex]\\,r&gt;0\\,[\/latex], then the formula represents continuous growth. If[latex]\\,r&lt;0\\,[\/latex], then the formula represents continuous decay.<\/p>\n<p id=\"fs-id1165137812323\">For business applications, the continuous growth formula is called the continuous compounding formula and takes the form<\/p>\n\n<div id=\"eip-id1165134324899\" class=\"unnumbered\">[latex]A\\left(t\\right)=P{e}^{rt}[\/latex]<\/div>\n<p id=\"eip-962\">where<\/p>\n\n<ul id=\"fs-id1165137827330\">\n \t<li>[latex]P\\,[\/latex]is the principal or the initial invested,<\/li>\n \t<li>[latex]r\\,[\/latex]is the growth or interest rate per unit time,<\/li>\n \t<li>and [latex]t\\,[\/latex]is the period or term of the investment.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165135411368\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165135411373\"><strong>Given the initial value, rate of growth or decay, and time[latex]\\,t,[\/latex] solve a continuous growth or decay function.<\/strong><\/p>\n\n<ol id=\"fs-id1165135511371\" type=\"1\">\n \t<li>Use the information in the problem to determine[latex]\\,a[\/latex], the initial value of the function.<\/li>\n \t<li>Use the information in the problem to determine the growth rate[latex]\\,r.[\/latex]\n<ol id=\"fs-id1165135188096\" type=\"a\">\n \t<li>If the problem refers to continuous growth, then[latex]\\,r&gt;0.[\/latex]<\/li>\n \t<li>If the problem refers to continuous decay, then[latex]\\,r&lt;0.[\/latex]<\/li>\n<\/ol>\n<\/li>\n \t<li>Use the information in the problem to determine the time[latex]\\,t.[\/latex]<\/li>\n \t<li>Substitute the given information into the continuous growth formula and solve for[latex]\\,A\\left(t\\right).[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_01_11\" class=\"textbox examples\">\n<div id=\"fs-id1165137835464\">\n<div id=\"fs-id1165137835466\">\n<h3>Calculating Continuous Growth<\/h3>\n<p id=\"fs-id1165137835472\">A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year?<\/p>\n\n<\/div>\n<div id=\"fs-id1165137694203\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137694203\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137694203\"]\n<p id=\"fs-id1165137694205\">Since the account is growing in value, this is a continuous compounding problem with growth rate[latex]\\,r=0.10.\\,[\/latex]The initial investment was $1,000, so[latex]\\,P=1000.\\,[\/latex]We use the continuous compounding formula to find the value after[latex]\\,t=1\\,[\/latex]year:<\/p>\n\n<div id=\"eip-id1165133351794\" class=\"unnumbered\">[latex]\\begin{array}{lll}A\\left(t\\right)\\hfill &amp; =P{e}^{rt}\\hfill &amp; \\text{Use the continuous compounding formula}.\\hfill \\\\ \\hfill &amp; =1000{\\left(e\\right)}^{0.1}\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; \\text{Substitute known values for }P, r,\\text{ and }t.\\hfill \\\\ \\hfill &amp; \\approx 1105.17\\hfill &amp; \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137895288\">The account is worth $1,105.17 after one year.[\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137895295\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_04_01_11\">\n<div id=\"fs-id1165137895303\">\n<p id=\"fs-id1165137895305\">A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years?<\/p>\n\n<\/div>\n<div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1165134389978\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134389978\"]\n<p id=\"fs-id1165134389978\">$3,659,823.44<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_01_12\" class=\"textbox examples\">\n<div id=\"fs-id1165134389988\">\n<div id=\"fs-id1165134389990\">\n<h3>Calculating Continuous Decay<\/h3>\n<p id=\"fs-id1165137803700\">Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days?<\/p>\n\n<\/div>\n<div id=\"fs-id1165137803706\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137803706\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137803706\"]Since the substance is decaying, the rate,[latex]\\,17.3%[\/latex], is negative. So,[latex]\\,r\\text{ }=\\text{ }-0.173.\\,[\/latex]The initial amount of radon-222 was[latex]\\,100\\,[\/latex]mg, so[latex]\\,a=100.\\,[\/latex]We use the continuous decay formula to find the value after[latex]\\,t=3\\,[\/latex]days:\n<div id=\"eip-id1165137779893\" class=\"unnumbered\">[latex]\\begin{array}{lll}A\\left(t\\right)\\hfill &amp; =a{e}^{rt}\\hfill &amp; \\text{Use the continuous growth formula}.\\hfill \\\\ \\hfill &amp; =100{e}^{-0.173\\left(3\\right)}\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; \\text{Substitute known values for }a, r,\\text{ and }t.\\hfill \\\\ \\hfill &amp; \\approx 59.5115\\hfill &amp; \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137697132\">So 59.5115 mg of radon-222 will remain.[\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135348462\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_04_01_12\">\n<div id=\"fs-id1165135348470\">\n<p id=\"fs-id1165135348472\">Using the data in <a class=\"autogenerated-content\" href=\"#Example_04_01_12\">(Figure)<\/a>, how much radon-222 will remain after one year?<\/p>\n\n<\/div>\n<div id=\"fs-id1165135245732\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135245732\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135245732\"]\n<p id=\"fs-id1165135245734\">3.77E-26 (This is calculator notation for the number written as[latex]\\,3.77\u00d7{10}^{-26}\\,[\/latex]in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191186\" class=\"precalculus media\">\n<p id=\"fs-id1165135191191\">Access these online resources for additional instruction and practice with exponential functions.<\/p>\n\n<ul id=\"fs-id1165135191194\">\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/expgrowth\">Exponential Growth Function<\/a><\/li>\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/compoundint\">Compound Interest<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135264762\" class=\"key-equations\">\n<h3>Key Equations<\/h3>\n<table id=\"fs-id2306479\" style=\"height: 248px\" summary=\"...\">\n<tbody>\n<tr style=\"height: 41px\">\n<td style=\"height: 41px\">definition of the exponential function<\/td>\n<td style=\"height: 41px\">[latex]f\\left(x\\right)={b}^{x}\\text{, where }b&gt;0, b\\ne 1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 41px\">\n<td style=\"height: 41px\">definition of exponential growth<\/td>\n<td style=\"height: 41px\">[latex]f\\left(x\\right)=a{b}^{x},\\text{ where }a&gt;0,b&gt;0,b\\ne 1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 83px\">\n<td style=\"height: 83px\">compound interest formula<\/td>\n<td style=\"height: 83px\">[latex]\\begin{array}{l}A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt} ,\\text{ where}\\hfill \\\\ A\\left(t\\right)\\text{ is the account value at time }t\\hfill \\\\ t\\text{ is the number of years}\\hfill \\\\ P\\text{ is the initial investment, often called the principal}\\hfill \\\\ r\\text{ is the annual percentage rate (APR), or nominal rate}\\hfill \\\\ n\\text{ is the number of compounding periods in one year}\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 83px\">\n<td style=\"height: 83px\">continuous growth formula<\/td>\n<td style=\"height: 83px\">[latex]A\\left(t\\right)=a{e}^{rt},\\text{ where}[\/latex] [latex]t[\/latex]is the number of unit time periods of growth\n[latex]a[\/latex]is the starting amount (in the continuous compounding formula a is replaced with P, the principal)\n[latex]e[\/latex]is the mathematical constant,[latex] \\text{ }e\\approx 2.718282[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165137846440\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1165137846446\">\n \t<li>An exponential function is defined as a function with a positive constant other than[latex]\\,1\\,[\/latex]raised to a variable exponent. See <a class=\"autogenerated-content\" href=\"#Example_04_01_01\">(Figure)<\/a>.<\/li>\n \t<li>A function is evaluated by solving at a specific value. See <a class=\"autogenerated-content\" href=\"#Example_04_01_02\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_04_01_03\">(Figure)<\/a>.<\/li>\n \t<li>An exponential model can be found when the growth rate and initial value are known. See <a class=\"autogenerated-content\" href=\"#Example_04_01_04\">(Figure)<\/a>.<\/li>\n \t<li>An exponential model can be found when the two data points from the model are known. See <a class=\"autogenerated-content\" href=\"#Example_04_01_05\">(Figure)<\/a>.<\/li>\n \t<li>An exponential model can be found using two data points from the graph of the model. See <a class=\"autogenerated-content\" href=\"#Example_04_01_06\">(Figure)<\/a>.<\/li>\n \t<li>An exponential model can be found using two data points from the graph and a calculator. See <a class=\"autogenerated-content\" href=\"#Example_04_01_07\">(Figure)<\/a>.<\/li>\n \t<li>The value of an account at any time[latex]\\,t\\,[\/latex]can be calculated using the compound interest formula when the principal, annual interest rate, and compounding periods are known. See <a class=\"autogenerated-content\" href=\"#Example_04_01_08\">(Figure)<\/a>.<\/li>\n \t<li>The initial investment of an account can be found using the compound interest formula when the value of the account, annual interest rate, compounding periods, and life span of the account are known. See <a class=\"autogenerated-content\" href=\"#Example_04_01_09\">(Figure)<\/a>.<\/li>\n \t<li>The number[latex]\\,e\\,[\/latex]is a mathematical constant often used as the base of real world exponential growth and decay models. Its decimal approximation is[latex]\\,e\\approx 2.718282.[\/latex]<\/li>\n \t<li>Scientific and graphing calculators have the key[latex]\\,\\left[{e}^{x}\\right]\\,[\/latex]or[latex]\\,\\left[\\mathrm{exp}\\left(x\\right)\\right]\\,[\/latex]for calculating powers of[latex]\\,e.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Example_04_01_10\">(Figure)<\/a><strong>.<\/strong><\/li>\n \t<li>Continuous growth or decay models are exponential models that use[latex]\\,e\\,[\/latex]as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known. See <a class=\"autogenerated-content\" href=\"#Example_04_01_11\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_04_01_12\">(Figure)<\/a>.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165137406918\" class=\"textbox exercises\">\n<h3>Section Exercises<\/h3>\n<div id=\"fs-id1165137406922\" class=\"bc-section section\">\n<h4>Verbal<\/h4>\n<div id=\"fs-id1165137406928\">\n<div id=\"fs-id1165137406930\">\n<p id=\"fs-id1165137406932\">Explain why the values of an increasing exponential function will eventually overtake the values of an increasing linear function.<\/p>\n\n<\/div>\n<div id=\"fs-id1165137406938\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137406938\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137406938\"]\n<p id=\"fs-id1165137406940\">Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135536244\">\n<div id=\"fs-id1165135536246\">\n<p id=\"fs-id1165135536249\">Given a formula for an exponential function, is it possible to determine whether the function grows or decays exponentially just by looking at the formula? Explain.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135536255\">\n<div>\n<p id=\"fs-id1165135536260\">The Oxford Dictionary defines the word <em>nominal<\/em> as a value that is \u201cstated or expressed but not necessarily corresponding exactly to the real value.\u201d[footnote]Oxford Dictionary. <a href=\"http:\/\/oxforddictionaries.com\/us\/definition\/american_english\/nomina.\">http:\/\/oxforddictionaries.com\/us\/definition\/american_english\/nomina.<\/a>[\/footnote] Develop a reasonable argument for why the term <em>nominal rate<\/em> is used to describe the annual percentage rate of an investment account that compounds interest.<\/p>\n\n<\/div>\n<div id=\"fs-id1165135189885\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135189885\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135189885\"]\n<p id=\"fs-id1165135189887\">When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of <em>nominal<\/em>.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134340071\" class=\"bc-section section\">\n<h4>Algebraic<\/h4>\n<p id=\"fs-id1165134340076\">For the following exercises, identify whether the statement represents an exponential function. Explain.<\/p>\n\n<div id=\"fs-id1165134340080\">\n<div id=\"fs-id1165134340082\">\n<p id=\"fs-id1165134340084\">The average annual population increase of a pack of wolves is 25.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137737593\">\n<div id=\"fs-id1165137737595\">\n<p id=\"fs-id1165137737598\">A population of bacteria decreases by a factor of[latex]\\,\\frac{1}{8}\\,[\/latex]every[latex]\\,24\\,[\/latex]hours.<\/p>\n\n<\/div>\n<div id=\"fs-id1165137416069\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137416069\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137416069\"]\n<p id=\"fs-id1165135582265\">exponential; the population decreases by a proportional rate.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135582271\">\n<div id=\"fs-id1165135582273\">\n<p id=\"fs-id1165135582275\">The value of a coin collection has increased by[latex]\\,3.25%\\,[\/latex]annually over the last[latex]\\,20\\,[\/latex]years.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135414352\">\n<div id=\"fs-id1165135414354\">\n<p id=\"fs-id1165135318974\">For each training session, a personal trainer charges his clients[latex]\\,\\text{\\$}5\\,[\/latex]\nless than the previous training session.<\/p>\n\n<\/div>\n<div id=\"fs-id1165135318995\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135318995\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135318995\"]\n<p id=\"fs-id1165135318997\">not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function. .<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191053\">\n<div id=\"fs-id1165135191056\">\n<p id=\"fs-id1165135191058\">The height of a projectile at time[latex]\\,t\\,[\/latex]is represented by the function[latex]\\,h\\left(t\\right)=-4.9{t}^{2}+18t+40.[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<p id=\"fs-id1165137767849\">For the following exercises, consider this scenario: For each year[latex]\\,t,[\/latex]the population of a forest of trees is represented by the function[latex]\\,A\\left(t\\right)=115{\\left(1.025\\right)}^{t}.\\,[\/latex]In a neighboring forest, the population of the same type of tree is represented by the function[latex]\\,B\\left(t\\right)=82{\\left(1.029\\right)}^{t}.\\,[\/latex](Round answers to the nearest whole number.)<\/p>\n\n<div id=\"fs-id1165135570343\">\n<div id=\"fs-id1165135570345\">\n<p id=\"fs-id1165135570347\">Which forest\u2019s population is growing at a faster rate?<\/p>\n\n<\/div>\n<div id=\"fs-id1165135570352\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135570352\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135570352\"]\n<p id=\"fs-id1165135570355\">The forest represented by the function[latex]\\,B\\left(t\\right)=82{\\left(1.029\\right)}^{t}.[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134049940\">\n<div id=\"fs-id1165134049942\">\n<p id=\"fs-id1165137755646\">Which forest had a greater number of trees initially? By how many?<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137755651\">\n<div id=\"fs-id1165137755653\">\n<p id=\"fs-id1165137755656\">Assuming the population growth models continue to represent the growth of the forests, which forest will have a greater number of trees after[latex]\\,20\\,[\/latex]years? By how many?<\/p>\n\n<\/div>\n<div id=\"fs-id1165137761058\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137761058\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137761058\"]\n<p id=\"fs-id1165137761061\">After[latex]\\,t=20\\,[\/latex]years, forest A will have[latex]\\,43\\,[\/latex]more trees than forest B.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135508291\">\n<div id=\"fs-id1165135508293\">\n<p id=\"fs-id1165135508295\">Assuming the population growth models continue to represent the growth of the forests, which forest will have a greater number of trees after[latex]\\,100\\,[\/latex]years? By how many?<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137828270\">\n<div id=\"fs-id1165137828272\">\n<p id=\"fs-id1165137828274\">Discuss the above results from the previous four exercises. Assuming the population growth models continue to represent the growth of the forests, which forest will have the greater number of trees in the long run? Why? What are some factors that might influence the long-term validity of the exponential growth model?<\/p>\n\n<\/div>\n<div id=\"fs-id1165137407578\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137407578\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137407578\"]\n<p id=\"fs-id1165137407580\">Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\nFor the following exercises, determine whether the equation represents exponential growth, exponential decay, or neither. Explain.\n<div>\n<div id=\"fs-id1165137407597\">\n<p id=\"fs-id1165135560752\">[latex]y=300{\\left(1-t\\right)}^{5}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135252048\">\n<div id=\"fs-id1165135252050\">\n<p id=\"fs-id1165135252052\">[latex]y=220{\\left(1.06\\right)}^{x}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135209944\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135209944\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135209944\"]\n<p id=\"fs-id1165135209946\">exponential growth; The growth factor,[latex]\\,1.06,[\/latex] is greater than[latex]\\,1.[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137806960\">\n<div id=\"fs-id1165137806962\">\n<p id=\"fs-id1165137806964\">[latex]y=16.5{\\left(1.025\\right)}^{\\frac{1}{x}}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135519282\">\n<div id=\"fs-id1165135519284\">\n<p id=\"fs-id1165135519286\">[latex]y=11,701{\\left(0.97\\right)}^{t}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135397288\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135397288\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135397288\"]exponential decay; The decay factor,[latex]\\,0.97,[\/latex] is between[latex]\\,0\\,[\/latex]and[latex]\\,1.[\/latex][\/hidden-answer]\n\n<\/div>\n<\/div>\n<p id=\"fs-id1165135176427\">For the following exercises, find the formula for an exponential function that passes through the two points given.<\/p>\n\n<div id=\"fs-id1165135176432\">\n<div id=\"fs-id1165135176434\">\n<p id=\"fs-id1165135176436\">[latex]\\left(0,6\\right)\\,[\/latex]and[latex]\\,\\left(3,750\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137770026\">\n<div id=\"fs-id1165137770029\">\n<p id=\"fs-id1165137770031\">[latex]\\left(0,2000\\right)\\,[\/latex]and[latex]\\,\\left(2,20\\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137735589\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137735589\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137735589\"]\n<p id=\"fs-id1165135478535\">[latex]f\\left(x\\right)=2000{\\left(0.1\\right)}^{x}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135189914\">\n<div id=\"fs-id1165135189916\">\n<p id=\"fs-id1165135189919\">[latex]\\left(-1,\\frac{3}{2}\\right)\\,[\/latex]and[latex]\\,\\left(3,24\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137400553\">\n<div id=\"fs-id1165137400555\">\n<p id=\"fs-id1165137400557\">[latex]\\left(-2,6\\right)\\,[\/latex]and[latex]\\,\\left(3,1\\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135187318\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135187318\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135187318\"]\n<p id=\"fs-id1165135187320\">[latex]f\\left(x\\right)={\\left(\\frac{1}{6}\\right)}^{-\\frac{3}{5}}{\\left(\\frac{1}{6}\\right)}^{\\frac{x}{5}}\\approx 2.93{\\left(0.699\\right)}^{x}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135180070\">\n<div id=\"fs-id1165135180073\">\n<p id=\"fs-id1165135180075\">[latex]\\left(3,1\\right)\\,[\/latex]and[latex]\\,\\left(5,4\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<p id=\"fs-id1165134154588\">For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points.<\/p>\n\n<div id=\"fs-id1165134154593\">\n<div id=\"fs-id1165134154595\">\n<table id=\"fs-id1165134154597\" class=\"unnumbered\" summary=\"\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>70<\/td>\n<td>40<\/td>\n<td>10<\/td>\n<td>-20<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165137680442\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137680442\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137680442\"]\n<p id=\"fs-id1165137680445\">Linear<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137680450\">\n<div id=\"fs-id1165137680452\">\n<table id=\"fs-id1165137680454\" class=\"unnumbered\" summary=\"\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]h\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>70<\/td>\n<td>49<\/td>\n<td>34.3<\/td>\n<td>24.01<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135237063\">\n<div id=\"fs-id1165135237065\">\n<table class=\"unnumbered\" summary=\"\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]m\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>80<\/td>\n<td>61<\/td>\n<td>42.9<\/td>\n<td>25.61<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165137761697\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137761697\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137761697\"]\n<p id=\"fs-id1165137761699\">Neither<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137761705\">\n<div id=\"fs-id1165137761707\">\n<table id=\"fs-id1165137761709\" class=\"unnumbered\" summary=\"\"><caption>&nbsp;<\/caption>\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>10<\/td>\n<td>20<\/td>\n<td>40<\/td>\n<td>80<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<div>\n<div>\n<table id=\"fs-id1165137749151\" class=\"unnumbered\" summary=\"\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]g\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>-3.25<\/td>\n<td>2<\/td>\n<td>7.25<\/td>\n<td>12.5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165137874537\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137874537\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137874537\"]\n<p id=\"fs-id1165137874539\">Linear<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<p id=\"fs-id1165137874544\">For the following exercises, use the compound interest formula,[latex]\\,A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}.[\/latex]<\/p>\n\n<div>\n<div id=\"fs-id1165135435599\">\n<p id=\"fs-id1165135435601\">After a certain number of years, the value of an investment account is represented by the equation[latex]\\,10,250{\\left(1+\\frac{0.04}{12}\\right)}^{120}.\\,[\/latex]What is the value of the account?<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135527075\">\n<div id=\"fs-id1165135527077\">\n<p id=\"fs-id1165135527079\">What was the initial deposit made to the account in the previous exercise?<\/p>\n\n<\/div>\n<div id=\"fs-id1165135527084\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135527084\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135527084\"]\n<p id=\"fs-id1165135527086\">[latex]$10,250[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134042860\">\n<div id=\"fs-id1165134042862\">\n<p id=\"fs-id1165134042865\">How many years had the account from the previous exercise been accumulating interest?<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134042870\">\n<div id=\"fs-id1165134042872\">\n<p id=\"fs-id1165134042874\">An account is opened with an initial deposit of $6,500 and earns[latex]\\,3.6%\\,[\/latex]interest compounded semi-annually. What will the account be worth in[latex]\\,20\\,[\/latex]years?<\/p>\n\n<\/div>\n<div id=\"fs-id1165137749553\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137749553\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137749553\"]\n<p id=\"fs-id1165137749555\">[latex]$13,268.58[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135196890\">\n<div id=\"fs-id1165135196892\">\n<p id=\"fs-id1165135196894\">How much more would the account in the previous exercise have been worth if the interest were compounding weekly?<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135196900\">\n<div id=\"fs-id1165135196902\">\n<p id=\"fs-id1165137831959\">Solve the compound interest formula for the principal,[latex]\\,P[\/latex].<\/p>\n\n<\/div>\n<div id=\"fs-id1165137831975\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137831975\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137831975\"]\n<p id=\"fs-id1165137831977\">[latex]P=A\\left(t\\right)\\cdot {\\left(1+\\frac{r}{n}\\right)}^{-nt}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134234548\">\n<div id=\"fs-id1165134234550\">\n<p id=\"fs-id1165134234552\">Use the formula found in the previous exercise to calculate the initial deposit of an account that is worth[latex]\\,$14,472.74\\,[\/latex]after earning[latex]\\,5.5%\\,[\/latex]interest compounded monthly for[latex]\\,5\\,[\/latex]years. (Round to the nearest dollar.)<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135415804\">\n<div id=\"fs-id1165135415806\">\n<p id=\"fs-id1165135415808\">How much more would the account in the previous two exercises be worth if it were earning interest for[latex]\\,5\\,[\/latex]more years?<\/p>\n\n<\/div>\n<div id=\"fs-id1165135415827\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135415827\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135415827\"]\n<p id=\"fs-id1165135415829\">[latex]$4,572.56[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135528933\">\n<div id=\"fs-id1165135528935\">\n<p id=\"fs-id1165135528937\">Use properties of rational exponents to solve the compound interest formula for the interest rate,[latex]\\,r.[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137836684\">\n<div id=\"fs-id1165137836687\">\n<p id=\"fs-id1165137836689\">Use the formula found in the previous exercise to calculate the interest rate for an account that was compounded semi-annually, had an initial deposit of $9,000 and was worth $13,373.53 after 10 years.<\/p>\n\n<\/div>\n<div id=\"fs-id1165135487075\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135487075\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135487075\"]\n<p id=\"fs-id1165135487077\">[latex]4%[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134071602\">\n<div id=\"fs-id1165134071604\">\n<p id=\"fs-id1165134071606\">Use the formula found in the previous exercise to calculate the interest rate for an account that was compounded monthly, had an initial deposit of $5,500, and was worth $38,455 after 30 years.<\/p>\n\n<\/div>\n<\/div>\n<p id=\"fs-id1165135441763\">For the following exercises, determine whether the equation represents continuous growth, continuous decay, or neither. Explain.<\/p>\n\n<div id=\"fs-id1165135441768\">\n<div id=\"fs-id1165135441770\">\n<p id=\"fs-id1165135441772\">[latex]y=3742{\\left(e\\right)}^{0.75t}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135336087\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135336087\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135336087\"]\n<p id=\"fs-id1165135336089\">continuous growth; the growth rate is greater than[latex]\\,0.[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135336105\">\n<div id=\"fs-id1165135336107\">\n<p id=\"fs-id1165135336109\">[latex]y=150{\\left(e\\right)}^{\\frac{3.25}{t}}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135545718\">\n<div id=\"fs-id1165135545721\">\n<p id=\"fs-id1165135545723\">[latex]y=2.25{\\left(e\\right)}^{-2t}[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135191882\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135191882\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135191882\"]\n<p id=\"fs-id1165135191884\">continuous decay; the growth rate is less than[latex]\\,0.[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191900\">\n<div id=\"fs-id1165135191902\">\n<p id=\"fs-id1165135191904\">Suppose an investment account is opened with an initial deposit of[latex]\\,$12,000\\,[\/latex]earning[latex]\\,7.2%\\,[\/latex]interest compounded continuously. How much will the account be worth after[latex]\\,30\\,[\/latex]years?<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135471130\">\n<div id=\"fs-id1165135471132\">\n<p id=\"fs-id1165135471134\">How much less would the account from Exercise 42 be worth after[latex]\\,30\\,[\/latex]years if it were compounded monthly instead?<\/p>\n\n<\/div>\n<div id=\"fs-id1165135190851\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135190851\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135190851\"]\n<p id=\"fs-id1165135190853\">[latex]$669.42[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135190869\" class=\"bc-section section\">\n<h4>Numeric<\/h4>\n<p id=\"fs-id1165135421696\">For the following exercises, evaluate each function. Round answers to four decimal places, if necessary.<\/p>\n\n<div id=\"fs-id1165135421700\">\n<div id=\"fs-id1165135421702\">\n<p id=\"fs-id1165135421704\">[latex]f\\left(x\\right)=2{\\left(5\\right)}^{x},[\/latex] for[latex]\\,f\\left(-3\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137811660\">\n<div id=\"fs-id1165137811662\">\n<p id=\"fs-id1165135245591\">[latex]f\\left(x\\right)=-{4}^{2x+3},[\/latex] for[latex]\\,f\\left(-1\\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135415781\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135415781\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135415781\"]\n<p id=\"fs-id1165135415783\">[latex]f\\left(-1\\right)=-4[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135593586\">\n<div id=\"fs-id1165135593588\">\n<p id=\"fs-id1165135593590\">[latex]f\\left(x\\right)={e}^{x},[\/latex] for[latex]\\,f\\left(3\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135185903\">\n<div id=\"fs-id1165135185905\">\n<p id=\"fs-id1165135185907\">[latex]f\\left(x\\right)=-2{e}^{x-1},[\/latex] for[latex]\\,f\\left(-1\\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135500970\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135500970\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135500970\"]\n<p id=\"fs-id1165135500972\">[latex]f\\left(-1\\right)\\approx -0.2707[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135508316\">\n<div id=\"fs-id1165135508318\">\n<p id=\"fs-id1165135508320\">[latex]f\\left(x\\right)=2.7{\\left(4\\right)}^{-x+1}+1.5,[\/latex] for[latex]f\\left(-2\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134222572\">\n<div id=\"fs-id1165134222574\">\n<p id=\"fs-id1165134222576\">[latex]f\\left(x\\right)=1.2{e}^{2x}-0.3,[\/latex] for[latex]\\,f\\left(3\\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135510681\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135510681\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135510681\"]\n<p id=\"fs-id1165135510683\">[latex]f\\left(3\\right)\\approx 483.8146[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137749955\">\n<div id=\"fs-id1165137749957\">\n<p id=\"fs-id1165137749959\">[latex]f\\left(x\\right)=-\\frac{3}{2}{\\left(3\\right)}^{-x}+\\frac{3}{2},[\/latex] for[latex]\\,f\\left(2\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137843692\" class=\"bc-section section\">\n<h4>Technology<\/h4>\n<p id=\"fs-id1165137843697\">For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.<\/p>\n\n<div id=\"fs-id1165137843702\">\n<div id=\"fs-id1165137843704\">\n<p id=\"fs-id1165135251278\">[latex]\\left(0,3\\right)\\,[\/latex]and[latex]\\,\\left(3,375\\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165137715431\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165137715431\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165137715431\"]\n<p id=\"fs-id1165137715433\">[latex]y=3\\cdot {5}^{x}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137740770\">\n<div id=\"fs-id1165137740772\">\n<p id=\"fs-id1165137740775\">[latex]\\left(3,222.62\\right)\\,[\/latex]and[latex]\\,\\left(10,77.456\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135209922\">\n<div id=\"fs-id1165135209924\">\n<p id=\"fs-id1165135209926\">[latex]\\left(20,29.495\\right)\\,[\/latex]and[latex]\\,\\left(150,730.89\\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135347282\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135347282\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135347282\"]\n<p id=\"fs-id1165135347284\">[latex]y\\approx 18\\cdot {1.025}^{x}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135409820\">\n<div id=\"fs-id1165135409822\">\n<p id=\"fs-id1165135409824\">[latex]\\left(5,2.909\\right)\\,[\/latex]and[latex]\\,\\left(13,0.005\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135545870\">\n<div id=\"fs-id1165135545872\">\n<p id=\"fs-id1165135545874\">[latex]\\left(11,310.035\\right)\\,[\/latex] and [latex]\\left(25,356.3652\\right)[\/latex]<\/p>\n\n<\/div>\n<div id=\"fs-id1165135255110\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135255110\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135255110\"]\n<p id=\"fs-id1165135255112\">[latex]y\\approx 0.2\\cdot {1.95}^{x}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134388205\" class=\"bc-section section\">\n<h4>Extensions<\/h4>\n<div id=\"fs-id1165134388210\">\n<div id=\"fs-id1165134388212\">\n\nThe <em>annual percentage yield<\/em> (APY) of an investment account is a representation of the actual interest rate earned on a compounding account. It is based on a compounding period of one year. Show that the APY of an account that compounds monthly can be found with the formula[latex]\\,\\text{APY}={\\left(1+\\frac{r}{12}\\right)}^{12}-1.[\/latex]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135499889\">\n<div id=\"fs-id1165135499891\">\n<p id=\"fs-id1165135499893\">Repeat the previous exercise to find the formula for the APY of an account that compounds daily. Use the results from this and the previous exercise to develop a function[latex]\\,I\\left(n\\right)\\,[\/latex]for the APY of any account that compounds[latex]\\,n\\,[\/latex]times per year.<\/p>\n\n<\/div>\n<div id=\"fs-id1165135530489\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135530489\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135530489\"]\n<p id=\"fs-id1165135530491\">[latex]\\text{APY}=\\frac{A\\left(t\\right)-a}{a}=\\frac{a{\\left(1+\\frac{r}{365}\\right)}^{365\\left(1\\right)}-a}{a}=\\frac{a\\left[{\\left(1+\\frac{r}{365}\\right)}^{365}-1\\right]}{a}={\\left(1+\\frac{r}{365}\\right)}^{365}-1;[\/latex][latex]I\\left(n\\right)={\\left(1+\\frac{r}{n}\\right)}^{n}-1[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135408530\">\n<div id=\"fs-id1165135408533\">\n<p id=\"fs-id1165135408535\">Recall that an exponential function is any equation written in the form[latex]\\,f\\left(x\\right)=a\\cdot {b}^{x}\\,[\/latex]such that[latex] a [\/latex]and[latex] b [\/latex]are positive numbers and[latex] b\\ne 1. [\/latex]Any positive number[latex] b [\/latex]can be written as[latex] b={e}^{n} [\/latex]for some value of[latex] n[\/latex]. Use this fact to rewrite the formula for an exponential function that uses the number[latex] e [\/latex]as a base.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165134108543\">\n<div id=\"fs-id1165134108546\">\n<p id=\"fs-id1165134108548\">In an exponential decay function, the base of the exponent is a value between 0 and 1. Thus, for some number[latex]\\,b&gt;1,[\/latex] the exponential decay function can be written as[latex]\\,f\\left(x\\right)=a\\cdot {\\left(\\frac{1}{b}\\right)}^{x}.\\,[\/latex]Use this formula, along with the fact that[latex]\\,b={e}^{n},[\/latex] to show that an exponential decay function takes the form[latex]\\,f\\left(x\\right)=a{\\left(e\\right)}^{-nx}\\,[\/latex]for some positive number[latex]\\,n\\,[\/latex].<\/p>\n\n<\/div>\n<div id=\"fs-id1165135661465\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135661465\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135661465\"]\n<p id=\"fs-id1165135661467\">Let[latex]\\,f\\,[\/latex]be the exponential decay function[latex]\\,f\\left(x\\right)=a\\cdot {\\left(\\frac{1}{b}\\right)}^{x}\\,[\/latex]such that[latex]\\,b&gt;1.\\,[\/latex]Then for some number[latex]\\,n&gt;0,[\/latex][latex]f\\left(x\\right)=a\\cdot {\\left(\\frac{1}{b}\\right)}^{x}=a{\\left({b}^{-1}\\right)}^{x}=a{\\left({\\left({e}^{n}\\right)}^{-1}\\right)}^{x}=a{\\left({e}^{-n}\\right)}^{x}=a{\\left(e\\right)}^{-nx}.[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165137848892\">\n<div id=\"fs-id1165137848894\">\n<p id=\"fs-id1165137848897\">The formula for the amount[latex]\\,A\\,[\/latex]in an investment account with a nominal interest rate[latex]\\,r\\,[\/latex]at any time[latex]\\,t\\,[\/latex]is given by[latex]\\,A\\left(t\\right)=a{\\left(e\\right)}^{rt},[\/latex]where[latex]\\,a\\,[\/latex]is the amount of principal initially deposited into an account that compounds continuously. Prove that the percentage of interest earned to principal at any time[latex]\\,t\\,[\/latex]can be calculated with the formula[latex]\\,I\\left(t\\right)={e}^{rt}-1.[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137705181\" class=\"bc-section section\">\n<h4>Real-World Applications<\/h4>\n<div id=\"fs-id1165137705187\">\n<div id=\"fs-id1165137705189\">\n<p id=\"fs-id1165137705191\">The fox population in a certain region has an annual growth rate of 9% per year. In the year 2012, there were 23,900 fox counted in the area. What is the fox population predicted to be in the year 2020?<\/p>\n\n<\/div>\n<div id=\"fs-id1165135205873\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135205873\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135205873\"]\n<p id=\"fs-id1165135434782\">[latex]47,622\\,[\/latex]fox<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135434802\">\n<div id=\"fs-id1165135434804\">\n<p id=\"fs-id1165135434806\">A scientist begins with 100 milligrams of a radioactive substance that decays exponentially. After 35 hours, 50mg of the substance remains. How many milligrams will remain after 54 hours?<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135236978\">\n<div id=\"fs-id1165135236980\">\n<p id=\"fs-id1165135236982\">In the year 1985, a house was valued at $110,000. By the year 2005, the value had appreciated to $145,000. What was the annual growth rate between 1985 and 2005? Assume that the value continued to grow by the same percentage. What was the value of the house in the year 2010?<\/p>\n\n<\/div>\n<div id=\"fs-id1165134086071\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165134086071\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165134086071\"]\n<p id=\"fs-id1165134086074\">[latex]1.39%;\\,[\/latex][latex]$155,368.09[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135432945\">\n<div id=\"fs-id1165135432947\">\n<p id=\"fs-id1165135432949\">A car was valued at $38,000 in the year 2007. By 2013, the value had depreciated to $11,000 If the car\u2019s value continues to drop by the same percentage, what will it be worth by 2017?<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135176224\">\n<div id=\"fs-id1165135176226\">\n<p id=\"fs-id1165135176228\">Jamal wants to save $54,000 for a down payment on a home. How much will he need to invest in an account with 8.2% APR, compounding daily, in order to reach his goal in 5 years?<\/p>\n\n<\/div>\n<div id=\"fs-id1165135208510\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135208510\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135208510\"]\n<p id=\"fs-id1165135208512\">[latex]$35,838.76[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135311577\">\n<div id=\"fs-id1165135311579\">\n\nKyoko has $10,000 that she wants to invest. Her bank has several investment accounts to choose from, all compounding daily. Her goal is to have $15,000 by the time she finishes graduate school in 6 years. To the nearest hundredth of a percent, what should her minimum annual interest rate be in order to reach her goal? (<em>Hint<\/em>: solve the compound interest formula for the interest rate.)\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135337738\">\n<div id=\"fs-id1165135337740\">\n<p id=\"fs-id1165135337742\">Alyssa opened a retirement account with 7.25% APR in the year 2000. Her initial deposit was $13,500. How much will the account be worth in 2025 if interest compounds monthly? How much more would she make if interest compounded continuously?<\/p>\n\n<\/div>\n<div id=\"fs-id1165135567423\" class=\"solution textbox shaded\">[reveal-answer q=\"fs-id1165135567423\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1165135567423\"]\n<p id=\"fs-id1165135567426\">[latex]$82,247.78;\\,[\/latex][latex]$449.75[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1165135567460\">\n<div id=\"fs-id1165135567462\">\n<p id=\"fs-id1165135567464\">An investment account with an annual interest rate of 7% was opened with an initial deposit of $4,000 Compare the values of the account after 9 years when the interest is compounded annually, quarterly, monthly, and continuously.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165135397912\">\n \t<dt>annual percentage rate (APR)<\/dt>\n \t<dd id=\"fs-id1165135397918\">the yearly interest rate earned by an investment account, also called <em>nominal rate<\/em><\/dd>\n<\/dl>\n<dl id=\"fs-id1165135397926\">\n \t<dt>compound interest<\/dt>\n \t<dd id=\"fs-id1165135397932\">interest earned on the total balance, not just the principal<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838635\">\n \t<dt>exponential growth<\/dt>\n \t<dd id=\"fs-id1165137838640\">a model that grows by a rate proportional to the amount present<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838644\">\n \t<dt>nominal rate<\/dt>\n \t<dd id=\"fs-id1165137838650\">the yearly interest rate earned by an investment account, also called <em>annual percentage rate<\/em><\/dd>\n<\/dl>\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Evaluate exponential functions.<\/li>\n<li>Find the equation of an exponential function.<\/li>\n<li>Use compound interest formulas.<\/li>\n<li>Evaluate exponential functions with base e.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137640062\">India is the second most populous country in the world with a population of about[latex]\\,1.25\\,[\/latex]billion people in 2013. The population is growing at a rate of about[latex]\\,1.2%\\,[\/latex]each year<a class=\"footnote\" title=\"http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014.\" id=\"return-footnote-98-1\" href=\"#footnote-98-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> . If this rate continues, the population of India will exceed China\u2019s population by the year[latex]\\,2031.[\/latex]When populations grow rapidly, we often say that the growth is \u201cexponential,\u201d meaning that something is growing very rapidly. To a mathematician, however, the term <em>exponential growth <\/em>has a very specific meaning. In this section, we will take a look at <em>exponential functions<\/em>, which model this kind of rapid growth.<\/p>\n<div id=\"fs-id1165135159940\" class=\"bc-section section\">\n<h3>Identifying Exponential Functions<\/h3>\n<p id=\"fs-id1165137446191\">When exploring linear growth, we observed a constant rate of change\u2014a constant number by which the output increased for each unit increase in input. For example, in the equation[latex]\\,f\\left(x\\right)=3x+4,[\/latex]the slope tells us the output increases by 3 each time the input increases by 1. The scenario in the India population example is different because we have a <em>percent<\/em> change per unit time (rather than a constant change) in the number of people.<\/p>\n<div id=\"fs-id1165137477143\" class=\"bc-section section\">\n<h4>Defining an Exponential Function<\/h4>\n<p id=\"fs-id1165137553324\">A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, 2.5% of the population was vegan, adhering to a diet that does not include any animal products\u2014no meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up 10% of the U.S. population in 2015, 40% in 2019, and 80% in 2021.<\/p>\n<p id=\"fs-id1165134069131\">What exactly does it mean to <em>grow exponentially<\/em>? What does the word <em>double <\/em>have in common with <em>percent increase<\/em>? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media.<\/p>\n<ul id=\"fs-id1165134042783\">\n<li><strong>Percent change <\/strong>refers to a <em>change<\/em> based on a <em>percent<\/em> of the original amount.<\/li>\n<li><strong>Exponential growth <\/strong>refers to an <em>increase<\/em> based on a constant multiplicative rate of change over equal increments of time, that is, a <em>percent<\/em> increase of the original amount over time.<\/li>\n<li>Exponential decay refers to a <em>decrease<\/em> based on a constant multiplicative rate of change over equal increments of time, that is, a <em>percent<\/em> decrease of the original amount over time.<\/li>\n<\/ul>\n<p id=\"fs-id1165137760753\">For us to gain a clear understanding of <span class=\"no-emphasis\">exponential growth<\/span>, let us contrast exponential growth with <span class=\"no-emphasis\">linear growth<\/span>. We will construct two functions. The first function is exponential. We will start with an input of 0, and increase each input by 1. We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of 0, and increase each input by 1. We will add 2 to the corresponding consecutive outputs. See <a class=\"autogenerated-content\" href=\"#Table_04_01_01\">(Figure)<\/a>.<\/p>\n<table id=\"Table_04_01_01\" summary=\"Eight rows and three columns. The first column is labeled, \u201cx\u201d, which goes from 0 to 6; the second column is labeled, \u201cf(x)=2^x\u201d; and the third column is labeled, \u201cg(x) = 2x\u201d. The following values are for the function f: (0, 1), (1, 2), (2, 4), (3, 8), (4, 16), (5, 32), and (6, 64). The following values are for the function g: (0, 0), (1, 2), (2, 4), (3, 6), (4, 8), (5, 10), and (6, 12).\">\n<thead>\n<tr>\n<th>[latex]x[\/latex]<\/th>\n<th>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/th>\n<th>[latex]g\\left(x\\right)=2x[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>1<\/td>\n<td>0<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>2<\/td>\n<td>2<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>4<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>8<\/td>\n<td>6<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>16<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>32<\/td>\n<td>10<\/td>\n<\/tr>\n<tr>\n<td>6<\/td>\n<td>64<\/td>\n<td>12<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165135190676\">From <a class=\"autogenerated-content\" href=\"#Table_04_01_01\">(Figure)<\/a> we can infer that for these two functions, exponential growth dwarfs linear growth.<\/p>\n<ul id=\"fs-id1165137725808\">\n<li><strong>Exponential growth <\/strong>refers to the original value from the range increases by the <em>same percentage<\/em> over equal increments found in the domain.<\/li>\n<li><strong>Linear growth<\/strong> refers to the original value from the range increases by the <em>same amount<\/em> over equal increments found in the domain.<\/li>\n<\/ul>\n<p id=\"fs-id1165137561507\">Apparently, the difference between \u201cthe same percentage\u201d and \u201cthe same amount\u201d is quite significant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding 2 to the output whenever the input was increased by one.<\/p>\n<p id=\"fs-id1165135445949\">The general form of the <span class=\"no-emphasis\">exponential function<\/span> is[latex]\\,f\\left(x\\right)=a{b}^{x},\\,[\/latex]where[latex]\\,a\\,[\/latex]is any nonzero number,[latex]\\,b\\,[\/latex]is a positive real number not equal to 1.<\/p>\n<ul id=\"fs-id1165137635065\">\n<li>If[latex]\\,b>1,[\/latex]the function grows at a rate proportional to its size.<\/li>\n<li>If[latex]\\,0<b<1,[\/latex] the function decays at a rate proportional to its size.<\/li>\n<\/ul>\n<p id=\"fs-id1165137465225\">Let\u2019s look at the function[latex]\\,f\\left(x\\right)={2}^{x}\\,[\/latex]from our example. We will create a table (<a class=\"autogenerated-content\" href=\"#Table_04_01_02\">(Figure)<\/a>) to determine the corresponding outputs over an interval in the domain from[latex]\\,-3\\,[\/latex]to[latex]\\,3.[\/latex]<\/p>\n<table summary=\"Two rows and eight columns. The first row is labeled, \u201cx\u201d, and the second row is labeled, \u201cf(x)=2^x\u201d. Reading the columns as ordered pairs, we have the following values: (-3, 2^(-3)=1\/8), (-2, 2^(-2)=1\/4), (-1, 2^(-1)=1\/2), (0, 2^(0)=1), (1, 2^(1)=2), (2, 2^(2)=4), and (3, 2^(3)=8).\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>[latex]-3[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\n<td>[latex]{2}^{-3}=\\frac{1}{8}[\/latex]<\/td>\n<td>[latex]{2}^{-2}=\\frac{1}{4}[\/latex]<\/td>\n<td>[latex]{2}^{-1}=\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]{2}^{0}=1[\/latex]<\/td>\n<td>[latex]{2}^{1}=2[\/latex]<\/td>\n<td>[latex]{2}^{2}=4[\/latex]<\/td>\n<td>[latex]{2}^{3}=8[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137640874\">Let us examine the graph of[latex]\\,f\\,[\/latex]by plotting the ordered pairs we observe on the table in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_006\">(Figure)<\/a>, and then make a few observations.<\/p>\n<div id=\"CNX_Precalc_Figure_04_01_006\" class=\"medium aligncenter\">\n<figure style=\"width: 731px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140059\/CNX_Precalc_Figure_04_01_006.jpg\" alt=\"Graph of Companies A and B\u2019s functions, which values are found in the previous table.\" width=\"731\" height=\"479\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 1.<\/strong><\/figcaption><\/figure>\n<\/div>\n<p id=\"fs-id1165137408862\">Let\u2019s define the behavior of the graph of the exponential function[latex]\\,f\\left(x\\right)={2}^{x}\\,[\/latex]and highlight some its key characteristics.<\/p>\n<ul id=\"fs-id1165137566018\">\n<li>the domain is[latex]\\,\\left(-\\infty ,\\infty \\right),[\/latex]<\/li>\n<li>the range is[latex]\\,\\left(0,\\infty \\right),[\/latex]<\/li>\n<li>as[latex]\\,x\\to \\infty ,f\\left(x\\right)\\to \\infty ,[\/latex]<\/li>\n<li>as [latex]\\,x\\to -\\infty ,f\\left(x\\right)\\to 0,[\/latex]<\/li>\n<li>[latex]\\,f\\left(x\\right)\\,[\/latex]is always increasing,<\/li>\n<li>the graph of[latex]\\,f\\left(x\\right)\\,[\/latex]will never touch the <em>x<\/em>-axis because base two raised to any exponent never has the result of zero.<\/li>\n<li>[latex]\\,y=0\\,[\/latex]is the horizontal asymptote.<\/li>\n<li>the <em>y<\/em>-intercept is 1.<\/li>\n<\/ul>\n<div id=\"fs-id1165137442472\" class=\"textbox key-takeaways\">\n<h3>Exponential Function<\/h3>\n<p id=\"fs-id1165137911387\">For any real number[latex]\\,x,[\/latex]an exponential function is a function with the form<\/p>\n<div id=\"Equation_4_1_1\">[latex]f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\n<p id=\"eip-751\">where<\/p>\n<ul id=\"fs-id1165137401680\">\n<li>[latex]\\,a\\,[\/latex]is a non-zero real number called the initial value and<\/li>\n<li>[latex]\\,b\\,[\/latex]is any positive real number such that[latex]\\,b\\ne 1.[\/latex]<\/li>\n<li>The domain of[latex]\\,f\\,[\/latex]is all real numbers.<\/li>\n<li>The range of[latex]\\,f\\,[\/latex]is all positive real numbers if[latex]\\,a>0.[\/latex]<\/li>\n<li>The range of[latex]\\,f\\,[\/latex]is all negative real numbers if[latex]\\,a<0.[\/latex]<\/li>\n<li>The <em>y<\/em>-intercept is[latex]\\,\\left(0,a\\right),[\/latex]and the horizontal asymptote is[latex]\\,y=0.[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165137387381\">\n<div id=\"fs-id1165137659178\">\n<h3>Identifying Exponential Functions<\/h3>\n<p id=\"fs-id1165137601478\">Which of the following equations are <em>not<\/em> exponential functions?<\/p>\n<ul id=\"fs-id1165135176602\">\n<li>[latex]f\\left(x\\right)={4}^{3\\left(x-2\\right)}[\/latex]<\/li>\n<li>[latex]g\\left(x\\right)={x}^{3}[\/latex]<\/li>\n<li>[latex]h\\left(x\\right)={\\left(\\frac{1}{3}\\right)}^{x}[\/latex]<\/li>\n<li>[latex]j\\left(x\\right)={\\left(-2\\right)}^{x}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165134108513\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137698136\">By definition, an exponential function has a constant as a base and an independent variable as an exponent. Thus,[latex]\\,g\\left(x\\right)={x}^{3}\\,[\/latex]does not represent an exponential function because the base is an independent variable. In fact,[latex]\\,g\\left(x\\right)={x}^{3}\\,[\/latex]is a power function.<\/p>\n<p id=\"fs-id1165137612252\">Recall that the base <em>b <\/em>of an exponential function is always a positive constant, and[latex]\\,b\\ne 1.\\,[\/latex]Thus,[latex]\\,j\\left(x\\right)={\\left(-2\\right)}^{x}\\,[\/latex]does not represent an exponential function because the base,[latex]\\,-2,[\/latex] is less than[latex]\\,0.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137436342\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1165137862673\">\n<p id=\"fs-id1165137424249\">Which of the following equations represent exponential functions?<\/p>\n<ul id=\"fs-id1165135161022\">\n<li>[latex]f\\left(x\\right)=2{x}^{2}-3x+1[\/latex]<\/li>\n<li>[latex]g\\left(x\\right)={0.875}^{x}[\/latex]<\/li>\n<li>[latex]h\\left(x\\right)=1.75x+2[\/latex]<\/li>\n<li>[latex]j\\left(x\\right)={1095.6}^{-2x}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165137597685\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137714559\">[latex]g\\left(x\\right)={0.875}^{x}\\,[\/latex]and[latex]j\\left(x\\right)={1095.6}^{-2x}\\,[\/latex]represent exponential functions.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137640042\" class=\"bc-section section\">\n<h3>Evaluating Exponential Functions<\/h3>\n<p id=\"fs-id1165137784783\">Recall that the base of an exponential function must be a positive real number other than[latex]\\,1.[\/latex]Why do we limit the base [latex]b\\,[\/latex]to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:<\/p>\n<ul id=\"fs-id1165137754880\">\n<li>Let[latex]\\,b=-9\\,[\/latex]and[latex]\\,x=\\frac{1}{2}.\\,[\/latex]Then[latex]\\,f\\left(x\\right)=f\\left(\\frac{1}{2}\\right)={\\left(-9\\right)}^{\\frac{1}{2}}=\\sqrt{-9},[\/latex]which is not a real number.<\/li>\n<\/ul>\n<p id=\"fs-id1165137563360\">Why do we limit the base to positive values other than [latex]1?[\/latex]Because base [latex]1\\,[\/latex]results in the constant function. Observe what happens if the base is [latex]1:[\/latex]<\/p>\n<ul id=\"fs-id1165137400268\">\n<li>Let[latex]\\,b=1.\\,[\/latex]Then[latex]\\,f\\left(x\\right)={1}^{x}=1\\,[\/latex]for any value of[latex]\\,x.[\/latex]<\/li>\n<\/ul>\n<p id=\"fs-id1165137459694\">To evaluate an exponential function with the form[latex]\\,f\\left(x\\right)={b}^{x},[\/latex]we simply substitute [latex]x\\,[\/latex]with the given value, and calculate the resulting power. For example:<\/p>\n<p id=\"fs-id1165135403544\">Let [latex]\\,f\\left(x\\right)={2}^{x}.\\,[\/latex]What is [latex]f\\left(3\\right)?[\/latex]<\/p>\n<div id=\"eip-id1165137643186\" class=\"unnumbered\">[latex]\\begin{array}{lll}f\\left(x\\right)\\hfill & ={2}^{x}\\hfill & \\hfill \\\\ f\\left(3\\right)\\hfill & ={2}^{3}\\text{ }\\hfill & \\text{Substitute }x=3.\\hfill \\\\ \\hfill & =8\\text{ }\\hfill & \\text{Evaluate the power}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137849020\">To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:<\/p>\n<p id=\"fs-id1165137849024\">Let[latex]\\,f\\left(x\\right)=30{\\left(2\\right)}^{x}.\\,[\/latex]What is[latex]\\,f\\left(3\\right)?[\/latex]<\/p>\n<div id=\"eip-id1165134086025\" class=\"unnumbered\">[latex]\\begin{array}{lll}f\\left(x\\right)\\hfill & =30{\\left(2\\right)}^{x}\\hfill & \\hfill \\\\ f\\left(3\\right)\\hfill & =30{\\left(2\\right)}^{3}\\hfill & \\text{Substitute }x=3.\\hfill \\\\ \\hfill & =30\\left(8\\right)\\text{ }\\hfill & \\text{Simplify the power first}\\text{.}\\hfill \\\\ \\hfill & =240\\hfill & \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137841073\">Note that if the order of operations were not followed, the result would be incorrect:<\/p>\n<div id=\"eip-id1165135320147\" class=\"unnumbered\">[latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165137455430\">\n<div id=\"fs-id1165137455432\">\n<h3>Evaluating Exponential Functions<\/h3>\n<p id=\"fs-id1165137767841\">Let [latex]\\,f\\left(x\\right)=5{\\left(3\\right)}^{x+1}.\\,[\/latex]Evaluate[latex]\\,f\\left(2\\right)\\,[\/latex]without using a calculator.<\/p>\n<\/div>\n<div id=\"fs-id1165135429364\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137598173\">Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\n<div id=\"eip-id1165135208555\" class=\"unnumbered\">[latex]\\begin{array}{lll}f\\left(x\\right)\\hfill & =5{\\left(3\\right)}^{x+1}\\hfill & \\hfill \\\\ f\\left(2\\right)\\hfill & =5{\\left(3\\right)}^{2+1}\\hfill & \\text{Substitute }x=2.\\hfill \\\\ \\hfill & =5{\\left(3\\right)}^{3}\\hfill & \\text{Add the exponents}.\\hfill \\\\ \\hfill & =5\\left(27\\right)\\hfill & \\text{Simplify the power}\\text{.}\\hfill \\\\ \\hfill & =135\\hfill & \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137843864\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1165137762786\">\n<p id=\"fs-id1165137762788\">Let[latex]f\\left(x\\right)=8{\\left(1.2\\right)}^{x-5}.\\,[\/latex]Evaluate[latex]\\,f\\left(3\\right)\\,[\/latex]using a calculator. Round to four decimal places.<\/p>\n<\/div>\n<div id=\"fs-id1165137637369\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>[latex]5.5556[\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137439018\" class=\"bc-section section\">\n<h4>Defining Exponential Growth<\/h4>\n<p id=\"fs-id1165137748523\">Because the output of exponential functions increases very rapidly, the term \u201cexponential growth\u201d is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.<\/p>\n<div id=\"fs-id1165137564690\" class=\"textbox key-takeaways\">\n<h3>Exponential Growth<\/h3>\n<p id=\"fs-id1165137834019\">A function that models exponential growth grows by a rate proportional to the amount present. For any real number[latex]\\,x\\,[\/latex]and any positive real numbers[latex]\\,a \\,[\/latex]and[latex]\\,b\\,[\/latex]such that[latex]\\,b\\ne 1,[\/latex]an exponential growth function has the form<\/p>\n<div id=\"fs-id1165137851784\" class=\"unnumbered aligncenter\">[latex]\\text{ }f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\n<p>where<\/p>\n<ul id=\"fs-id1165137863819\">\n<li>[latex]a\\,[\/latex]is the initial or starting value of the function.<\/li>\n<li>[latex]b\\,[\/latex]is the growth factor or growth multiplier per unit[latex]\\,x[\/latex].<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137644244\">In more general terms, we have an <em>exponential function<\/em>, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let\u2019s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function[latex]\\,A\\left(x\\right)=100+50x.\\,[\/latex]Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function [latex]\\,B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}.[\/latex]<\/p>\n<p id=\"fs-id1165135512493\">A few years of growth for these companies are illustrated in <a class=\"autogenerated-content\" href=\"#Table_04_01_05\">(Figure)<\/a>.<\/p>\n<table summary=\"Six rows and three columns. The first column is labeled, \u201cYear, x\u201d, which goes from 0 to 3; the second column is labeled, \u201cStores, Company A\u201d, which has a function of A(x) = 100+50x; and the third column is labeled, \u201cStores, Company B\u201d, which has a function of B(x)=100(1+0.5)^x. The following values are for Company A\u2019s function: (0, 100), (1, 150), (2, 200), and (3, 250). The following values are for the function Company B\u2019s function: (0, 100), (1, 150), (2, 225), and (3, 337.5).\">\n<thead>\n<tr>\n<th>Year, [latex]x[\/latex]<\/th>\n<th>Stores, Company A<\/th>\n<th>Stores, Company B<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]100+50\\left(0\\right)=100[\/latex]<\/td>\n<td>[latex]100{\\left(1+0.5\\right)}^{0}=100[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]100+50\\left(1\\right)=150[\/latex]<\/td>\n<td>[latex]100{\\left(1+0.5\\right)}^{1}=150[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]100+50\\left(2\\right)=200[\/latex]<\/td>\n<td>[latex]100{\\left(1+0.5\\right)}^{2}=225[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]100+50\\left(3\\right)=250[\/latex]<\/td>\n<td>[latex]100{\\left(1+0.5\\right)}^{3}=337.5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]x[\/latex]<\/td>\n<td>[latex]A\\left(x\\right)=100+50x[\/latex]<\/td>\n<td>[latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137653733\">The graphs comparing the number of stores for each company over a five-year period are shown in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_001\">(Figure)<\/a><strong>.<\/strong> We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.<\/p>\n<div id=\"CNX_Precalc_Figure_04_01_001\" class=\"small aligncenter\">\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140101\/CNX_Precalc_Figure_04_01_001.jpg\" alt=\"Graph of Companies A and B\u2019s functions, which values are found in the previous table.\" width=\"487\" height=\"845\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 2. <\/strong>The graph shows the numbers of stores Companies A and B opened over a five-year period.<\/figcaption><\/figure>\n<\/div>\n<p id=\"fs-id1165135209682\">Notice that the domain for both functions is[latex]\\,\\left[0,\\infty \\right),[\/latex]and the range for both functions is[latex]\\,\\left[100,\\infty \\right).\\,[\/latex]After year 1, Company B always has more stores than Company A.<\/p>\n<p id=\"fs-id1165137836429\">Now we will turn our attention to the function representing the number of stores for Company B,[latex]\\,B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}.\\,[\/latex]In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and[latex]\\,1+0.5=1.5\\,[\/latex]represents the growth factor. Generalizing further, we can write this function as[latex]\\,B\\left(x\\right)=100{\\left(1.5\\right)}^{x},[\/latex]where 100 is the initial value,[latex]\\,1.5\\,[\/latex]is called the <em>base<\/em>, and[latex]\\,x\\,[\/latex]is called the <em>exponent<\/em>.<\/p>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165137535640\">\n<div id=\"fs-id1165137535642\">\n<h3>Evaluating a Real-World Exponential Model<\/h3>\n<p id=\"fs-id1165135541867\">At the beginning of this section, we learned that the population of India was about[latex]\\,1.25\\,[\/latex]billion in the year 2013, with an annual growth rate of about[latex]\\,1.2%.\\,[\/latex]This situation is represented by the growth function[latex]\\,P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t},[\/latex] where [latex]\\,t\\,[\/latex] is the number of years since[latex]\\,2013.\\,[\/latex]To the nearest thousandth, what will the population of India be in[latex]\\,\\text{2031?}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137786632\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137786635\">To estimate the population in 2031, we evaluate the models for[latex]\\,t=18,[\/latex] because 2031 is[latex]\\,18[\/latex] years after 2013. Rounding to the nearest thousandth,<\/p>\n<div id=\"eip-id1165135657117\" class=\"unnumbered\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/div>\n<p id=\"fs-id1165135394343\">There will be about 1.549 billion people in India in the year 2031.<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135536569\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_04_01_03\">\n<div id=\"fs-id1165137635312\">\n<p id=\"fs-id1165137635314\">The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about[latex]\\,0.6%.\\,[\/latex]This situation is represented by the growth function[latex]\\,P\\left(t\\right)=1.39{\\left(1.006\\right)}^{t},[\/latex] where [latex]\\,t\\,[\/latex] is the number of years since[latex]\\,2013.[\/latex]To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in <a class=\"autogenerated-content\" href=\"#Example_04_01_03\">(Figure)<\/a>?<\/p>\n<\/div>\n<div id=\"fs-id1165134200184\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134200186\">About[latex]\\,1.548\\,[\/latex]billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135526980\" class=\"bc-section section\">\n<h3>Finding Equations of Exponential Functions<\/h3>\n<p id=\"fs-id1165135526985\">In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants[latex]\\,a\\,[\/latex]and[latex]\\,b,[\/latex]and evaluate the function.<\/p>\n<div id=\"fs-id1165135369632\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165135180102\"><strong>Given two data points, write an exponential model.<\/strong><\/p>\n<ol id=\"fs-id1165135180107\" type=\"1\">\n<li>If one of the data points has the form[latex]\\,\\left(0,a\\right),[\/latex] then[latex]\\,a\\,[\/latex]is the initial value. Using[latex]\\,a,[\/latex] substitute the second point into the equation[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x},[\/latex] and solve for[latex]\\,b.[\/latex]<\/li>\n<li>If neither of the data points have the form[latex]\\,\\left(0,a\\right),[\/latex] substitute both points into two equations with the form[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x}.\\,[\/latex]Solve the resulting system of two equations in two unknowns to find[latex]\\,a\\,[\/latex]and[latex]\\,b.[\/latex]<\/li>\n<li>Using the[latex]\\,a\\,[\/latex]and[latex]\\,b\\,[\/latex]found in the steps above, write the exponential function in the form[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x}.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165137580876\">\n<div id=\"fs-id1165137580878\">\n<h3>Writing an Exponential Model When the Initial Value Is Known<\/h3>\n<p id=\"fs-id1165137667588\">In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function[latex]\\,N\\left(t\\right)\\,[\/latex]representing the population[latex]\\,\\left(N\\right)\\,[\/latex]of deer over time[latex]\\,t.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135188416\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135188418\">We let our independent variable[latex]\\,t\\,[\/latex]be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function,[latex]\\,a=80.\\,[\/latex]We can now substitute the second point into the equation[latex]\\,N\\left(t\\right)=80{b}^{t}\\,[\/latex]to find[latex]\\,b:[\/latex]<\/p>\n<div id=\"eip-id1165135432669\" class=\"unnumbered\">[latex]\\begin{array}{lll}N\\left(t\\right)\\hfill & =80{b}^{t}\\hfill & \\hfill \\\\ \\,\\,180\\hfill & =80{b}^{6}\\hfill & \\text{Substitute using point }\\left(6, 180\\right).\\hfill \\\\ \\,\\,\\,\\,\\,\\frac{9}{4}\\hfill & ={b}^{6}\\hfill & \\text{Divide and write in lowest terms}.\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,b\\hfill & ={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}}\\hfill & \\text{Isolate }b\\text{ using properties of exponents}.\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,b\\hfill & \\approx 1.1447\\begin{array}{cccc}& & & \\end{array}\\hfill & \\text{Round to 4 decimal places}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135193230\"><strong>NOTE:<\/strong><em>Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.<\/em><\/p>\n<p id=\"fs-id1165137705073\">The exponential model for the population of deer is[latex]\\,N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}.\\,[\/latex](Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)<\/p>\n<p>We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_002\">(Figure)<\/a> passes through the initial points given in the problem,[latex]\\,\\left(0,\\text{ 8}0\\right)\\,[\/latex]and[latex]\\,\\left(\\text{6},\\text{ 18}0\\right).\\,[\/latex]We can also see that the domain for the function is[latex]\\,\\left[0,\\infty \\right),[\/latex]and the range for the function is[latex]\\,\\left[80,\\infty \\right).[\/latex]<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140104\/CNX_Precalc_Figure_04_01_002.jpg\" alt=\"Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).\" width=\"487\" height=\"700\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 3.<\/strong> Graph showing the population of deer over time,[latex]\\,N\\left(t\\right)=80{\\left(1.1447\\right)}^{t},[\/latex][latex]t\\,[\/latex]years after 2006<\/figcaption><\/figure>\n<p id=\"fs-id1165137724117\"><\/details>\n<\/p>\n<div id=\"CNX_Precalc_Figure_04_01_002\" class=\"small aligncenter\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137761908\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1165135496544\">\n<p id=\"fs-id1165135496547\">A wolf population is growing exponentially. In 2011,[latex]\\,129\\,[\/latex]wolves were counted. By[latex]\\,\\text{2013,}\\,[\/latex]the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population[latex]\\,N\\,[\/latex]of wolves over time[latex]\\,t.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137737094\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137737097\">[latex]\\left(0,129\\right)\\,[\/latex]and[latex]\\,\\left(2,236\\right);\\,\\,\\,N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165135411399\">\n<div id=\"fs-id1165135411402\">\n<h3>Writing an Exponential Model When the Initial Value is Not Known<\/h3>\n<p id=\"fs-id1165135411407\">Find an exponential function that passes through the points[latex]\\,\\left(-2,6\\right)\\,[\/latex]and[latex]\\,\\left(2,1\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135646190\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135646192\">Because we don\u2019t have the initial value, we substitute both points into an equation of the form[latex]\\,f\\left(x\\right)=a{b}^{x},[\/latex] and then solve the system for[latex]\\,a\\,[\/latex]and[latex]\\,b.[\/latex]<\/p>\n<ul id=\"fs-id1165134044679\">\n<li>Substituting[latex]\\,\\left(-2,6\\right)\\,[\/latex]gives[latex]\\,6=a{b}^{-2}\\,[\/latex]<\/li>\n<li>Substituting[latex]\\,\\left(2,1\\right)\\,[\/latex]gives[latex]\\,1=a{b}^{2}\\,[\/latex]<\/li>\n<\/ul>\n<p id=\"fs-id1165135361777\">Use the first equation to solve for[latex]\\,a\\,[\/latex]in terms of[latex]\\,b:[\/latex]<\/p>\n<p><span id=\"eip-id1619372\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140111\/CNX_Precalc_Figure_04_01_007.jpg\" alt=\"..\" \/><\/span><\/p>\n<p id=\"fs-id1165135180085\">Substitute[latex]\\,a\\,[\/latex]in the second equation, and solve for[latex]\\,b:[\/latex]<\/p>\n<p><span id=\"eip-id2591037\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140118\/CNX_Precalc_Figure_04_01_008.jpg\" alt=\"..\" \/><\/span><\/p>\n<p id=\"fs-id1165137761627\">Use the value of[latex]\\,b\\,[\/latex]in the first equation to solve for the value of[latex]\\,a:[\/latex]<span id=\"eip-id2811781\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140126\/CNX_Precalc_Figure_04_01_009.jpg\" alt=\"..\" \/><\/span><\/p>\n<p id=\"fs-id1165135168188\">Thus, the equation is[latex]\\,f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}.[\/latex]<\/p>\n<p>We can graph our model to check our work. Notice that the graph in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_003\">(Figure)<\/a> passes through the initial points given in the problem,[latex]\\,\\left(-2,\\text{ 6}\\right)\\,[\/latex]and[latex]\\,\\left(2,\\text{ 1}\\right).\\,[\/latex]The graph is an example of an <span class=\"no-emphasis\">exponential decay<\/span> function.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140137\/CNX_Precalc_Figure_04_01_003.jpg\" alt=\"Graph of the exponential function, f(x)=2.4492(0.6389)^x, with labeled points at (-2, 6) and (2, 1).\" width=\"487\" height=\"445\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 4. <\/strong>The graph of[latex]\\,f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}\\,[\/latex]models exponential decay.<\/figcaption><\/figure>\n<p id=\"fs-id1165135482016\"><\/details>\n<\/p>\n<div id=\"CNX_Precalc_Figure_04_01_003\" class=\"small aligncenter\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"ti_11_04_01\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"fs-id1165135169306\">\n<div id=\"fs-id1165135169308\">\n<p id=\"fs-id1165135169310\">Given the two points[latex]\\,\\left(1,3\\right)\\,[\/latex]and[latex]\\,\\left(2,4.5\\right),[\/latex]find the equation of the exponential function that passes through these two points.<\/p>\n<\/div>\n<div id=\"fs-id1165137732255\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137732257\">[latex]f\\left(x\\right)=2{\\left(1.5\\right)}^{x}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137737999\" class=\"precalculus qa textbox shaded\">\n<p id=\"fs-id1165137738004\"><strong>Do two points always determine a unique exponential function?<\/strong><\/p>\n<p id=\"fs-id1165137663981\"><em>Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in[latex]\\,x,[\/latex] which in many real world cases involves time.<\/em><\/p>\n<\/div>\n<div id=\"fs-id1165137771573\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137771578\"><strong>Given the graph of an exponential function, write its equation.<\/strong><\/p>\n<ol id=\"fs-id1165137771583\" type=\"1\">\n<li>First, identify two points on the graph. Choose the <em>y<\/em>-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error.<\/li>\n<li>If one of the data points is the <em>y-<\/em>intercept[latex]\\,\\left(0,a\\right)[\/latex], then[latex]\\,a\\,[\/latex]is the initial value. Using[latex]\\,a,[\/latex] substitute the second point into the equation[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x},[\/latex] and solve for[latex]\\,b.[\/latex]<\/li>\n<li>If neither of the data points have the form[latex]\\,\\left(0,a\\right),[\/latex] substitute both points into two equations with the form[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x}.\\,[\/latex]Solve the resulting system of two equations in two unknowns to find[latex]\\,a\\,[\/latex]and[latex]\\,b.[\/latex]<\/li>\n<li>Write the exponential function,[latex]\\,f\\left(x\\right)=a{\\left(b\\right)}^{x}.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165137806415\">\n<div id=\"fs-id1165137806417\">\n<h3>Writing an Exponential Function Given Its Graph<\/h3>\n<p id=\"fs-id1165135524507\">Find an equation for the exponential function graphed in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_004\">(Figure)<\/a>.<\/p>\n<div id=\"CNX_Precalc_Figure_04_01_004\" class=\"medium aligncenter\">\n<figure style=\"width: 731px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140140\/CNX_Precalc_Figure_04_01_004.jpg\" alt=\"Graph of an increasing exponential function with notable points at (0, 3) and (2, 12).\" width=\"731\" height=\"369\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 5.<\/strong><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135415647\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135415649\">We can choose the <em>y<\/em>-intercept of the graph,[latex]\\,\\left(0,3\\right),[\/latex] as our first point. This gives us the initial value,[latex]\\,a=3.\\,[\/latex]Next, choose a point on the curve some distance away from[latex]\\,\\left(0,3\\right)\\,[\/latex]that has integer coordinates. One such point is[latex]\\,\\left(2,12\\right).[\/latex]<\/p>\n<div id=\"eip-id1165134216216\" class=\"unnumbered\">[latex]\\begin{array}{ll}\\text{ }y=a{b}^{x}\\hfill & \\text{Write the general form of an exponential equation}.\\hfill \\\\ \\text{ }y=3{b}^{x}\\hfill & \\text{Substitute the initial value 3 for }a.\\hfill \\\\ 12=3{b}^{2}\\hfill & \\text{Substitute in 12 for }y\\text{ and 2 for }x.\\hfill \\\\ \\text{ }4={b}^{2}\\hfill & \\text{Divide by 3}.\\hfill \\\\ \\text{ }b=\u00b12\\begin{array}{cccc}& & & \\end{array}\\hfill & \\text{Take the square root}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137769996\">Because we restrict ourselves to positive values of[latex]\\,b,[\/latex] we will use[latex]\\,b=2.\\,[\/latex]Substitute[latex]\\,a\\,[\/latex]and[latex]\\,b\\,[\/latex]into the standard form to yield the equation[latex]\\,f\\left(x\\right)=3{\\left(2\\right)}^{x}.[\/latex]<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135169326\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1165135431682\">\n<p id=\"fs-id1165135431684\">Find an equation for the exponential function graphed in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_005\">(Figure)<\/a>.<\/p>\n<div id=\"CNX_Precalc_Figure_04_01_005\" class=\"small aligncenter\">\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19140142\/CNX_Precalc_Figure_04_01_005.jpg\" alt=\"Graph of an increasing function with a labeled point at (0, sqrt(2)).\" width=\"487\" height=\"294\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 6.<\/strong><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191210\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135191213\">[latex]f\\left(x\\right)=\\sqrt{2}{\\left(\\sqrt{2}\\right)}^{x}.\\,[\/latex]Answers may vary due to round-off error. The answer should be very close to[latex]\\,1.4142{\\left(1.4142\\right)}^{x}.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137936929\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137936934\"><strong>Given two points on the curve of an exponential function, use a graphing calculator to find the equation.<\/strong><\/p>\n<ol id=\"fs-id1165135369319\" type=\"1\">\n<li>Press <strong>[STAT].<\/strong><\/li>\n<li>Clear any existing entries in columns <strong>L1<\/strong> or <strong>L2.<\/strong><\/li>\n<li>In <strong>L1<\/strong>, enter the <em>x<\/em>-coordinates given.<\/li>\n<li>In <strong>L2<\/strong>, enter the corresponding <em>y<\/em>-coordinates.<\/li>\n<li>Press <strong>[STAT] <\/strong>again. Cursor right to <strong>CALC<\/strong>, scroll down to <strong>ExpReg (Exponential Regression)<\/strong>, and press <strong>[ENTER].<\/strong><\/li>\n<li>The screen displays the values of <em>a<\/em> and <em>b<\/em> in the exponential equation[latex]\\,y=a\\cdot {b}^{x}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div>\n<div>\n<h3>Using a Graphing Calculator to Find an Exponential Function<\/h3>\n<p id=\"fs-id1165137748687\">Use a graphing calculator to find the exponential equation that includes the points[latex]\\,\\left(2,24.8\\right)\\,[\/latex]and[latex]\\,\\left(5,198.4\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137738185\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137843706\">Follow the guidelines above. First press <strong>[STAT]<\/strong>, <strong>[EDIT]<\/strong>, <strong>[1: Edit\u2026], <\/strong>and clear the lists <strong>L1<\/strong> and <strong>L2<\/strong>. Next, in the <strong>L1<\/strong> column, enter the <em>x<\/em>-coordinates, 2 and 5. Do the same in the <strong>L2<\/strong> column for the <em>y<\/em>-coordinates, 24.8 and 198.4.<\/p>\n<p id=\"fs-id1165137639756\">Now press <strong>[STAT]<\/strong>, <strong>[CALC]<\/strong>, <strong>[0: ExpReg] <\/strong>and press <strong>[ENTER]<\/strong>. The values[latex]\\,a=6.2\\,[\/latex]and[latex]\\,b=2\\,[\/latex]will be displayed. The exponential equation is[latex]\\,y=6.2\\cdot {2}^{x}.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137771710\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_07_01_07\">\n<div id=\"fs-id1165137656116\">\n<p id=\"fs-id1165137656119\">Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6, 481.07).<\/p>\n<\/div>\n<div id=\"fs-id1165137656124\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137651684\">[latex]y\\approx 12\\cdot {1.85}^{x}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137871708\" class=\"bc-section section\">\n<h3>Applying the Compound-Interest Formula<\/h3>\n<p id=\"fs-id1165137447026\">Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest. The term <em>compounding<\/em> refers to interest earned not only on the original value, but on the accumulated value of the account.<\/p>\n<p id=\"fs-id1165137447037\">The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an investment account. The term <em>nominal<\/em> is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being <em>greater<\/em> than the nominal rate! This is a powerful tool for investing.<\/p>\n<p id=\"fs-id1165135160118\">We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time[latex]\\,t,[\/latex] principal[latex]\\,P,[\/latex] APR[latex]\\,r,[\/latex] and number of compounding periods in a year[latex]\\,n:[\/latex]<\/p>\n<div id=\"eip-986\" class=\"unnumbered aligncenter\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/div>\n<p id=\"fs-id1165137935717\">For example, observe <a class=\"autogenerated-content\" href=\"#Table_04_01_03\">(Figure)<\/a>, which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases.<\/p>\n<table summary=\"Six rows and two columns. The first column is labeled, \u201cFrequency\u201d, and the second column is labeled, \u201cValue after 1 Year\u201d. Reading the rows from left to right, we have that Annually is valued at 100, Semiannually at 102.50, Quarterly at 103.81, Monthly at 104.71, and Daily at 105.16.\">\n<thead>\n<tr>\n<th>Frequency<\/th>\n<th>Value after 1 year<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Annually<\/td>\n<td>$1100<\/td>\n<\/tr>\n<tr>\n<td>Semiannually<\/td>\n<td>$1102.50<\/td>\n<\/tr>\n<tr>\n<td>Quarterly<\/td>\n<td>$1103.81<\/td>\n<\/tr>\n<tr>\n<td>Monthly<\/td>\n<td>$1104.71<\/td>\n<\/tr>\n<tr>\n<td>Daily<\/td>\n<td>$1105.16<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1165137793679\" class=\"textbox key-takeaways\">\n<h3>The Compound Interest Formula<\/h3>\n<p id=\"fs-id1165135184167\"><strong>Compound interest<\/strong> can be calculated using the formula<\/p>\n<div id=\"fs-id1165135184172\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/div>\n<p id=\"eip-237\">where<\/p>\n<ul id=\"fs-id1165137448453\">\n<li>[latex]A\\left(t\\right)\\,[\/latex]is the account value,<\/li>\n<li>[latex]t\\,[\/latex]is measured in years,<\/li>\n<li>[latex]P\\,[\/latex]is the starting amount of the account, often called the principal, or more generally present value,<\/li>\n<li>[latex]r\\,[\/latex]is the annual percentage rate (APR) expressed as a decimal, and<\/li>\n<li>[latex]n\\,[\/latex]is the number of compounding periods in one year.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<div>\n<div id=\"fs-id1165137812820\">\n<h3>Calculating Compound Interest<\/h3>\n<p id=\"fs-id1165137812825\">If we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years?<\/p>\n<\/div>\n<div id=\"fs-id1165137812830\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137812832\">Because we are starting with $3,000,[latex]\\,P=3000.\\,[\/latex]Our interest rate is 3%, so[latex]\\,r\\text{ }=\\text{ }0.03.\\,[\/latex]Because we are compounding quarterly, we are compounding 4 times per year, so[latex]\\,n=4.\\,[\/latex]We want to know the value of the account in 10 years, so we are looking for[latex]\\,A\\left(10\\right),[\/latex]the value when[latex]\\,t\\text{ }=\\text{ }10.[\/latex]<\/p>\n<div id=\"eip-id1402796\" class=\"unnumbered\">[latex]\\begin{array}{lll}A\\left(t\\right)\\hfill & =P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\hfill & \\text{Use the compound interest formula}.\\hfill \\\\ A\\left(10\\right)\\hfill & =3000{\\left(1+\\frac{0.03}{4}\\right)}^{4\\cdot 10}\\begin{array}{cccc}& & & \\end{array}\\hfill & \\text{Substitute using given values}.\\hfill \\\\ \\hfill & \\approx \\text{\\$}4045.05\\hfill & \\text{Round to two decimal places}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137694040\">The account will be worth about $4,045.05 in 10 years.<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137694046\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1165137694055\">\n<p id=\"fs-id1165135180428\">An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What will the investment be worth in 30 years?<\/p>\n<details>\n<summary>Show Solution<\/summary>\n<p>about $3,644,675.88<\/p>\n<\/details>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1165135180446\">\n<div>\n<h3>Using the Compound Interest Formula to Solve for the Principal<\/h3>\n<p id=\"fs-id1165135175327\">A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child\u2019s future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to $40,000 over 18 years. She believes the account will earn 6% compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now?<\/p>\n<\/div>\n<div id=\"fs-id1165135175338\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137664627\">The nominal interest rate is 6%, so[latex]\\,r=0.06.\\,[\/latex]Interest is compounded twice a year, so[latex]\\,k=2.[\/latex]<\/p>\n<p>We want to find the initial investment,[latex]\\,P,[\/latex] needed so that the value of the account will be worth $40,000 in[latex]\\,18\\,[\/latex]years. Substitute the given values into the compound interest formula, and solve for[latex]\\,P.[\/latex]<\/p>\n<div id=\"eip-id1165131884554\" class=\"unnumbered\">[latex][\/latex]\\begin{array}{lll}\\,\\,\\,\\,\\,\\,\\,A\\left(t\\right)\\hfill &amp; =P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\hfill &amp; \\text{Use the compound interest formula}.\\hfill \\\\ 40,000\\hfill &amp; =P{\\left(1+\\frac{0.06}{2}\\right)}^{2\\left(18\\right)}\\begin{array}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; \\text{Substitute using given values }A\\text{, }r, n\\text{, and }t.\\hfill \\\\ 40,000\\hfill &amp; =P{\\left(1.03\\right)}^{36}\\hfill &amp; \\text{Simplify}.\\hfill \\\\ \\frac{40,000}{{\\left(1.03\\right)}^{36}}\\hfill &amp; =P\\hfill &amp; \\text{Isolate }P.\\hfill \\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,P\\hfill &amp; \\approx \\text{\\$}13,801\\hfill &amp; \\text{Divide and round to the nearest dollar}.\\hfill \\end{array}[\/latex<\/div>\n<p id=\"fs-id1165137937589\">Lily will need to invest $13,801 to have $40,000 in 18 years.]<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_04_01_9\">\n<div id=\"fs-id1165135176727\">\n<p id=\"fs-id1165135176729\">Refer to <a class=\"autogenerated-content\" href=\"#Example_04_01_09\">(Figure)<\/a>. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?<\/p>\n<\/div>\n<div id=\"fs-id1165135176736\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135176738\">$13,693<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137724961\" class=\"bc-section section\">\n<h3>Evaluating Functions with Base <em>e<\/em><\/h3>\n<p id=\"fs-id1165135684369\">As we saw earlier, the amount earned on an account increases as the compounding frequency increases. <a class=\"autogenerated-content\" href=\"#Table_04_01_04\">(Figure)<\/a> shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.<\/p>\n<p id=\"fs-id1165135684377\">Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies, listed in <a class=\"autogenerated-content\" href=\"#Table_04_01_04\">(Figure)<\/a>.<\/p>\n<table summary=\"Nine rows and three columns. The first column is labeled, \u201cFrequency\u201d, the second column is labeled, \u201cA(t)=(1+1\/n)^x\u201d, and the third column is labeled, \u201cValue\u201d. Reading the rows from left to right, we have that Annually has the input value of (1+1\/1)^1 which equals to \ud83d\udcb22, and Semiannually has the input value of (1+1\/2)^2 which equals to \ud83d\udcb22.25, Quarterly has the input value of (1+1\/4)^4 which equals to \ud83d\udcb22.441406, Monthly has the input value of (1+1\/12)^12 which equals to \ud83d\udcb22.613035, Daily has the input value of (1+1\/365)^365 which equals to \ud83d\udcb22.714567, Hourly has the input value of (1+1\/8766)^8766 which equals to \ud83d\udcb22.718127, One per minute has the input value of (1+1\/525960)^525960 which equals to \ud83d\udcb22.718279, and Once per second has the input value of (1+1\/31557600)^31557600 which equals to \ud83d\udcb22.718282.\">\n<caption>&nbsp;<\/caption>\n<thead>\n<tr>\n<th>Frequency<\/th>\n<th>[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\n<th>Value<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Annually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\n<td>$2<\/td>\n<\/tr>\n<tr>\n<td>Semiannually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\n<td>$2.25<\/td>\n<\/tr>\n<tr>\n<td>Quarterly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\n<td>$2.441406<\/td>\n<\/tr>\n<tr>\n<td>Monthly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\n<td>$2.613035<\/td>\n<\/tr>\n<tr>\n<td>Daily<\/td>\n<td>[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\n<td>$2.714567<\/td>\n<\/tr>\n<tr>\n<td>Hourly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{\\text{8760}}\\right)}^{\\text{8760}}[\/latex]<\/td>\n<td>$2.718127<\/td>\n<\/tr>\n<tr>\n<td>Once per minute<\/td>\n<td>[latex]{\\left(1+\\frac{1}{\\text{525600}}\\right)}^{\\text{525600}}[\/latex]<\/td>\n<td>$2.718279<\/td>\n<\/tr>\n<tr>\n<td>Once per second<\/td>\n<td>[latex]{\\left(1+\\frac{1}{31536000}\\right)}^{31536000}[\/latex]<\/td>\n<td>$2.718282<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137828146\">These values appear to be approaching a limit as[latex]\\,n\\,[\/latex]increases without bound. In fact, as[latex]\\,n\\,[\/latex]gets larger and larger, the expression[latex]\\,{\\left(1+\\frac{1}{n}\\right)}^{n}\\,[\/latex]approaches a number used so frequently in mathematics that it has its own name: the letter[latex]\\,e.\\,[\/latex]This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below.<\/p>\n<div id=\"fs-id1165135511324\" class=\"textbox key-takeaways\">\n<h3>The Number <em>e<\/em><\/h3>\n<p id=\"fs-id1165135511335\">The letter <em>e<\/em> represents the irrational number<\/p>\n<div id=\"eip-id1165135378658\" class=\"unnumbered\">[latex]{\\left(1+\\frac{1}{n}\\right)}^{n},\\text{as}\\,n\\,\\text{increases without bound}[\/latex]<\/div>\n<p id=\"fs-id1165135369344\">The letter <em>e <\/em>is used as a base for many real-world exponential models. To work with base <em>e<\/em>, we use the approximation,[latex]\\,e\\approx 2.718282.\\,[\/latex]The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.<\/p>\n<\/div>\n<div id=\"Example_04_01_10\" class=\"textbox examples\">\n<div id=\"fs-id1165135344893\">\n<div id=\"fs-id1165135344895\">\n<h3>Using a Calculator to Find Powers of <em>e<\/em><\/h3>\n<p id=\"fs-id1165135545973\">Calculate[latex]\\,{e}^{3.14}.\\,[\/latex]Round to five decimal places.<\/p>\n<\/div>\n<div id=\"fs-id1165135192743\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135192745\">On a calculator, press the button labeled[latex]\\,\\left[{e}^{x}\\right].\\,[\/latex]The window shows[latex]\\,\\left[e^(\\text{ }\\right].\\,[\/latex]Type[latex]\\,3.14\\,[\/latex]and then close parenthesis,[latex]\\,\\left[)\\right].\\,[\/latex]Press [ENTER]. Rounding to[latex]\\,5\\,[\/latex]decimal places,[latex]\\,{e}^{3.14}\\approx 23.10387.\\,[\/latex]Caution: Many scientific calculators have an \u201cExp\u201d button, which is used to enter numbers in scientific notation. It is not used to find powers of[latex]\\,e.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137447188\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_04_01_10\">\n<div id=\"fs-id1165135348435\">\n<p id=\"fs-id1165135348437\">Use a calculator to find[latex]\\,{e}^{-0.5}.\\,[\/latex]Round to five decimal places.<\/p>\n<\/div>\n<div id=\"fs-id1165135176286\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135176288\">[latex]{e}^{-0.5}\\approx 0.60653[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137827923\" class=\"bc-section section\">\n<h3>Investigating Continuous Growth<\/h3>\n<p id=\"fs-id1165137827929\">So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, <em>e <\/em>is used as the base for exponential functions. Exponential models that use[latex]\\,e\\,[\/latex]as the base are called <em>continuous growth or decay models<\/em>. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.<\/p>\n<div id=\"fs-id1165137664673\" class=\"textbox key-takeaways\">\n<h3>The Continuous Growth\/Decay Formula<\/h3>\n<p id=\"fs-id1165135453868\">For all real numbers[latex]\\,t,[\/latex]and all positive numbers[latex]\\,a\\,[\/latex]and[latex]\\,r,[\/latex]continuous growth or decay is represented by the formula<\/p>\n<div id=\"fs-id1165135536370\">[latex]A\\left(t\\right)=a{e}^{rt}[\/latex]<\/div>\n<p id=\"eip-101\">where<\/p>\n<ul id=\"fs-id1165135152052\">\n<li>[latex]a\\,[\/latex]is the initial value,<\/li>\n<li>[latex]r\\,[\/latex]is the continuous growth rate per unit time,<\/li>\n<li>and[latex]\\,t\\,[\/latex]is the elapsed time.<\/li>\n<\/ul>\n<p id=\"fs-id1165135560686\">If[latex]\\,r>0\\,[\/latex], then the formula represents continuous growth. If[latex]\\,r<0\\,[\/latex], then the formula represents continuous decay.<\/p>\n<p id=\"fs-id1165137812323\">For business applications, the continuous growth formula is called the continuous compounding formula and takes the form<\/p>\n<div id=\"eip-id1165134324899\" class=\"unnumbered\">[latex]A\\left(t\\right)=P{e}^{rt}[\/latex]<\/div>\n<p id=\"eip-962\">where<\/p>\n<ul id=\"fs-id1165137827330\">\n<li>[latex]P\\,[\/latex]is the principal or the initial invested,<\/li>\n<li>[latex]r\\,[\/latex]is the growth or interest rate per unit time,<\/li>\n<li>and [latex]t\\,[\/latex]is the period or term of the investment.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165135411368\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165135411373\"><strong>Given the initial value, rate of growth or decay, and time[latex]\\,t,[\/latex] solve a continuous growth or decay function.<\/strong><\/p>\n<ol id=\"fs-id1165135511371\" type=\"1\">\n<li>Use the information in the problem to determine[latex]\\,a[\/latex], the initial value of the function.<\/li>\n<li>Use the information in the problem to determine the growth rate[latex]\\,r.[\/latex]\n<ol id=\"fs-id1165135188096\" type=\"a\">\n<li>If the problem refers to continuous growth, then[latex]\\,r>0.[\/latex]<\/li>\n<li>If the problem refers to continuous decay, then[latex]\\,r<0.[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Use the information in the problem to determine the time[latex]\\,t.[\/latex]<\/li>\n<li>Substitute the given information into the continuous growth formula and solve for[latex]\\,A\\left(t\\right).[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_01_11\" class=\"textbox examples\">\n<div id=\"fs-id1165137835464\">\n<div id=\"fs-id1165137835466\">\n<h3>Calculating Continuous Growth<\/h3>\n<p id=\"fs-id1165137835472\">A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year?<\/p>\n<\/div>\n<div id=\"fs-id1165137694203\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137694205\">Since the account is growing in value, this is a continuous compounding problem with growth rate[latex]\\,r=0.10.\\,[\/latex]The initial investment was $1,000, so[latex]\\,P=1000.\\,[\/latex]We use the continuous compounding formula to find the value after[latex]\\,t=1\\,[\/latex]year:<\/p>\n<div id=\"eip-id1165133351794\" class=\"unnumbered\">[latex]\\begin{array}{lll}A\\left(t\\right)\\hfill & =P{e}^{rt}\\hfill & \\text{Use the continuous compounding formula}.\\hfill \\\\ \\hfill & =1000{\\left(e\\right)}^{0.1}\\begin{array}{cccc}& & & \\end{array}\\hfill & \\text{Substitute known values for }P, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 1105.17\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137895288\">The account is worth $1,105.17 after one year.<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137895295\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_04_01_11\">\n<div id=\"fs-id1165137895303\">\n<p id=\"fs-id1165137895305\">A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years?<\/p>\n<\/div>\n<div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134389978\">$3,659,823.44<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_01_12\" class=\"textbox examples\">\n<div id=\"fs-id1165134389988\">\n<div id=\"fs-id1165134389990\">\n<h3>Calculating Continuous Decay<\/h3>\n<p id=\"fs-id1165137803700\">Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days?<\/p>\n<\/div>\n<div id=\"fs-id1165137803706\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>Since the substance is decaying, the rate,[latex]\\,17.3%[\/latex], is negative. So,[latex]\\,r\\text{ }=\\text{ }-0.173.\\,[\/latex]The initial amount of radon-222 was[latex]\\,100\\,[\/latex]mg, so[latex]\\,a=100.\\,[\/latex]We use the continuous decay formula to find the value after[latex]\\,t=3\\,[\/latex]days:<\/p>\n<div id=\"eip-id1165137779893\" class=\"unnumbered\">[latex]\\begin{array}{lll}A\\left(t\\right)\\hfill & =a{e}^{rt}\\hfill & \\text{Use the continuous growth formula}.\\hfill \\\\ \\hfill & =100{e}^{-0.173\\left(3\\right)}\\begin{array}{cccc}& & & \\end{array}\\hfill & \\text{Substitute known values for }a, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 59.5115\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137697132\">So 59.5115 mg of radon-222 will remain.<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135348462\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div id=\"ti_04_01_12\">\n<div id=\"fs-id1165135348470\">\n<p id=\"fs-id1165135348472\">Using the data in <a class=\"autogenerated-content\" href=\"#Example_04_01_12\">(Figure)<\/a>, how much radon-222 will remain after one year?<\/p>\n<\/div>\n<div id=\"fs-id1165135245732\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135245734\">3.77E-26 (This is calculator notation for the number written as[latex]\\,3.77\u00d7{10}^{-26}\\,[\/latex]in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191186\" class=\"precalculus media\">\n<p id=\"fs-id1165135191191\">Access these online resources for additional instruction and practice with exponential functions.<\/p>\n<ul id=\"fs-id1165135191194\">\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/expgrowth\">Exponential Growth Function<\/a><\/li>\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/compoundint\">Compound Interest<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135264762\" class=\"key-equations\">\n<h3>Key Equations<\/h3>\n<table id=\"fs-id2306479\" style=\"height: 248px\" summary=\"...\">\n<tbody>\n<tr style=\"height: 41px\">\n<td style=\"height: 41px\">definition of the exponential function<\/td>\n<td style=\"height: 41px\">[latex]f\\left(x\\right)={b}^{x}\\text{, where }b>0, b\\ne 1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 41px\">\n<td style=\"height: 41px\">definition of exponential growth<\/td>\n<td style=\"height: 41px\">[latex]f\\left(x\\right)=a{b}^{x},\\text{ where }a>0,b>0,b\\ne 1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 83px\">\n<td style=\"height: 83px\">compound interest formula<\/td>\n<td style=\"height: 83px\">[latex]\\begin{array}{l}A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt} ,\\text{ where}\\hfill \\\\ A\\left(t\\right)\\text{ is the account value at time }t\\hfill \\\\ t\\text{ is the number of years}\\hfill \\\\ P\\text{ is the initial investment, often called the principal}\\hfill \\\\ r\\text{ is the annual percentage rate (APR), or nominal rate}\\hfill \\\\ n\\text{ is the number of compounding periods in one year}\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 83px\">\n<td style=\"height: 83px\">continuous growth formula<\/td>\n<td style=\"height: 83px\">[latex]A\\left(t\\right)=a{e}^{rt},\\text{ where}[\/latex] [latex]t[\/latex]is the number of unit time periods of growth<br \/>\n[latex]a[\/latex]is the starting amount (in the continuous compounding formula a is replaced with P, the principal)<br \/>\n[latex]e[\/latex]is the mathematical constant,[latex]\\text{ }e\\approx 2.718282[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165137846440\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1165137846446\">\n<li>An exponential function is defined as a function with a positive constant other than[latex]\\,1\\,[\/latex]raised to a variable exponent. See <a class=\"autogenerated-content\" href=\"#Example_04_01_01\">(Figure)<\/a>.<\/li>\n<li>A function is evaluated by solving at a specific value. See <a class=\"autogenerated-content\" href=\"#Example_04_01_02\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_04_01_03\">(Figure)<\/a>.<\/li>\n<li>An exponential model can be found when the growth rate and initial value are known. See <a class=\"autogenerated-content\" href=\"#Example_04_01_04\">(Figure)<\/a>.<\/li>\n<li>An exponential model can be found when the two data points from the model are known. See <a class=\"autogenerated-content\" href=\"#Example_04_01_05\">(Figure)<\/a>.<\/li>\n<li>An exponential model can be found using two data points from the graph of the model. See <a class=\"autogenerated-content\" href=\"#Example_04_01_06\">(Figure)<\/a>.<\/li>\n<li>An exponential model can be found using two data points from the graph and a calculator. See <a class=\"autogenerated-content\" href=\"#Example_04_01_07\">(Figure)<\/a>.<\/li>\n<li>The value of an account at any time[latex]\\,t\\,[\/latex]can be calculated using the compound interest formula when the principal, annual interest rate, and compounding periods are known. See <a class=\"autogenerated-content\" href=\"#Example_04_01_08\">(Figure)<\/a>.<\/li>\n<li>The initial investment of an account can be found using the compound interest formula when the value of the account, annual interest rate, compounding periods, and life span of the account are known. See <a class=\"autogenerated-content\" href=\"#Example_04_01_09\">(Figure)<\/a>.<\/li>\n<li>The number[latex]\\,e\\,[\/latex]is a mathematical constant often used as the base of real world exponential growth and decay models. Its decimal approximation is[latex]\\,e\\approx 2.718282.[\/latex]<\/li>\n<li>Scientific and graphing calculators have the key[latex]\\,\\left[{e}^{x}\\right]\\,[\/latex]or[latex]\\,\\left[\\mathrm{exp}\\left(x\\right)\\right]\\,[\/latex]for calculating powers of[latex]\\,e.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Example_04_01_10\">(Figure)<\/a><strong>.<\/strong><\/li>\n<li>Continuous growth or decay models are exponential models that use[latex]\\,e\\,[\/latex]as the base. Continuous growth and decay models can be found when the initial value and growth or decay rate are known. See <a class=\"autogenerated-content\" href=\"#Example_04_01_11\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_04_01_12\">(Figure)<\/a>.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165137406918\" class=\"textbox exercises\">\n<h3>Section Exercises<\/h3>\n<div id=\"fs-id1165137406922\" class=\"bc-section section\">\n<h4>Verbal<\/h4>\n<div id=\"fs-id1165137406928\">\n<div id=\"fs-id1165137406930\">\n<p id=\"fs-id1165137406932\">Explain why the values of an increasing exponential function will eventually overtake the values of an increasing linear function.<\/p>\n<\/div>\n<div id=\"fs-id1165137406938\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137406940\">Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135536244\">\n<div id=\"fs-id1165135536246\">\n<p id=\"fs-id1165135536249\">Given a formula for an exponential function, is it possible to determine whether the function grows or decays exponentially just by looking at the formula? Explain.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135536255\">\n<div>\n<p id=\"fs-id1165135536260\">The Oxford Dictionary defines the word <em>nominal<\/em> as a value that is \u201cstated or expressed but not necessarily corresponding exactly to the real value.\u201d<a class=\"footnote\" title=\"Oxford Dictionary. http:\/\/oxforddictionaries.com\/us\/definition\/american_english\/nomina.\" id=\"return-footnote-98-2\" href=\"#footnote-98-2\" aria-label=\"Footnote 2\"><sup class=\"footnote\">[2]<\/sup><\/a> Develop a reasonable argument for why the term <em>nominal rate<\/em> is used to describe the annual percentage rate of an investment account that compounds interest.<\/p>\n<\/div>\n<div id=\"fs-id1165135189885\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135189887\">When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of <em>nominal<\/em>.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134340071\" class=\"bc-section section\">\n<h4>Algebraic<\/h4>\n<p id=\"fs-id1165134340076\">For the following exercises, identify whether the statement represents an exponential function. Explain.<\/p>\n<div id=\"fs-id1165134340080\">\n<div id=\"fs-id1165134340082\">\n<p id=\"fs-id1165134340084\">The average annual population increase of a pack of wolves is 25.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137737593\">\n<div id=\"fs-id1165137737595\">\n<p id=\"fs-id1165137737598\">A population of bacteria decreases by a factor of[latex]\\,\\frac{1}{8}\\,[\/latex]every[latex]\\,24\\,[\/latex]hours.<\/p>\n<\/div>\n<div id=\"fs-id1165137416069\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135582265\">exponential; the population decreases by a proportional rate.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135582271\">\n<div id=\"fs-id1165135582273\">\n<p id=\"fs-id1165135582275\">The value of a coin collection has increased by[latex]\\,3.25%\\,[\/latex]annually over the last[latex]\\,20\\,[\/latex]years.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135414352\">\n<div id=\"fs-id1165135414354\">\n<p id=\"fs-id1165135318974\">For each training session, a personal trainer charges his clients[latex]\\,\\text{\\$}5\\,[\/latex]<br \/>\nless than the previous training session.<\/p>\n<\/div>\n<div id=\"fs-id1165135318995\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135318997\">not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function. .<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191053\">\n<div id=\"fs-id1165135191056\">\n<p id=\"fs-id1165135191058\">The height of a projectile at time[latex]\\,t\\,[\/latex]is represented by the function[latex]\\,h\\left(t\\right)=-4.9{t}^{2}+18t+40.[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137767849\">For the following exercises, consider this scenario: For each year[latex]\\,t,[\/latex]the population of a forest of trees is represented by the function[latex]\\,A\\left(t\\right)=115{\\left(1.025\\right)}^{t}.\\,[\/latex]In a neighboring forest, the population of the same type of tree is represented by the function[latex]\\,B\\left(t\\right)=82{\\left(1.029\\right)}^{t}.\\,[\/latex](Round answers to the nearest whole number.)<\/p>\n<div id=\"fs-id1165135570343\">\n<div id=\"fs-id1165135570345\">\n<p id=\"fs-id1165135570347\">Which forest\u2019s population is growing at a faster rate?<\/p>\n<\/div>\n<div id=\"fs-id1165135570352\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135570355\">The forest represented by the function[latex]\\,B\\left(t\\right)=82{\\left(1.029\\right)}^{t}.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134049940\">\n<div id=\"fs-id1165134049942\">\n<p id=\"fs-id1165137755646\">Which forest had a greater number of trees initially? By how many?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137755651\">\n<div id=\"fs-id1165137755653\">\n<p id=\"fs-id1165137755656\">Assuming the population growth models continue to represent the growth of the forests, which forest will have a greater number of trees after[latex]\\,20\\,[\/latex]years? By how many?<\/p>\n<\/div>\n<div id=\"fs-id1165137761058\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137761061\">After[latex]\\,t=20\\,[\/latex]years, forest A will have[latex]\\,43\\,[\/latex]more trees than forest B.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135508291\">\n<div id=\"fs-id1165135508293\">\n<p id=\"fs-id1165135508295\">Assuming the population growth models continue to represent the growth of the forests, which forest will have a greater number of trees after[latex]\\,100\\,[\/latex]years? By how many?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137828270\">\n<div id=\"fs-id1165137828272\">\n<p id=\"fs-id1165137828274\">Discuss the above results from the previous four exercises. Assuming the population growth models continue to represent the growth of the forests, which forest will have the greater number of trees in the long run? Why? What are some factors that might influence the long-term validity of the exponential growth model?<\/p>\n<\/div>\n<div id=\"fs-id1165137407578\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137407580\">Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<p>For the following exercises, determine whether the equation represents exponential growth, exponential decay, or neither. Explain.<\/p>\n<div>\n<div id=\"fs-id1165137407597\">\n<p id=\"fs-id1165135560752\">[latex]y=300{\\left(1-t\\right)}^{5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135252048\">\n<div id=\"fs-id1165135252050\">\n<p id=\"fs-id1165135252052\">[latex]y=220{\\left(1.06\\right)}^{x}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135209944\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135209946\">exponential growth; The growth factor,[latex]\\,1.06,[\/latex] is greater than[latex]\\,1.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137806960\">\n<div id=\"fs-id1165137806962\">\n<p id=\"fs-id1165137806964\">[latex]y=16.5{\\left(1.025\\right)}^{\\frac{1}{x}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135519282\">\n<div id=\"fs-id1165135519284\">\n<p id=\"fs-id1165135519286\">[latex]y=11,701{\\left(0.97\\right)}^{t}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135397288\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p>exponential decay; The decay factor,[latex]\\,0.97,[\/latex] is between[latex]\\,0\\,[\/latex]and[latex]\\,1.[\/latex]<\/details>\n<\/div>\n<\/div>\n<p id=\"fs-id1165135176427\">For the following exercises, find the formula for an exponential function that passes through the two points given.<\/p>\n<div id=\"fs-id1165135176432\">\n<div id=\"fs-id1165135176434\">\n<p id=\"fs-id1165135176436\">[latex]\\left(0,6\\right)\\,[\/latex]and[latex]\\,\\left(3,750\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137770026\">\n<div id=\"fs-id1165137770029\">\n<p id=\"fs-id1165137770031\">[latex]\\left(0,2000\\right)\\,[\/latex]and[latex]\\,\\left(2,20\\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137735589\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135478535\">[latex]f\\left(x\\right)=2000{\\left(0.1\\right)}^{x}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135189914\">\n<div id=\"fs-id1165135189916\">\n<p id=\"fs-id1165135189919\">[latex]\\left(-1,\\frac{3}{2}\\right)\\,[\/latex]and[latex]\\,\\left(3,24\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137400553\">\n<div id=\"fs-id1165137400555\">\n<p id=\"fs-id1165137400557\">[latex]\\left(-2,6\\right)\\,[\/latex]and[latex]\\,\\left(3,1\\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135187318\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135187320\">[latex]f\\left(x\\right)={\\left(\\frac{1}{6}\\right)}^{-\\frac{3}{5}}{\\left(\\frac{1}{6}\\right)}^{\\frac{x}{5}}\\approx 2.93{\\left(0.699\\right)}^{x}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135180070\">\n<div id=\"fs-id1165135180073\">\n<p id=\"fs-id1165135180075\">[latex]\\left(3,1\\right)\\,[\/latex]and[latex]\\,\\left(5,4\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165134154588\">For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points.<\/p>\n<div id=\"fs-id1165134154593\">\n<div id=\"fs-id1165134154595\">\n<table id=\"fs-id1165134154597\" class=\"unnumbered\" summary=\"\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>70<\/td>\n<td>40<\/td>\n<td>10<\/td>\n<td>-20<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165137680442\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137680445\">Linear<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137680450\">\n<div id=\"fs-id1165137680452\">\n<table id=\"fs-id1165137680454\" class=\"unnumbered\" summary=\"\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]h\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>70<\/td>\n<td>49<\/td>\n<td>34.3<\/td>\n<td>24.01<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135237063\">\n<div id=\"fs-id1165135237065\">\n<table class=\"unnumbered\" summary=\"\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]m\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>80<\/td>\n<td>61<\/td>\n<td>42.9<\/td>\n<td>25.61<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165137761697\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137761699\">Neither<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137761705\">\n<div id=\"fs-id1165137761707\">\n<table id=\"fs-id1165137761709\" class=\"unnumbered\" summary=\"\">\n<caption>&nbsp;<\/caption>\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>10<\/td>\n<td>20<\/td>\n<td>40<\/td>\n<td>80<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<div>\n<div>\n<table id=\"fs-id1165137749151\" class=\"unnumbered\" summary=\"\">\n<tbody>\n<tr>\n<td><strong>[latex]x[\/latex]<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]g\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>-3.25<\/td>\n<td>2<\/td>\n<td>7.25<\/td>\n<td>12.5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165137874537\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137874539\">Linear<\/p>\n<\/details>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137874544\">For the following exercises, use the compound interest formula,[latex]\\,A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}.[\/latex]<\/p>\n<div>\n<div id=\"fs-id1165135435599\">\n<p id=\"fs-id1165135435601\">After a certain number of years, the value of an investment account is represented by the equation[latex]\\,10,250{\\left(1+\\frac{0.04}{12}\\right)}^{120}.\\,[\/latex]What is the value of the account?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135527075\">\n<div id=\"fs-id1165135527077\">\n<p id=\"fs-id1165135527079\">What was the initial deposit made to the account in the previous exercise?<\/p>\n<\/div>\n<div id=\"fs-id1165135527084\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135527086\">[latex]$10,250[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134042860\">\n<div id=\"fs-id1165134042862\">\n<p id=\"fs-id1165134042865\">How many years had the account from the previous exercise been accumulating interest?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134042870\">\n<div id=\"fs-id1165134042872\">\n<p id=\"fs-id1165134042874\">An account is opened with an initial deposit of $6,500 and earns[latex]\\,3.6%\\,[\/latex]interest compounded semi-annually. What will the account be worth in[latex]\\,20\\,[\/latex]years?<\/p>\n<\/div>\n<div id=\"fs-id1165137749553\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137749555\">[latex]$13,268.58[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135196890\">\n<div id=\"fs-id1165135196892\">\n<p id=\"fs-id1165135196894\">How much more would the account in the previous exercise have been worth if the interest were compounding weekly?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135196900\">\n<div id=\"fs-id1165135196902\">\n<p id=\"fs-id1165137831959\">Solve the compound interest formula for the principal,[latex]\\,P[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165137831975\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137831977\">[latex]P=A\\left(t\\right)\\cdot {\\left(1+\\frac{r}{n}\\right)}^{-nt}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134234548\">\n<div id=\"fs-id1165134234550\">\n<p id=\"fs-id1165134234552\">Use the formula found in the previous exercise to calculate the initial deposit of an account that is worth[latex]\\,$14,472.74\\,[\/latex]after earning[latex]\\,5.5%\\,[\/latex]interest compounded monthly for[latex]\\,5\\,[\/latex]years. (Round to the nearest dollar.)<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135415804\">\n<div id=\"fs-id1165135415806\">\n<p id=\"fs-id1165135415808\">How much more would the account in the previous two exercises be worth if it were earning interest for[latex]\\,5\\,[\/latex]more years?<\/p>\n<\/div>\n<div id=\"fs-id1165135415827\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135415829\">[latex]$4,572.56[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135528933\">\n<div id=\"fs-id1165135528935\">\n<p id=\"fs-id1165135528937\">Use properties of rational exponents to solve the compound interest formula for the interest rate,[latex]\\,r.[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137836684\">\n<div id=\"fs-id1165137836687\">\n<p id=\"fs-id1165137836689\">Use the formula found in the previous exercise to calculate the interest rate for an account that was compounded semi-annually, had an initial deposit of $9,000 and was worth $13,373.53 after 10 years.<\/p>\n<\/div>\n<div id=\"fs-id1165135487075\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135487077\">[latex]4%[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134071602\">\n<div id=\"fs-id1165134071604\">\n<p id=\"fs-id1165134071606\">Use the formula found in the previous exercise to calculate the interest rate for an account that was compounded monthly, had an initial deposit of $5,500, and was worth $38,455 after 30 years.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165135441763\">For the following exercises, determine whether the equation represents continuous growth, continuous decay, or neither. Explain.<\/p>\n<div id=\"fs-id1165135441768\">\n<div id=\"fs-id1165135441770\">\n<p id=\"fs-id1165135441772\">[latex]y=3742{\\left(e\\right)}^{0.75t}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135336087\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135336089\">continuous growth; the growth rate is greater than[latex]\\,0.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135336105\">\n<div id=\"fs-id1165135336107\">\n<p id=\"fs-id1165135336109\">[latex]y=150{\\left(e\\right)}^{\\frac{3.25}{t}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135545718\">\n<div id=\"fs-id1165135545721\">\n<p id=\"fs-id1165135545723\">[latex]y=2.25{\\left(e\\right)}^{-2t}[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135191882\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135191884\">continuous decay; the growth rate is less than[latex]\\,0.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191900\">\n<div id=\"fs-id1165135191902\">\n<p id=\"fs-id1165135191904\">Suppose an investment account is opened with an initial deposit of[latex]\\,$12,000\\,[\/latex]earning[latex]\\,7.2%\\,[\/latex]interest compounded continuously. How much will the account be worth after[latex]\\,30\\,[\/latex]years?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135471130\">\n<div id=\"fs-id1165135471132\">\n<p id=\"fs-id1165135471134\">How much less would the account from Exercise 42 be worth after[latex]\\,30\\,[\/latex]years if it were compounded monthly instead?<\/p>\n<\/div>\n<div id=\"fs-id1165135190851\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135190853\">[latex]$669.42[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135190869\" class=\"bc-section section\">\n<h4>Numeric<\/h4>\n<p id=\"fs-id1165135421696\">For the following exercises, evaluate each function. Round answers to four decimal places, if necessary.<\/p>\n<div id=\"fs-id1165135421700\">\n<div id=\"fs-id1165135421702\">\n<p id=\"fs-id1165135421704\">[latex]f\\left(x\\right)=2{\\left(5\\right)}^{x},[\/latex] for[latex]\\,f\\left(-3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137811660\">\n<div id=\"fs-id1165137811662\">\n<p id=\"fs-id1165135245591\">[latex]f\\left(x\\right)=-{4}^{2x+3},[\/latex] for[latex]\\,f\\left(-1\\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135415781\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135415783\">[latex]f\\left(-1\\right)=-4[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135593586\">\n<div id=\"fs-id1165135593588\">\n<p id=\"fs-id1165135593590\">[latex]f\\left(x\\right)={e}^{x},[\/latex] for[latex]\\,f\\left(3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135185903\">\n<div id=\"fs-id1165135185905\">\n<p id=\"fs-id1165135185907\">[latex]f\\left(x\\right)=-2{e}^{x-1},[\/latex] for[latex]\\,f\\left(-1\\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135500970\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135500972\">[latex]f\\left(-1\\right)\\approx -0.2707[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135508316\">\n<div id=\"fs-id1165135508318\">\n<p id=\"fs-id1165135508320\">[latex]f\\left(x\\right)=2.7{\\left(4\\right)}^{-x+1}+1.5,[\/latex] for[latex]f\\left(-2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134222572\">\n<div id=\"fs-id1165134222574\">\n<p id=\"fs-id1165134222576\">[latex]f\\left(x\\right)=1.2{e}^{2x}-0.3,[\/latex] for[latex]\\,f\\left(3\\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135510681\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135510683\">[latex]f\\left(3\\right)\\approx 483.8146[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137749955\">\n<div id=\"fs-id1165137749957\">\n<p id=\"fs-id1165137749959\">[latex]f\\left(x\\right)=-\\frac{3}{2}{\\left(3\\right)}^{-x}+\\frac{3}{2},[\/latex] for[latex]\\,f\\left(2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137843692\" class=\"bc-section section\">\n<h4>Technology<\/h4>\n<p id=\"fs-id1165137843697\">For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.<\/p>\n<div id=\"fs-id1165137843702\">\n<div id=\"fs-id1165137843704\">\n<p id=\"fs-id1165135251278\">[latex]\\left(0,3\\right)\\,[\/latex]and[latex]\\,\\left(3,375\\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137715431\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165137715433\">[latex]y=3\\cdot {5}^{x}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137740770\">\n<div id=\"fs-id1165137740772\">\n<p id=\"fs-id1165137740775\">[latex]\\left(3,222.62\\right)\\,[\/latex]and[latex]\\,\\left(10,77.456\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135209922\">\n<div id=\"fs-id1165135209924\">\n<p id=\"fs-id1165135209926\">[latex]\\left(20,29.495\\right)\\,[\/latex]and[latex]\\,\\left(150,730.89\\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135347282\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135347284\">[latex]y\\approx 18\\cdot {1.025}^{x}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135409820\">\n<div id=\"fs-id1165135409822\">\n<p id=\"fs-id1165135409824\">[latex]\\left(5,2.909\\right)\\,[\/latex]and[latex]\\,\\left(13,0.005\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135545870\">\n<div id=\"fs-id1165135545872\">\n<p id=\"fs-id1165135545874\">[latex]\\left(11,310.035\\right)\\,[\/latex] and [latex]\\left(25,356.3652\\right)[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135255110\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135255112\">[latex]y\\approx 0.2\\cdot {1.95}^{x}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134388205\" class=\"bc-section section\">\n<h4>Extensions<\/h4>\n<div id=\"fs-id1165134388210\">\n<div id=\"fs-id1165134388212\">\n<p>The <em>annual percentage yield<\/em> (APY) of an investment account is a representation of the actual interest rate earned on a compounding account. It is based on a compounding period of one year. Show that the APY of an account that compounds monthly can be found with the formula[latex]\\,\\text{APY}={\\left(1+\\frac{r}{12}\\right)}^{12}-1.[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135499889\">\n<div id=\"fs-id1165135499891\">\n<p id=\"fs-id1165135499893\">Repeat the previous exercise to find the formula for the APY of an account that compounds daily. Use the results from this and the previous exercise to develop a function[latex]\\,I\\left(n\\right)\\,[\/latex]for the APY of any account that compounds[latex]\\,n\\,[\/latex]times per year.<\/p>\n<\/div>\n<div id=\"fs-id1165135530489\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135530491\">[latex]\\text{APY}=\\frac{A\\left(t\\right)-a}{a}=\\frac{a{\\left(1+\\frac{r}{365}\\right)}^{365\\left(1\\right)}-a}{a}=\\frac{a\\left[{\\left(1+\\frac{r}{365}\\right)}^{365}-1\\right]}{a}={\\left(1+\\frac{r}{365}\\right)}^{365}-1;[\/latex][latex]I\\left(n\\right)={\\left(1+\\frac{r}{n}\\right)}^{n}-1[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135408530\">\n<div id=\"fs-id1165135408533\">\n<p id=\"fs-id1165135408535\">Recall that an exponential function is any equation written in the form[latex]\\,f\\left(x\\right)=a\\cdot {b}^{x}\\,[\/latex]such that[latex]a[\/latex]and[latex]b[\/latex]are positive numbers and[latex]b\\ne 1.[\/latex]Any positive number[latex]b[\/latex]can be written as[latex]b={e}^{n}[\/latex]for some value of[latex]n[\/latex]. Use this fact to rewrite the formula for an exponential function that uses the number[latex]e[\/latex]as a base.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134108543\">\n<div id=\"fs-id1165134108546\">\n<p id=\"fs-id1165134108548\">In an exponential decay function, the base of the exponent is a value between 0 and 1. Thus, for some number[latex]\\,b>1,[\/latex] the exponential decay function can be written as[latex]\\,f\\left(x\\right)=a\\cdot {\\left(\\frac{1}{b}\\right)}^{x}.\\,[\/latex]Use this formula, along with the fact that[latex]\\,b={e}^{n},[\/latex] to show that an exponential decay function takes the form[latex]\\,f\\left(x\\right)=a{\\left(e\\right)}^{-nx}\\,[\/latex]for some positive number[latex]\\,n\\,[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1165135661465\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135661467\">Let[latex]\\,f\\,[\/latex]be the exponential decay function[latex]\\,f\\left(x\\right)=a\\cdot {\\left(\\frac{1}{b}\\right)}^{x}\\,[\/latex]such that[latex]\\,b>1.\\,[\/latex]Then for some number[latex]\\,n>0,[\/latex][latex]f\\left(x\\right)=a\\cdot {\\left(\\frac{1}{b}\\right)}^{x}=a{\\left({b}^{-1}\\right)}^{x}=a{\\left({\\left({e}^{n}\\right)}^{-1}\\right)}^{x}=a{\\left({e}^{-n}\\right)}^{x}=a{\\left(e\\right)}^{-nx}.[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137848892\">\n<div id=\"fs-id1165137848894\">\n<p id=\"fs-id1165137848897\">The formula for the amount[latex]\\,A\\,[\/latex]in an investment account with a nominal interest rate[latex]\\,r\\,[\/latex]at any time[latex]\\,t\\,[\/latex]is given by[latex]\\,A\\left(t\\right)=a{\\left(e\\right)}^{rt},[\/latex]where[latex]\\,a\\,[\/latex]is the amount of principal initially deposited into an account that compounds continuously. Prove that the percentage of interest earned to principal at any time[latex]\\,t\\,[\/latex]can be calculated with the formula[latex]\\,I\\left(t\\right)={e}^{rt}-1.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137705181\" class=\"bc-section section\">\n<h4>Real-World Applications<\/h4>\n<div id=\"fs-id1165137705187\">\n<div id=\"fs-id1165137705189\">\n<p id=\"fs-id1165137705191\">The fox population in a certain region has an annual growth rate of 9% per year. In the year 2012, there were 23,900 fox counted in the area. What is the fox population predicted to be in the year 2020?<\/p>\n<\/div>\n<div id=\"fs-id1165135205873\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135434782\">[latex]47,622\\,[\/latex]fox<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135434802\">\n<div id=\"fs-id1165135434804\">\n<p id=\"fs-id1165135434806\">A scientist begins with 100 milligrams of a radioactive substance that decays exponentially. After 35 hours, 50mg of the substance remains. How many milligrams will remain after 54 hours?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135236978\">\n<div id=\"fs-id1165135236980\">\n<p id=\"fs-id1165135236982\">In the year 1985, a house was valued at $110,000. By the year 2005, the value had appreciated to $145,000. What was the annual growth rate between 1985 and 2005? Assume that the value continued to grow by the same percentage. What was the value of the house in the year 2010?<\/p>\n<\/div>\n<div id=\"fs-id1165134086071\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165134086074\">[latex]1.39%;\\,[\/latex][latex]$155,368.09[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135432945\">\n<div id=\"fs-id1165135432947\">\n<p id=\"fs-id1165135432949\">A car was valued at $38,000 in the year 2007. By 2013, the value had depreciated to $11,000 If the car\u2019s value continues to drop by the same percentage, what will it be worth by 2017?<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135176224\">\n<div id=\"fs-id1165135176226\">\n<p id=\"fs-id1165135176228\">Jamal wants to save $54,000 for a down payment on a home. How much will he need to invest in an account with 8.2% APR, compounding daily, in order to reach his goal in 5 years?<\/p>\n<\/div>\n<div id=\"fs-id1165135208510\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135208512\">[latex]$35,838.76[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135311577\">\n<div id=\"fs-id1165135311579\">\n<p>Kyoko has $10,000 that she wants to invest. Her bank has several investment accounts to choose from, all compounding daily. Her goal is to have $15,000 by the time she finishes graduate school in 6 years. To the nearest hundredth of a percent, what should her minimum annual interest rate be in order to reach her goal? (<em>Hint<\/em>: solve the compound interest formula for the interest rate.)<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135337738\">\n<div id=\"fs-id1165135337740\">\n<p id=\"fs-id1165135337742\">Alyssa opened a retirement account with 7.25% APR in the year 2000. Her initial deposit was $13,500. How much will the account be worth in 2025 if interest compounds monthly? How much more would she make if interest compounded continuously?<\/p>\n<\/div>\n<div id=\"fs-id1165135567423\" class=\"solution textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1165135567426\">[latex]$82,247.78;\\,[\/latex][latex]$449.75[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135567460\">\n<div id=\"fs-id1165135567462\">\n<p id=\"fs-id1165135567464\">An investment account with an annual interest rate of 7% was opened with an initial deposit of $4,000 Compare the values of the account after 9 years when the interest is compounded annually, quarterly, monthly, and continuously.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165135397912\">\n<dt>annual percentage rate (APR)<\/dt>\n<dd id=\"fs-id1165135397918\">the yearly interest rate earned by an investment account, also called <em>nominal rate<\/em><\/dd>\n<\/dl>\n<dl id=\"fs-id1165135397926\">\n<dt>compound interest<\/dt>\n<dd id=\"fs-id1165135397932\">interest earned on the total balance, not just the principal<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838635\">\n<dt>exponential growth<\/dt>\n<dd id=\"fs-id1165137838640\">a model that grows by a rate proportional to the amount present<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838644\">\n<dt>nominal rate<\/dt>\n<dd id=\"fs-id1165137838650\">the yearly interest rate earned by an investment account, also called <em>annual percentage rate<\/em><\/dd>\n<\/dl>\n<\/div>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-98-1\"><a href=\"http:\/\/www.worldometers.info\/world-population\/\">http:\/\/www.worldometers.info\/world-population\/<\/a>. Accessed February 24, 2014. <a href=\"#return-footnote-98-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-98-2\">Oxford Dictionary. <a href=\"http:\/\/oxforddictionaries.com\/us\/definition\/american_english\/nomina.\">http:\/\/oxforddictionaries.com\/us\/definition\/american_english\/nomina.<\/a> <a href=\"#return-footnote-98-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":291,"menu_order":2,"template":"","meta":{"pb_show_title":null,"pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-98","chapter","type-chapter","status-publish","hentry"],"part":95,"_links":{"self":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/98","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/users\/291"}],"version-history":[{"count":1,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/98\/revisions"}],"predecessor-version":[{"id":99,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/98\/revisions\/99"}],"part":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/parts\/95"}],"metadata":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/98\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/media?parent=98"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapter-type?post=98"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/contributor?post=98"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/license?post=98"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}