{"id":238,"date":"2019-02-08T21:02:56","date_gmt":"2019-02-08T21:02:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/exponential-functions\/"},"modified":"2025-03-07T22:21:18","modified_gmt":"2025-03-07T22:21:18","slug":"exponential-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/exponential-functions\/","title":{"raw":"2.2 Exponential Functions","rendered":"2.2 Exponential Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nIn this section, you will:\r\n<ul>\r\n \t<li>Compare exponential functions to linear functions.<\/li>\r\n \t<li>Evaluate exponential functions.<\/li>\r\n \t<li>Determine the characteristics of exponential functions.<\/li>\r\n \t<li>Find the equation of an exponential function.<\/li>\r\n \t<li>Find equations that use continuous growth and decay rates.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIndia is the second most populous country in the world with a population of about 1.37 billion people in 2019. The population is growing at a rate of about 1.08% each year[footnote]http:\/\/www.worldometers.info\/world-population\/. Accessed April 22, 2019.[\/footnote]. If this rate continues, the population of India will exceed China\u2019s population by the year 2031.\u00a0 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.\r\n<div id=\"fs-id1165135159940\" class=\"bc-section section\">\r\n<h3>Identifying Exponential Functions<\/h3>\r\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 the 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>\r\n\r\n<div id=\"fs-id1165137477143\" class=\"bc-section section\">\r\n<h4>Defining an Exponential Function<\/h4>\r\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>\r\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>\r\n\r\n<ul id=\"fs-id1165134042783\">\r\n \t<li><strong>Percent change <\/strong>refers to a <em>change<\/em> based on a <em>percent<\/em> of the original amount.<\/li>\r\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>\r\n \t<li><strong>Exponential decay<\/strong> 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>\r\n<\/ul>\r\n<p id=\"fs-id1165137760753\">For us to gain a clear understanding of <strong><span class=\"no-emphasis\">exponential growth<\/span><\/strong>, 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\">Table 1<\/a>.<\/p>\r\n\r\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).\"><caption>Table 1<\/caption>\r\n<thead>\r\n<tr>\r\n<th class=\"border\">[latex]x[\/latex]<\/th>\r\n<th class=\"border\" style=\"text-align: center;\">[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/th>\r\n<th class=\"border\" style=\"text-align: center;\">[latex]g\\left(x\\right)=2x[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">0<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">1<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">0<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">1<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">4<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">8<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">4<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">16<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">5<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">32<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">10<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\">6<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">64<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">12<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165135190676\">From <a class=\"autogenerated-content\" href=\"#Table_04_01_01\">Table 1<\/a>\u00a0we can infer that for these two functions, exponential growth dominates linear growth.<\/p>\r\n\r\n<ul id=\"fs-id1165137725808\">\r\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. Another way to say this that there is a constant multiplier over equal increments in the domain.<\/li>\r\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>\r\n<\/ul>\r\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>\r\n<p id=\"fs-id1165135445949\">The general form of the <strong><span class=\"no-emphasis\">exponential function<\/span><\/strong> is [latex]f\\left(x\\right)=a{b}^{x},[\/latex] where [latex]a[\/latex] is any nonzero number, and [latex]b[\/latex] is a positive real number not equal to 1.<\/p>\r\n\r\n<ul id=\"fs-id1165137635065\">\r\n \t<li>If [latex]b&gt;1,[\/latex] the function grows at a rate proportional to its size.<\/li>\r\n \t<li>If [latex]0\\lt b&lt;1,[\/latex] the function decays at a rate proportional to its size.<\/li>\r\n<\/ul>\r\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\u00a0to determine the corresponding outputs over an interval in the domain from [latex]-3[\/latex] to [latex]3.[\/latex]\u00a0See <a class=\"autogenerated-content\" href=\"#Table_04_01_02\">Table 2.<\/a><\/p>\r\n\r\n<table id=\"Table_04_01_02\" 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).\"><caption>Table 2<\/caption>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"text-align: center;\"><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]-3[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]-2[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]-1[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]0[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]1[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]2[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center;\">[latex]3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\"><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\r\n<td class=\"border\">[latex]{2}^{-3}=\\frac{1}{8}[\/latex]<\/td>\r\n<td class=\"border\">[latex]{2}^{-2}=\\frac{1}{4}[\/latex]<\/td>\r\n<td class=\"border\">[latex]{2}^{-1}=\\frac{1}{2}[\/latex]<\/td>\r\n<td class=\"border\">[latex]{2}^{0}=1[\/latex]<\/td>\r\n<td class=\"border\">[latex]{2}^{1}=2[\/latex]<\/td>\r\n<td class=\"border\">[latex]{2}^{2}=4[\/latex]<\/td>\r\n<td class=\"border\">[latex]{2}^{3}=8[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165137640874\">Let us examine the graph of [latex]f[\/latex] by plotting the ordered pairs we observe on the table, and then make a few observations. See\u00a0<a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_006\">Figure 1.<\/a><\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_01_006\" class=\"medium\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"490\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210230\/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=\"490\" height=\"321\" \/> <strong>Figure 1.<\/strong>[\/caption]\r\n\r\n<\/div>\r\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>\r\n\r\n<ul id=\"fs-id1165137566018\">\r\n \t<li>The domain is [latex]\\left(-\\infty ,\\infty \\right).[\/latex]<\/li>\r\n \t<li>The range is [latex]\\left(0,\\infty \\right).[\/latex]<\/li>\r\n \t<li>As x gets larger and larger so does [latex]f\\left(x\\right).[\/latex]<\/li>\r\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>\r\n \t<li>[latex]f\\left(x\\right)[\/latex] is always increasing.<\/li>\r\n \t<li>The vertical intercept is [latex]\\left(0,1\\right)[\/latex].<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1165137442472\">\r\n<div class=\"textbox shaded\">\r\n<h3>Exponential Function<\/h3>\r\n<p id=\"fs-id1165137911387\">For any real number [latex]x,[\/latex] an exponential function is a function with the form<\/p>\r\n\r\n<div id=\"Equation_4_1_1\" style=\"text-align: center;\">[latex]f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\r\n<p id=\"eip-751\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137401680\">\r\n \t<li>[latex]a[\/latex] is a non-zero real number called the initial value or initial condition and<\/li>\r\n \t<li>[latex]b[\/latex] is any positive real number such that [latex]b\\ne 1.[\/latex]<\/li>\r\n \t<li>The domain of [latex]f[\/latex] is all real numbers.<\/li>\r\n \t<li>The range of [latex]f[\/latex] is all positive real numbers if [latex]a&gt;0.[\/latex]<\/li>\r\n \t<li>The range of [latex]f[\/latex] is all negative real numbers if [latex]a&lt;0.[\/latex]<\/li>\r\n \t<li>The vertical intercept is [latex]\\left(0,a\\right).[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_01_01\" class=\"textbox examples\">\r\n<div>\r\n<div id=\"fs-id1165137659178\">\r\n<h3>Example 1:\u00a0 Identifying Exponential Functions<\/h3>\r\n<p id=\"fs-id1165137601478\">Which of the following equations are <em>not<\/em> exponential functions?<\/p>\r\n\r\n<ul id=\"fs-id1165135176602\">\r\n \t<li>[latex]f\\left(x\\right)={4}^{3\\left(x-2\\right)}[\/latex]<\/li>\r\n \t<li>[latex]g\\left(x\\right)={x}^{3}[\/latex]<\/li>\r\n \t<li>[latex]h\\left(x\\right)={\\left(\\frac{1}{3}\\right)}^{x}[\/latex]<\/li>\r\n \t<li>[latex]j\\left(x\\right)={\\left(-2\\right)}^{x}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165134108513\">[reveal-answer q=\"fs-id1165134108513\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134108513\"]\r\n<p id=\"fs-id1165137698136\">By definition, an exponential function has a constant 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 which will be studied later.<\/p>\r\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>\r\nThe function\u00a0[latex]f\\left(x\\right)={4}^{3\\left(x-2\\right)}[\/latex] is a transformation of [latex]{4}^{x}[\/latex].\u00a0 [latex]{4}^{x}[\/latex] is compressed by a factor of 1\/3 followed by a shift\u00a0 two units right.\u00a0 Therefore, it is an exponential function.\u00a0 The function [latex]h\\left(x\\right)={\\left(\\frac{1}{3}\\right)}^{x}[\/latex] is also an exponential function with base 1\/3 and an initial condition of 1.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137436342\" class=\"precalculus tryit\">\r\n<h3>Try it #1<\/h3>\r\n<div id=\"ti_04_01_01\">\r\n<div id=\"fs-id1165137862673\">\r\n<p id=\"fs-id1165137424249\">Which of the following equations represent exponential functions?<\/p>\r\n\r\n<ul id=\"fs-id1165135161022\">\r\n \t<li>[latex]f\\left(x\\right)=2{x}^{2}-3x+1[\/latex]<\/li>\r\n \t<li>[latex]g\\left(x\\right)={0.875}^{x}[\/latex]<\/li>\r\n \t<li>[latex]h\\left(x\\right)=1.75x+2[\/latex]<\/li>\r\n \t<li>[latex]j\\left(x\\right)={1095.6}^{-2x}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165137597685\">[reveal-answer q=\"fs-id1165137597685\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137597685\"]\r\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>\r\n[latex]f\\left(x\\right)[\/latex] is a quadratic function and\u00a0[latex]h\\left(x\\right)[\/latex] is a linear function and therefore they are not exponential functions.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137640042\" class=\"bc-section section\">\r\n<h3>Writing Formulas for Exponential Functions and Evaluating Them<\/h3>\r\n<p id=\"fs-id1165137784783\">Recall that the base of an exponential function must be a positive real number other than [latex]1.[\/latex]\u00a0 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>\r\n\r\n<ul id=\"fs-id1165137754880\">\r\n \t<li>Let [latex]b=-9[\/latex] and [latex]x=\\frac{1}{2}.[\/latex] Then [latex]f\\left(\\frac{1}{2}\\right)={\\left(-9\\right)}^{\\frac{1}{2}}=\\sqrt[\\leftroot{1}\\uproot{2} ]{-9},[\/latex] which is not a real number.<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137563360\">Why do we limit the base to positive values other than [latex]1?[\/latex]\u00a0 Because base [latex]1[\/latex] results in the constant function. Observe what happens if the base is [latex]1:[\/latex]<\/p>\r\n\r\n<ul id=\"fs-id1165137400268\">\r\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>\r\n<\/ul>\r\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>\r\n<p id=\"fs-id1165135403544\">Let [latex]f\\left(x\\right)={2}^{x}.[\/latex] What is [latex]f\\left(3\\right)?[\/latex]<\/p>\r\n\r\n<div id=\"eip-id1165137643186\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(x\\right)&amp;={2}^{x}&amp;&amp; \\hfill \\\\ f\\left(3\\right)&amp;={2}^{3}&amp;&amp; \\text{Substitute }x=3.\\\\ \\hfill &amp;=8&amp;&amp; \\text{Evaluate the power}\\text{.}\\end{align*}[\/latex]<\/div>\r\n&nbsp;\r\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>\r\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>\r\n\r\n<div id=\"eip-id1165134086025\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(x\\right)&amp;=30{\\left(2\\right)}^{x}&amp;&amp; \\hfill \\\\ f\\left(3\\right)&amp;=30{\\left(2\\right)}^{3}&amp;&amp; \\textrm{Substitute }x=3.\\hfill \\\\ \\hfill &amp;=30\\left(8\\right)\\text{ }&amp;&amp; \\textrm{Simplify the power first}\\text{.}\\hfill \\\\ \\hfill &amp;=240&amp;&amp; \\textrm{Multiply}\\text{.}\\hfill \\end{align*}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137841073\">Note that if the order of operations were not followed, the result would be incorrect:<\/p>\r\n\r\n<div id=\"eip-id1165135320147\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div id=\"Example_04_01_02\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137455430\">\r\n<div id=\"fs-id1165137455432\">\r\n<h3>Example 2:\u00a0 Evaluating Exponential Functions<\/h3>\r\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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135429364\">[reveal-answer q=\"fs-id1165135429364\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135429364\"]\r\n<p id=\"fs-id1165137598173\">Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\r\n\r\n<div id=\"eip-id1165135208555\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(x\\right)&amp;=5{\\left(3\\right)}^{x+1}&amp;&amp; \\hfill \\\\ f\\left(2\\right)&amp;=5{\\left(3\\right)}^{2+1}&amp;&amp; \\text{Substitute }x=2.\\hfill \\\\ \\hfill &amp;=5{\\left(3\\right)}^{3}&amp;&amp; \\text{Add the exponents}.\\hfill \\\\ \\hfill &amp;=5\\left(27\\right)&amp;&amp; \\text{Simplify the power}\\text{.}\\hfill \\\\ \\hfill &amp;=135&amp;&amp; \\text{Multiply}\\text{.}\\hfill \\end{align*}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137843864\" class=\"precalculus tryit\">\r\n<h3>Try it #2<\/h3>\r\n<div id=\"ti_04_01_02\">\r\n<div id=\"fs-id1165137762786\">\r\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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137637369\">[reveal-answer q=\"fs-id1165137637369\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137637369\"][latex]5.5556[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137439018\" class=\"bc-section section\">\r\n<h4>Defining Exponential Growth and Decay<\/h4>\r\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 or decay.<\/p>\r\n\r\n<div id=\"fs-id1165137564690\">\r\n<div class=\"textbox shaded\">\r\n<h3>Exponential Growth or Decay<\/h3>\r\n<p id=\"fs-id1165137834019\">A function that models <strong>exponential growth or decay<\/strong> 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>\r\n\r\n<div id=\"fs-id1165137851784\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=a{b}^{x}=a\\left(1+r\\right)^x[\/latex]<\/div>\r\nwhere\r\n<ul id=\"fs-id1165137863819\">\r\n \t<li>[latex]a[\/latex] is the initial or starting value of the function.<\/li>\r\n \t<li>[latex]b=1+r[\/latex] is the growth factor or growth multiplier per unit [latex]x[\/latex].<\/li>\r\n \t<li>[latex]r = b - 1[\/latex] is the percent increase or decrease expressed as a decimal.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\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>\r\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\">Table 3<\/a>.<\/p>\r\n\r\n<table id=\"Table_04_01_05\" style=\"height: 100px; width: 757px;\" 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).\"><caption>Table 3<\/caption>\r\n<thead>\r\n<tr style=\"height: 26px;\">\r\n<th class=\"border\" style=\"height: 26px; width: 86.5px; text-align: center;\">Year, [latex]x[\/latex]<\/th>\r\n<th class=\"border\" style=\"height: 26px; width: 208.5px; text-align: center;\">Stores, Company A<\/th>\r\n<th class=\"border\" style=\"width: 155.5px;\"><\/th>\r\n<th class=\"border\" style=\"height: 26px; width: 255.5px; text-align: center;\">Stores, Company B<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 12px;\">\r\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\">[latex]0[\/latex]<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 208.5px; text-align: center;\">[latex]100+50\\left(0\\right)=100[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 155.5px;\">Starting with 100 each<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 255.5px; text-align: center;\">[latex]100{\\left(1+0.5\\right)}^{0}=100[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\">[latex]1[\/latex]<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 208.5px; text-align: center;\">[latex]100+50\\left(1\\right)=150[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 155.5px;\">Both grow by 50 stores in the first year.<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 255.5px; text-align: center;\">[latex]100{\\left(1+0.5\\right)}^{1}=150[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\">[latex]2[\/latex]<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 208.5px; text-align: center;\">[latex]100+50\\left(2\\right)=200[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 155.5px;\">Company A grows by 50 stores and company B by 75 stores.<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 255.5px; text-align: center;\">[latex]100{\\left(1+0.5\\right)}^{2}=225[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\">[latex]3[\/latex]<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 208.5px; text-align: center;\">[latex]100+50\\left(3\\right)=250[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 155.5px;\">Company A grows by 50 stores and company B by 112.5 stores.<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 255.5px; text-align: center;\">[latex]100{\\left(1+0.5\\right)}^{3}=337.5[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 26px;\">\r\n<td class=\"border\" style=\"height: 26px; width: 86.5px; text-align: center;\">[latex]x[\/latex]<\/td>\r\n<td class=\"border\" style=\"height: 26px; width: 208.5px; text-align: center;\">[latex]A\\left(x\\right)=100+50x[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 155.5px;\"><\/td>\r\n<td class=\"border\" style=\"height: 26px; width: 255.5px; text-align: center;\">[latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\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 2<\/a><strong>.<\/strong> We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_01_001\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"227\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210233\/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=\"227\" height=\"394\" \/> <strong>Figure 2.\u00a0<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<div class=\"wp-caption-text\"><\/div>\r\n<p id=\"fs-id1165135209682\">Notice that the contextual 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>\r\n<p id=\"fs-id1165137836429\">Now we will turn our attention to the function representing the number of stores for Company B,<\/p>\r\n<p style=\"text-align: center;\">[latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}.[\/latex]<\/p>\r\nIn this exponential function, 100 represents the initial number of stores, 0.50 represents the <strong>growth rate<\/strong>, [latex]r,[\/latex] and [latex]b=1+0.5=1.5[\/latex] represents the <strong>growth factor<\/strong>. Generalizing further, we can write this function as\r\n<p style=\"text-align: center;\">[latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}.[\/latex]<\/p>\r\n&nbsp;\r\n<div id=\"Example_04_01_03\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137535640\">\r\n<div id=\"fs-id1165137535642\">\r\n<h3>Example 3:\u00a0 Evaluating a Real-World Exponential Model<\/h3>\r\n<p id=\"fs-id1165135541867\">At the beginning of this section, we learned that the population of India was about 1.37 billion in the year 2019, with an annual growth rate of about 1.08%.\u00a0 Let [latex]t[\/latex] be the number of years since 2019. Find an exponential function to model this growth.\u00a0 Then, to the nearest thousandth, determine what will the population of India be in 2031.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137786632\">[reveal-answer q=\"fs-id1165137786632\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137786632\"]The growth rate is given as 1.08% so [latex]r=0.0108.[\/latex]\u00a0 Since the growth factor is [latex]b=1+r[\/latex], we have that [latex]b=1+0.0108=1.0108.[\/latex]\u00a0 There are approximately 1.37 billion people at our initial time [latex]t=0[\/latex] which represents 2019 so [latex]a=1.37.[\/latex]\u00a0 Note that we do not add a bunch of zeros to the equation but rather keep the billions unit in mind when we interpret the function's output.\u00a0 Finally our function is [latex]P\\left(t\\right)=1.37{\\left(1.0108\\right)}^{t}.[\/latex]\r\n<div><\/div>\r\n<div>\r\n\r\nTo\u00a0estimate the population in 2031, we evaluate the model for [latex]t=12[\/latex] because 2031 is 12 years after 2019. Rounding to the nearest thousandth,\r\n<div id=\"eip-id1165135657117\" class=\"unnumbered\" style=\"text-align: center;\">[latex]P\\left(12\\right)=1.37{\\left(1.0108\\right)}^{12}\\approx 1.558[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165135394343\">There will be about 1.558 billion people in India in the year 2031.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 4:\u00a0 Real World Exponential Models<\/h3>\r\nBismuth-210 is an isotope that radioactively decays by about 13% each day, meaning 13% of the remaining Bismuth-210 transforms into another atom (polonium-210 in this case) each day. If you begin with 100 mg of Bismuth-210, how much remains after one week?\r\n\r\n[reveal-answer q=\"685223\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"685223\"]\r\n<p style=\"text-align: left;\">With radioactive decay, instead of the quantity increasing at a percent rate, the quantity is decreasing at a percent rate. Our initial quantity is [latex]a=100[\/latex] mg, and our growth rate will be negative 13%, since we are decreasing: [latex]r=-0.13.[\/latex] This gives the equation:<\/p>\r\n<p style=\"text-align: center;\">[latex]Q\\left(d\\right)=100\\left(1\u2212 0.13\\right)^{d}=100\\left(0.87\\right)^{d}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">This can also be explained by recognizing that if 13% decays, then 87% remains.<\/p>\r\nNext, we are asked to find how much remains after one week so we evaluate the function when [latex]d=7.[\/latex] Therefore,\u00a0[latex]Q\\left(7\\right)=100\\left(0.87\\right)^{7}=37.73.[\/latex]\u00a0 After one week, 37.73 mg of Bismuth-210 remains.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135536569\" class=\"precalculus tryit\">\r\n<h3>Try it #3<\/h3>\r\n<div id=\"ti_04_01_03\">\r\n<div id=\"fs-id1165137635312\">\r\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 0.6%. Find an exponential function that models this situation where [latex]t[\/latex] represents the number of years since 2013.\u00a0 To the nearest thousandth, what will the population of China be for the year 2031?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134200184\">[reveal-answer q=\"fs-id1165134200184\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134200184\"]\r\nThis 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 2013.\r\n<p id=\"fs-id1165134200186\">China's population will be about 1.548 billion people in 2031.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135526980\" class=\"bc-section section\">\r\n<div class=\"textbox tryit\">\r\n<h3>Try It #4<\/h3>\r\nA population of 1000 animals is decreasing 3% each year. Find the population in 30 years.\r\n\r\n[reveal-answer q=\"972350\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"972350\"]\r\n\r\nThe function modeling this situation is [latex]P\\left(t\\right)=1000(0.97)^{t}[\/latex] and in 30 years the population will be 401 animals.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Finding Equations of Exponential Functions From Data<\/h3>\r\n<p id=\"fs-id1165135526985\">In the previous examples, we were given an exponential growth or decay rate, which we used to find a formula to model the situation and then evaluated for a given input. Sometimes we are given data points for an exponential function and we must use them to first find the growth factor and then the formula for the function before we can answer questions about the situation.<\/p>\r\n\r\n<div id=\"fs-id1165135369632\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165135180102\"><strong>Given two data points with the initial value known, write an exponential model.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165135180107\" type=\"1\">\r\n \t<li>Identify the initial value from the data point of the form [latex]\\left(0,a\\right).[\/latex] Then [latex]a[\/latex] is the initial value.<\/li>\r\n \t<li>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]\u00a0 To do this, divide both sides by [latex]a[\/latex] and then raise both sides to the appropriate power to solve the equation.<\/li>\r\n \t<li>Write the equation using the values you found for [latex]a[\/latex] and [latex]b[\/latex] in the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}.[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_01_04\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137580876\">\r\n<div id=\"fs-id1165137580878\">\r\n<h3>Example 5:\u00a0 Writing an Exponential Model When the Initial Value Is Known<\/h3>\r\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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135188416\">[reveal-answer q=\"fs-id1165135188416\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135188416\"]\r\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>\r\n\r\n<div id=\"eip-id1165135432669\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}N\\left(t\\right)&amp;=80{b}^{t}&amp;&amp;\\hfill \\\\180\\hfill &amp;=80{b}^{6}&amp;&amp; \\text{Substitute using point }\\left(6, 180\\right).\\hfill \\\\\\frac{9}{4}&amp;={b}^{6}&amp;&amp; \\text{Divide and write in lowest terms}.\\hfill \\\\b&amp;={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}}&amp;&amp; \\text{Isolate }b\\text{ using properties of exponents}.\\hfill \\\\b&amp;\\approx 1.1447&amp;&amp; \\text{Round to 4 decimal places}. \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1165135193230\"><strong>NOTE:\u00a0<\/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>\r\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>\r\n<p id=\"fs-id1165137724117\">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 3<\/a>\u00a0passes 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>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_01_002\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"244\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210236\/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=\"244\" height=\"350\" \/> <strong>Figure 3.\u00a0<\/strong>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137761908\" class=\"precalculus tryit\">\r\n<h3>Try it #5<\/h3>\r\n<div id=\"ti_04_01_04\">\r\n<div id=\"fs-id1165135496544\">\r\n<p id=\"fs-id1165135496547\">A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013, the population had reached 236 wolves. Let [latex]t=0[\/latex] represent the year 2011.\u00a0 \u00a0What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population, [latex]N\\text{, }[\/latex]of wolves over time [latex]t.[\/latex]<\/p>\r\n[reveal-answer q=\"189152\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"189152\"][latex]\\left(0,129\\right)[\/latex] and [latex]\\left(2,236\\right);\\text{ }\\text{ }\\text{ }N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165135180102\"><strong>Given two data points, write an exponential model with the initial value unknown.<\/strong><\/p>\r\n<strong><em>Method 1: Substitution<\/em><\/strong>\r\n<ol id=\"fs-id1165135180107\" type=\"1\">\r\n \t<li>Substitute both points into two equations with the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}.[\/latex]<\/li>\r\n \t<li>Solve the first equation for [latex]a[\/latex] in terms of [latex]b.[\/latex]<\/li>\r\n \t<li>Substitute this expression for [latex]a[\/latex] into the second equation and then solve for [latex]b.[\/latex]<\/li>\r\n \t<li>Once you have found [latex]b,[\/latex] substitute\u00a0a data point and [latex]b[\/latex] into the general equation and solve for [latex]a[\/latex].<\/li>\r\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>\r\n<\/ol>\r\n<em><strong>Method 2: Ratios<\/strong><\/em>\r\n<ol id=\"fs-id1165135180107\" type=\"1\">\r\n \t<li>Choose two points to work with:\u00a0 [latex]\\left(c, f\\left(c\\right)\\right)[\/latex] and [latex]\\left(d, f\\left(d\\right)\\right).[\/latex] Order your points so [latex]c &lt; d.[\/latex]<\/li>\r\n \t<li>Set up your ratio: [latex]\\frac{f\\left(d\\right)}{f\\left(c\\right)}=\\frac{ab^d}{ab^c}.[\/latex] Note that the values on the left are gotten from your output values of the data points and the values on the right come from the general equation with [latex]c[\/latex] and [latex]d[\/latex] filled in.<\/li>\r\n \t<li>Simplify. Note that [latex]\\frac{a}{a}=1[\/latex] so essentially they always cancel and you will always find [latex]b[\/latex] first.\u00a0 Remember that [latex]\\frac{b^d}{b^c}=b^{d-c}.[\/latex] You will then need to find the (d-c)<sup>th<\/sup> root of both sides.<\/li>\r\n \t<li>Once you have found [latex]b,[\/latex] substitute\u00a0a data point and [latex]b[\/latex] into the general equation and solve for [latex]a[\/latex].<\/li>\r\n \t<li>Write your equation using the values of [latex]a[\/latex] and [latex]b[\/latex]\u00a0in the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}.[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_01_05\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135411399\">\r\n<div id=\"fs-id1165135411402\">\r\n<h3>Example 6:\u00a0 Writing an Exponential Model When the Initial Value is Not Known<\/h3>\r\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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135646190\">[reveal-answer q=\"fs-id1165135646190\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135646190\"]\r\n<p id=\"fs-id1165135646192\"><em><strong>Method 1:<\/strong> <\/em>Because we don\u2019t have the initial value, we substitute both points into the 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>\r\n\r\n<ul id=\"fs-id1165134044679\">\r\n \t<li>Substituting [latex]\\left(-2,6\\right)[\/latex] gives [latex]6=a{b}^{-2}[\/latex]\u00a0 \u00a0(Equation 1)<\/li>\r\n \t<li>Substituting [latex]\\left(2,1\\right)[\/latex] gives [latex]1=a{b}^{2}[\/latex]\u00a0 \u00a0(Equation 2)<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165135361777\">Use the first equation to solve for [latex]a[\/latex] in terms of [latex]b:[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align*}6&amp;=ab^{-2}\\\\\\frac{6}{b^{-2}}&amp;=a&amp;&amp;\\text{Divide by }b^{-2}.\\\\a&amp;=6b^2&amp;&amp;\\text{Use properties of exponents to rewrite the denominator.}\\end{align*}[\/latex]<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1165135180085\">Substitute [latex]a[\/latex] in the second equation, and solve for [latex]b:[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align*}1&amp;=ab^2\\\\1&amp;=\\left(6b^2\\right)b^2=6b^4&amp;&amp;\\text{Substitute }a.\\\\b&amp;=\\left(\\frac{1}{6}\\right)^{\\frac{1}{4}}&amp;&amp;\\text{Use properties of exponents to isolate }b.\\\\b&amp;\\approx0.6389&amp;&amp;\\text{Round 4 decimal places.}\\end{align*}[\/latex]<\/p>\r\n&nbsp;\r\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]<\/p>\r\n\r\n<div style=\"text-align: center;\">[latex]a=6b^2\\approx6\\left(0.6389\\right)^2\\approx2.4492[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165135168188\">Thus, the equation is [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}.[\/latex]<\/p>\r\n<em><strong>Method 2:<\/strong><\/em>\u00a0 The two points\u00a0[latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(2,1\\right)[\/latex] were given so we start with step 2; setting up the ratio.\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{6}=\\frac{ab^2}{ab^{-2}}.[\/latex][latex]\\\\[\/latex]<\/p>\r\nNext we simplify using the properties of exponents.\r\n<p style=\"text-align: center;\">[latex]\\begin{align*}\\frac{1}{6}&amp;=b^{2-(-2)}\\\\\\frac{1}{6}&amp;=b^{4}\\\\\\left(\\frac{1}{6}\\right)^{\\frac{1}{4}}&amp;=\\left(b^4\\right)^{\\frac{1}{4}}\\\\b&amp;\\approx0.6389\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\r\nThen we find [latex]a[\/latex] by substituting in a point [latex]\\left(-2, 6\\right)[\/latex] and the known value of [latex]b\\approx0.6389[\/latex] into [latex]f\\left(x\\right)=ab^x.[\/latex]\u00a0 We get\r\n<p style=\"text-align: center;\">[latex]\\begin{align*}6&amp;=a\\left(0.6389\\right)^{-2}\\\\6&amp;=2.44982027a\\\\a&amp;\\approx2.4492.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\r\nFinally, the equation is [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}.[\/latex]\r\n<p id=\"fs-id1165135482016\"><em><strong>Check:<\/strong><\/em> 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 4<\/a>\u00a0passes 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 <strong><span class=\"no-emphasis\">exponential decay<\/span><\/strong> function.<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_01_003\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"291\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210248\/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=\"291\" height=\"266\" \/> <strong>Figure 4.\u00a0<\/strong>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"ti_11_04_01\" class=\"precalculus tryit\">\r\n<h3>Try it #6<\/h3>\r\n<div id=\"fs-id1165135169306\">\r\n<div id=\"fs-id1165135169308\">\r\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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137732255\">[reveal-answer q=\"fs-id1165137732255\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137732255\"]\r\n<p id=\"fs-id1165137732257\">[latex]f\\left(x\\right)=2{\\left(1.5\\right)}^{x}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137737999\" class=\"precalculus qa key-takeaways\">\r\n<h3>Q&amp;A<\/h3>\r\n<p id=\"fs-id1165137738004\"><strong>Do two points always determine a unique exponential function?<\/strong><\/p>\r\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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137771573\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137771578\"><strong>Given a graph or a table of an exponential function, write its equation.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137771583\" type=\"1\">\r\n \t<li>First, identify two points on the graph or table. Choose the vertical 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>\r\n \t<li>Use one of the methods for finding an equation given two points above.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_01_06\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137806415\">\r\n<div id=\"fs-id1165137806417\">\r\n<h3>\u00a0Example 7:\u00a0 Writing an Exponential Function Given Its Graph<\/h3>\r\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 5<\/a>.<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_01_004\" class=\"medium\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210251\/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]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135415647\">[reveal-answer q=\"fs-id1165135415647\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135415647\"]\r\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>\r\n\r\n<div id=\"eip-id1165134216216\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(x\\right)&amp;=a{b}^{x}&amp;&amp; \\text{Write the general form of an exponential equation}.\\hfill \\\\ f\\left(x\\right)&amp;=3{b}^{x}&amp;&amp; \\text{Substitute the initial value 3 for }a.\\hfill \\\\ 12&amp;=3{b}^{2}&amp;&amp; \\text{Substitute in 12 for }y\\text{ and 2 for }x.\\hfill \\\\4&amp;={b}^{2}&amp;&amp; \\text{Divide by 3}.\\hfill \\\\ b&amp;=\u00b12&amp;&amp; \\text{Take the square root}.\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\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]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135169326\" class=\"precalculus tryit\">\r\n<h3>Try it #7<\/h3>\r\n<div id=\"ti_04_01_06\">\r\n<div id=\"fs-id1165135431682\">\r\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 6<\/a>.<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_01_005\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210254\/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]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135191210\">[reveal-answer q=\"fs-id1165135191210\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135191210\"]\r\n<p id=\"fs-id1165135191213\">[latex]f\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2}2]{2}{\\left(\\sqrt[\\leftroot{1}\\uproot{2}2]{2}\\right)}^{x}.[\/latex] Answers may vary due to round-off error. The answer should be very close to [latex]f\\left(x\\right)=1.4142{\\left(1.4142\\right)}^{x}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137871708\" class=\"bc-section section\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 8: Writing an Exponential Function Given a Table of Values<\/h3>\r\nFind an equation for the exponential function given by Table 4.\r\n<table style=\"border-collapse: collapse;\" border=\"1\"><caption>Table 4<\/caption>\r\n<tbody>\r\n<tr style=\"height: 11px;\">\r\n<td class=\"border\" style=\"width: 117.656px; height: 11px; text-align: center;\">[latex]x[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 100.656px; height: 11px; text-align: center;\">1<\/td>\r\n<td class=\"border\" style=\"width: 100.656px; height: 11px; text-align: center;\">3<\/td>\r\n<td class=\"border\" style=\"width: 100.656px; height: 11px; text-align: center;\">6<\/td>\r\n<td class=\"border\" style=\"width: 101.656px; height: 11px; text-align: center;\">8<\/td>\r\n<td class=\"border\" style=\"width: 101.656px; height: 11px; text-align: center;\">12<\/td>\r\n<\/tr>\r\n<tr style=\"height: 23px;\">\r\n<td class=\"border\" style=\"width: 117.656px; height: 23px; text-align: center;\">[latex]f\\left(x\\right)[\/latex]<\/td>\r\n<td class=\"border\" style=\"width: 100.656px; height: 23px; text-align: center;\">4.8<\/td>\r\n<td class=\"border\" style=\"width: 100.656px; height: 23px; text-align: center;\">6.912<\/td>\r\n<td class=\"border\" style=\"width: 100.656px; height: 23px; text-align: center;\">11.943936<\/td>\r\n<td class=\"border\" style=\"width: 101.656px; height: 23px; text-align: center;\">17.19926784<\/td>\r\n<td class=\"border\" style=\"width: 101.656px; height: 23px; text-align: center;\">35.664401793<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"841908\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"841908\"]\r\n\r\nFirst we choose any two points to use in finding our equation.\u00a0 Let's use (1, 4.8) and (3, 6.912).\r\n\r\nFor this example, we will use ratios to find the equation.\r\n<p style=\"text-align: center;\">[latex]\\frac{6.912}{4.8}=\\frac{ab^3}{ab^1}[\/latex]<\/p>\r\nNext, we simplify and solve.\r\n<p style=\"text-align: center;\">[latex]\\begin{align*}1.44&amp;=b^{3-1}\\\\1.44&amp;=b^2\\\\1.44^{\\frac{1}{2}}&amp;=\\left(b^2\\right)^{\\frac{1}{2}}\\\\b&amp;=1.2\\end{align*}[\/latex]<\/p>\r\nFinally, we find [latex]a[\/latex] by plugging in 1.2 for [latex]b[\/latex] and 1 for [latex]x[\/latex] and 4.8 for [latex]f\\left(x\\right)[\/latex] in [latex]f\\left(x\\right)=ab^x[\/latex] to get\r\n<p style=\"text-align: center;\">[latex]4.8=a\\left(1.2\\right)^1[\/latex] or [latex]a=4.[\/latex]<\/p>\r\nThe final equation is [latex]f\\left(x\\right)=4\\left(1.2\\right)^x.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Applying the Compound-Interest Formula<\/h3>\r\n<p id=\"fs-id1165137447026\">Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use <strong>compound interest<\/strong>. The term <em>compounding<\/em> refers to interest earned not only on the original value, but on the accumulated value of the account.<\/p>\r\n<p id=\"fs-id1165137447037\">The <strong>annual percentage rate (APR)<\/strong> of an account, also called the <strong>nominal rate<\/strong>, 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>\r\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>\r\n\r\n<div id=\"eip-986\" class=\"unnumbered\" style=\"text-align: center;\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}.[\/latex]<\/div>\r\n<p id=\"fs-id1165137935717\">For example, observe <a class=\"autogenerated-content\" href=\"#Table_04_01_03\">Table 5<\/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>\r\n\r\n<table id=\"Table_04_01_03\" 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.\"><caption>Table 5<\/caption>\r\n<thead>\r\n<tr>\r\n<th class=\"border\">Frequency<\/th>\r\n<th class=\"border\">Value after 1 year<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\">Annually<\/td>\r\n<td class=\"border\">$1100<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">Semiannually<\/td>\r\n<td class=\"border\">$1102.50<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">Quarterly<\/td>\r\n<td class=\"border\">$1103.81<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">Monthly<\/td>\r\n<td class=\"border\">$1104.71<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">Daily<\/td>\r\n<td class=\"border\">$1105.16<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-id1165137793679\">\r\n<div class=\"textbox shaded\">\r\n<h3>The Compound Interest Formula<\/h3>\r\n<p id=\"fs-id1165135184167\"><strong>Compound interest<\/strong> can be calculated using the formula<\/p>\r\n\r\n<div id=\"fs-id1165135184172\" style=\"text-align: center;\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/div>\r\n<p id=\"eip-237\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137448453\">\r\n \t<li>[latex]A\\left(t\\right)[\/latex] is the account value,<\/li>\r\n \t<li>[latex]t[\/latex] is measured in years,<\/li>\r\n \t<li>[latex]P[\/latex] is the starting amount of the account, often called the principal, or more generally present value,<\/li>\r\n \t<li>[latex]r[\/latex] is the annual percentage rate (APR) expressed as a decimal, and<\/li>\r\n \t<li>[latex]n[\/latex] is the number of compounding periods in one year.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_01_08\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137812820\">\r\n<h3>Example 9: Calculating Compound Interest<\/h3>\r\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>\r\n[reveal-answer q=\"318094\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"318094\"]\r\n\r\nBecause 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=10.[\/latex]\r\n<div id=\"eip-id1402796\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}A\\left(t\\right)&amp;=P{\\left(1+\\frac{r}{n}\\right)}^{nt}&amp;&amp;\\text{Use the compound interest formula}.\\\\A\\left(10\\right)&amp;=3000{\\left(1+\\frac{0.03}{4}\\right)}^{4\\cdot10}&amp;&amp;\\text{Substitute using given values}.\\\\&amp;\\approx4045.05.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div class=\"unnumbered\" style=\"text-align: center;\"><\/div>\r\n<p id=\"fs-id1165137694040\">The account will be worth about $4,045.05 in 10 years.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137694046\" class=\"precalculus tryit\">\r\n<h3>Try it #8<\/h3>\r\n<div id=\"ti_04_01_08\">\r\n<div id=\"fs-id1165137694055\">\r\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>\r\n\r\n<\/div>\r\n<div>\r\n\r\n[reveal-answer q=\"89519\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"89519\"]\r\n\r\nabout $3,644,675.88[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_01_09\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135180446\">\r\n<div>\r\n<h3>Example 10:\u00a0 Using the Compound Interest Formula to Solve for the Principal<\/h3>\r\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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135175338\">[reveal-answer q=\"fs-id1165135175338\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135175338\"]\r\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]n=2.[\/latex]<\/p>\r\n<p id=\"fs-id1165135209414\">We want to find the initial investment, [latex]P,[\/latex] needed so that the value of the account will be worth $40,000 in 18 years. Substitute the given values into the compound interest formula, and solve for [latex]P.[\/latex]<\/p>\r\n\r\n<div id=\"eip-id1165131884554\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}A\\left(t\\right)&amp;=P{\\left(1+\\frac{r}{n}\\right)}^{nt}&amp;&amp;\\text{Use the compound interest formula}.\\\\40,000&amp;=P{\\left(1+\\frac{0.06}{2}\\right)}^{2\\left(18\\right)}&amp;&amp;\\text{Substitute using given values }A\\text{, }r, n\\text{, and }t.\\\\40,000&amp;=P{\\left(1.03\\right)}^{36}&amp;&amp;\\text{Simplify}.\\\\\\frac{40,000}{{\\left(1.03\\right)}^{36}}&amp;=P&amp;&amp;\\text{Isolate }P.\\\\P&amp;\\approx13,801&amp;&amp;\\text{Divide and round to the nearest dollar}.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1165137937589\">Lily will need to invest $13,801 to have $40,000 in 18 years.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"precalculus tryit\">\r\n<h3>Try it #9<\/h3>\r\n<div id=\"ti_04_01_9\">\r\n<div id=\"fs-id1165135176727\">\r\n<p id=\"fs-id1165135176729\">Refer to <a class=\"autogenerated-content\" href=\"#Example_04_01_09\">Example 10<\/a>. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135176736\">[reveal-answer q=\"fs-id1165135176736\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135176736\"]\r\n<p id=\"fs-id1165135176738\">$13,693<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137724961\" class=\"bc-section section\">\r\n<h3>Evaluating Functions with Base <em><span class=\"e2\">e<\/span><\/em><\/h3>\r\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\">Table 6<\/a>\u00a0shows 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>\r\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\">Table 6<\/a>.<\/p>\r\n\r\n<table id=\"Table_04_01_04\" 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 2, and Semiannually has the input value of (1+1\/2)^2 which equals to 2.25, Quarterly has the input value of (1+1\/4)^4 which equals to 2.441406, Monthly has the input value of (1+1\/12)^12 which equals to 2.613035, Daily has the input value of (1+1\/365)^365 which equals to 2.714567, Hourly has the input value of (1+1\/8766)^8766 which equals to 2.718127, One per minute has the input value of (1+1\/525960)^525960 which equals to 2.718279, and Once per second has the input value of (1+1\/31557600)^31557600 which equals to 2.718282.\"><caption>Table 6<\/caption>\r\n<thead>\r\n<tr>\r\n<th class=\"border\">Frequency<\/th>\r\n<th class=\"border\">[latex]A\\left(n\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\r\n<th class=\"border\">Value<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\">Annually<\/td>\r\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\r\n<td class=\"border\">$2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">Semiannually<\/td>\r\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\r\n<td class=\"border\">$2.25<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">Quarterly<\/td>\r\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\r\n<td class=\"border\">$2.441406<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">Monthly<\/td>\r\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\r\n<td class=\"border\">$2.613035<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">Daily<\/td>\r\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\r\n<td class=\"border\">$2.714567<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">Hourly<\/td>\r\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{\\text{8760}}\\right)}^{\\text{8760}}[\/latex]<\/td>\r\n<td class=\"border\">$2.718127<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">Once per minute<\/td>\r\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{\\text{525600}}\\right)}^{\\text{525600}}[\/latex]<\/td>\r\n<td class=\"border\">$2.718279<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">Once per second<\/td>\r\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{31536000}\\right)}^{31536000}[\/latex]<\/td>\r\n<td class=\"border\">$2.718282<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\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>\r\n\r\n<div id=\"fs-id1165135511324\">\r\n<div class=\"textbox shaded\">\r\n<h3>Euler's Number<\/h3>\r\n<p id=\"fs-id1165135511335\">The letter <em><span class=\"e2\">e<\/span><\/em> represents the irrational number [latex]{\\left(1+\\frac{1}{n}\\right)}^{n},[\/latex] as [latex]n[\/latex] increases without bound.<\/p>\r\n&nbsp;\r\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><span class=\"e2\">e<\/span><\/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>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_01_10\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135344893\">\r\n<div id=\"fs-id1165135344895\">\r\n<h3>Example 11:\u00a0 Using a Calculator to Find Powers of <em><span class=\"e2\">e<\/span><\/em><\/h3>\r\n<p id=\"fs-id1165135545973\">Calculate [latex]{e}^{3.14}.[\/latex] Round to five decimal places.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135192743\">[reveal-answer q=\"fs-id1165135192743\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135192743\"]\r\n<p id=\"fs-id1165135192745\">On a calculator, press the button labeled [latex]\\left[{e}^{x}\\right].[\/latex] The window shows [latex]\\left[e^{\\wedge}\\text{ ( }\\right].[\/latex]Type [latex]3.14[\/latex] and then close parenthesis, [latex]\\left[\\text{ ) }\\right].[\/latex] Press [ENTER]. Rounding to\u00a0 5 decimal places, [latex]{e}^{3.14}\\approx 23.10387.[\/latex]<\/p>\r\nCaution: 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]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137447188\" class=\"precalculus tryit\">\r\n<h3>Try it #10<\/h3>\r\n<div id=\"ti_04_01_10\">\r\n<div id=\"fs-id1165135348435\">\r\n<p id=\"fs-id1165135348437\">Use a calculator to find [latex]{e}^{-0.5}.[\/latex] Round to five decimal places.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135176286\">[reveal-answer q=\"fs-id1165135176286\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135176286\"]\r\n<p id=\"fs-id1165135176288\">[latex]{e}^{-0.5}\\approx 0.60653[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137827923\" class=\"bc-section section\">\r\n<h3>Investigating Continuous Growth<\/h3>\r\n<p id=\"fs-id1165137827929\">So far we have worked with rational bases for exponential functions. For many 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<strong> <em>continuous growth or decay models<\/em><\/strong>. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.<\/p>\r\n\r\n<div id=\"fs-id1165137664673\">\r\n<div class=\"textbox shaded\">\r\n<h3>The Continuous Growth\/Decay Formula<\/h3>\r\n<p id=\"fs-id1165135453868\">For all real numbers [latex]t,[\/latex] and all positive numbers [latex]a[\/latex] and [latex]k,[\/latex] continuous growth or decay is represented by the formula<\/p>\r\n\r\n<div id=\"fs-id1165135536370\" style=\"text-align: center;\">[latex]A\\left(t\\right)=a{e}^{kt}[\/latex]<\/div>\r\n<p id=\"eip-101\">where<\/p>\r\n\r\n<ul id=\"fs-id1165135152052\">\r\n \t<li>[latex]a[\/latex] is the initial value,<\/li>\r\n \t<li>[latex]k[\/latex] is the <strong>continuous rate<\/strong> per unit time, and<\/li>\r\n \t<li>[latex]t[\/latex] is the elapsed time.<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165135560686\">If [latex]k&gt;0[\/latex], then the formula represents continuous growth. If [latex]k&lt;0[\/latex], then the formula represents continuous decay.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135411368\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165135411373\"><strong>Given the initial value, continuous rate of growth or decay, and time, solve a continuous growth or decay function.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165135511371\" type=\"1\">\r\n \t<li>Use the information in the problem to determine [latex]a[\/latex], the initial value of the function.<\/li>\r\n \t<li>Use the information in the problem to determine the growth rate [latex]k.[\/latex]\r\n<ul id=\"fs-id1165135188096\" type=\"a\">\r\n \t<li>If the problem refers to continuous growth, then [latex]k&gt;0.[\/latex]<\/li>\r\n \t<li>If the problem refers to continuous decay, then [latex]k&lt;0.[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Use the information in the problem to determine the time [latex]t.[\/latex]<\/li>\r\n \t<li>Substitute the given information into the continuous growth formula and simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_01_11\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137835464\">\r\n<div id=\"fs-id1165137835466\">\r\n<h3>Example 12: Calculating Continuous Growth<\/h3>\r\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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137694203\">[reveal-answer q=\"fs-id1165137694203\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137694203\"]The rate given is continuous so we use the formula [latex]A\\left(t\\right)=a{e}^{kt}.[\/latex] Since the account is growing in value, this is a continuous compounding problem with growth rate [latex]k=0.10.[\/latex] The initial investment was $1,000, so [latex]a=1000.[\/latex] To find the value after [latex]t=1[\/latex] year:\r\n<div id=\"eip-id1165133351794\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll}A\\left(t\\right)\\hfill &amp; =a{e}^{kt}\\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 }a, k,\\text{ and }t.\\hfill \\\\ \\hfill &amp; \\approx 1105.17\\hfill &amp; \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137895288\">The account is worth $1,105.17 after one year.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137895295\" class=\"precalculus tryit\">\r\n<h3>Try it #11<\/h3>\r\n<div id=\"ti_04_01_11\">\r\n<div id=\"fs-id1165137895303\">\r\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>\r\n\r\n<\/div>\r\n<div>\r\n<p id=\"fs-id1165134389978\">[reveal-answer q=\"105348\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"105348\"]<\/p>\r\n$3,659,823.44[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_01_12\" class=\"textbox examples\">\r\n<div id=\"fs-id1165134389988\">\r\n<div id=\"fs-id1165134389990\">\r\n<h3>Example 13: Calculating Continuous Decay<\/h3>\r\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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137803706\">[reveal-answer q=\"fs-id1165137803706\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137803706\"]Since the substance is decaying, the rate, 17.3%, is negative. So, [latex]k=\\text{ }-0.173.[\/latex] The initial amount of radon-222 was [latex] 100\\text{ }[\/latex]mg, so [latex]a=100.[\/latex] We use the continuous decay formula to find the value after [latex]t=3[\/latex] days:\r\n<div id=\"eip-id1165137779893\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll}A\\left(t\\right)\\hfill &amp; =a{e}^{kt}\\hfill &amp; \\text{Use the continuous decay 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, k,\\text{ and }t.\\hfill \\\\ \\hfill &amp; \\approx 59.5115\\hfill &amp; \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165137697132\">So 59.5115 mg of radon-222 will remain after 3 days.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135348462\" class=\"precalculus tryit\">\r\n<h3>Try it #12<\/h3>\r\n<div id=\"ti_04_01_12\">\r\n<div id=\"fs-id1165135348470\">\r\n<p id=\"fs-id1165135348472\">Using the data in <a class=\"autogenerated-content\" href=\"#Example_04_01_12\">Example 13<\/a>, how much radon-222 will remain after one year?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135245732\">[reveal-answer q=\"fs-id1165135245732\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135245732\"]\r\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>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135191186\" class=\"precalculus media\">\r\n<div class=\"textbox shaded\">\r\n<div id=\"fs-id1165137827923\" class=\"bc-section section\">\r\n<div id=\"fs-id1165135191186\" class=\"precalculus media\">\r\n<h3>Media<\/h3>\r\n<p id=\"fs-id1165135191191\">Access these online resources for additional instruction and practice with exponential functions.<\/p>\r\n\r\n<ul id=\"fs-id1165135191194\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/expgrowth\">Exponential Growth Function<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/compoundint\">Compound Interest<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165135191191\"><span style=\"color: #6c64ad; font-size: 1em; font-weight: 600;\">Key Equations<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135264762\" class=\"key-equations\">\r\n<table id=\"fs-id2306479\" summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td class=\"border\">definition of exponential growth or decay<\/td>\r\n<td class=\"border\">[latex]f\\left(x\\right)=a{b}^{x},\\text{ where }a&gt;0,b&gt;0,b\\ne 1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">compound interest formula<\/td>\r\n<td class=\"border\">[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>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">continuous growth formula<\/td>\r\n<td class=\"border\">[latex]A\\left(t\\right)=a{e}^{kt},\\text{ where}[\/latex]\r\n<div><\/div>\r\n[latex]t[\/latex] is the number of unit time periods of growth\r\n<div><\/div>\r\n[latex]a[\/latex] is the starting amount (in the continuous compounding formula [latex]a[\/latex] may be replaced with P, the principal)\r\n<div><\/div>\r\n[latex]e[\/latex] is the mathematical constant,[latex] \\text{ }e\\approx 2.718282[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"fs-id1165137846440\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\r\n<ul id=\"fs-id1165137846446\">\r\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.<\/li>\r\n \t<li>A function is evaluated by solving at a specific value.<\/li>\r\n \t<li>An exponential model can be found when the growth rate and initial value are known.<\/li>\r\n \t<li>An exponential model can be found when the two data points from the model are known.<\/li>\r\n \t<li>An exponential model can be found using two data points from a graph or a table of the model.<\/li>\r\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.<\/li>\r\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.<\/li>\r\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>\r\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 continuous growth or decay rate are known.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1165135397912\">\r\n \t<dt>annual percentage rate (APR)<\/dt>\r\n \t<dd id=\"fs-id1165135397918\">the yearly interest rate earned by an investment account, also called <em>nominal rate<\/em><\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135397926\">\r\n \t<dt>compound interest<\/dt>\r\n \t<dd id=\"fs-id1165135397932\">interest earned on the total balance, not just the principal<\/dd>\r\n<\/dl>\r\n<dl>\r\n \t<dt>exponential growth<\/dt>\r\n \t<dd id=\"fs-id1165137838640\">a model that grows by a rate proportional to the amount present<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137838644\">\r\n \t<dt>nominal rate<\/dt>\r\n \t<dd id=\"fs-id1165137838650\">the yearly interest rate earned by an investment account, also called <em>annual percentage rate<\/em><\/dd>\r\n<\/dl>\r\n<\/div>\r\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>Compare exponential functions to linear functions.<\/li>\n<li>Evaluate exponential functions.<\/li>\n<li>Determine the characteristics of exponential functions.<\/li>\n<li>Find the equation of an exponential function.<\/li>\n<li>Find equations that use continuous growth and decay rates.<\/li>\n<\/ul>\n<\/div>\n<p>India is the second most populous country in the world with a population of about 1.37 billion people in 2019. The population is growing at a rate of about 1.08% each year<a class=\"footnote\" title=\"http:\/\/www.worldometers.info\/world-population\/. Accessed April 22, 2019.\" id=\"return-footnote-238-1\" href=\"#footnote-238-1\" 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 2031.\u00a0 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 the 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><strong>Exponential decay<\/strong> 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 <strong><span class=\"no-emphasis\">exponential growth<\/span><\/strong>, 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\">Table 1<\/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<caption>Table 1<\/caption>\n<thead>\n<tr>\n<th class=\"border\">[latex]x[\/latex]<\/th>\n<th class=\"border\" style=\"text-align: center;\">[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/th>\n<th class=\"border\" style=\"text-align: center;\">[latex]g\\left(x\\right)=2x[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">0<\/td>\n<td class=\"border\" style=\"text-align: center;\">1<\/td>\n<td class=\"border\" style=\"text-align: center;\">0<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">1<\/td>\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">2<\/td>\n<td class=\"border\" style=\"text-align: center;\">4<\/td>\n<td class=\"border\" style=\"text-align: center;\">4<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">3<\/td>\n<td class=\"border\" style=\"text-align: center;\">8<\/td>\n<td class=\"border\" style=\"text-align: center;\">6<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">4<\/td>\n<td class=\"border\" style=\"text-align: center;\">16<\/td>\n<td class=\"border\" style=\"text-align: center;\">8<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">5<\/td>\n<td class=\"border\" style=\"text-align: center;\">32<\/td>\n<td class=\"border\" style=\"text-align: center;\">10<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"text-align: center;\">6<\/td>\n<td class=\"border\" style=\"text-align: center;\">64<\/td>\n<td class=\"border\" style=\"text-align: center;\">12<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165135190676\">From <a class=\"autogenerated-content\" href=\"#Table_04_01_01\">Table 1<\/a>\u00a0we can infer that for these two functions, exponential growth dominates 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. Another way to say this that there is a constant multiplier over equal increments 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 <strong><span class=\"no-emphasis\">exponential function<\/span><\/strong> is [latex]f\\left(x\\right)=a{b}^{x},[\/latex] where [latex]a[\/latex] is any nonzero number, and [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\\lt 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\u00a0to determine the corresponding outputs over an interval in the domain from [latex]-3[\/latex] to [latex]3.[\/latex]\u00a0See <a class=\"autogenerated-content\" href=\"#Table_04_01_02\">Table 2.<\/a><\/p>\n<table id=\"Table_04_01_02\" 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<caption>Table 2<\/caption>\n<tbody>\n<tr>\n<td class=\"border\" style=\"text-align: center;\"><strong>[latex]x[\/latex]<\/strong><\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]-3[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]-2[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]-1[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]0[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]1[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]2[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center;\">[latex]3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td class=\"border\"><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\n<td class=\"border\">[latex]{2}^{-3}=\\frac{1}{8}[\/latex]<\/td>\n<td class=\"border\">[latex]{2}^{-2}=\\frac{1}{4}[\/latex]<\/td>\n<td class=\"border\">[latex]{2}^{-1}=\\frac{1}{2}[\/latex]<\/td>\n<td class=\"border\">[latex]{2}^{0}=1[\/latex]<\/td>\n<td class=\"border\">[latex]{2}^{1}=2[\/latex]<\/td>\n<td class=\"border\">[latex]{2}^{2}=4[\/latex]<\/td>\n<td class=\"border\">[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, and then make a few observations. See\u00a0<a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_01_006\">Figure 1.<\/a><\/p>\n<div id=\"CNX_Precalc_Figure_04_01_006\" class=\"medium\">\n<div style=\"width: 500px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210230\/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=\"490\" height=\"321\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 1.<\/strong><\/p>\n<\/div>\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 x gets larger and larger so does [latex]f\\left(x\\right).[\/latex]<\/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]f\\left(x\\right)[\/latex] is always increasing.<\/li>\n<li>The vertical intercept is [latex]\\left(0,1\\right)[\/latex].<\/li>\n<\/ul>\n<div id=\"fs-id1165137442472\">\n<div class=\"textbox shaded\">\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\" style=\"text-align: center;\">[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 or initial condition 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 vertical intercept is [latex]\\left(0,a\\right).[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"Example_04_01_01\" class=\"textbox examples\">\n<div>\n<div id=\"fs-id1165137659178\">\n<h3>Example 1:\u00a0 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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134108513\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134108513\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137698136\">By definition, an exponential function has a constant 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 which will be studied later.<\/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<p>The function\u00a0[latex]f\\left(x\\right)={4}^{3\\left(x-2\\right)}[\/latex] is a transformation of [latex]{4}^{x}[\/latex].\u00a0 [latex]{4}^{x}[\/latex] is compressed by a factor of 1\/3 followed by a shift\u00a0 two units right.\u00a0 Therefore, it is an exponential function.\u00a0 The function [latex]h\\left(x\\right)={\\left(\\frac{1}{3}\\right)}^{x}[\/latex] is also an exponential function with base 1\/3 and an initial condition of 1.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137436342\" class=\"precalculus tryit\">\n<h3>Try it #1<\/h3>\n<div id=\"ti_04_01_01\">\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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137597685\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137597685\" class=\"hidden-answer\" style=\"display: none\">\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<p>[latex]f\\left(x\\right)[\/latex] is a quadratic function and\u00a0[latex]h\\left(x\\right)[\/latex] is a linear function and therefore they are not exponential functions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137640042\" class=\"bc-section section\">\n<h3>Writing Formulas for Exponential Functions and Evaluating Them<\/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]\u00a0 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(\\frac{1}{2}\\right)={\\left(-9\\right)}^{\\frac{1}{2}}=\\sqrt[\\leftroot{1}\\uproot{2} ]{-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]\u00a0 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\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(x\\right)&={2}^{x}&& \\hfill \\\\ f\\left(3\\right)&={2}^{3}&& \\text{Substitute }x=3.\\\\ \\hfill &=8&& \\text{Evaluate the power}\\text{.}\\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\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\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(x\\right)&=30{\\left(2\\right)}^{x}&& \\hfill \\\\ f\\left(3\\right)&=30{\\left(2\\right)}^{3}&& \\textrm{Substitute }x=3.\\hfill \\\\ \\hfill &=30\\left(8\\right)\\text{ }&& \\textrm{Simplify the power first}\\text{.}\\hfill \\\\ \\hfill &=240&& \\textrm{Multiply}\\text{.}\\hfill \\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\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\" style=\"text-align: center;\">[latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<div id=\"Example_04_01_02\" class=\"textbox examples\">\n<div id=\"fs-id1165137455430\">\n<div id=\"fs-id1165137455432\">\n<h3>Example 2:\u00a0 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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135429364\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135429364\" class=\"hidden-answer\" style=\"display: none\">\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\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(x\\right)&=5{\\left(3\\right)}^{x+1}&& \\hfill \\\\ f\\left(2\\right)&=5{\\left(3\\right)}^{2+1}&& \\text{Substitute }x=2.\\hfill \\\\ \\hfill &=5{\\left(3\\right)}^{3}&& \\text{Add the exponents}.\\hfill \\\\ \\hfill &=5\\left(27\\right)&& \\text{Simplify the power}\\text{.}\\hfill \\\\ \\hfill &=135&& \\text{Multiply}\\text{.}\\hfill \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137843864\" class=\"precalculus tryit\">\n<h3>Try it #2<\/h3>\n<div id=\"ti_04_01_02\">\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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137637369\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137637369\" class=\"hidden-answer\" style=\"display: none\">[latex]5.5556[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137439018\" class=\"bc-section section\">\n<h4>Defining Exponential Growth and Decay<\/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 or decay.<\/p>\n<div id=\"fs-id1165137564690\">\n<div class=\"textbox shaded\">\n<h3>Exponential Growth or Decay<\/h3>\n<p id=\"fs-id1165137834019\">A function that models <strong>exponential growth or decay<\/strong> 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\" style=\"text-align: center;\">[latex]f\\left(x\\right)=a{b}^{x}=a\\left(1+r\\right)^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=1+r[\/latex] is the growth factor or growth multiplier per unit [latex]x[\/latex].<\/li>\n<li>[latex]r = b - 1[\/latex] is the percent increase or decrease expressed as a decimal.<\/li>\n<\/ul>\n<\/div>\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\">Table 3<\/a>.<\/p>\n<table id=\"Table_04_01_05\" style=\"height: 100px; width: 757px;\" 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<caption>Table 3<\/caption>\n<thead>\n<tr style=\"height: 26px;\">\n<th class=\"border\" style=\"height: 26px; width: 86.5px; text-align: center;\">Year, [latex]x[\/latex]<\/th>\n<th class=\"border\" style=\"height: 26px; width: 208.5px; text-align: center;\">Stores, Company A<\/th>\n<th class=\"border\" style=\"width: 155.5px;\"><\/th>\n<th class=\"border\" style=\"height: 26px; width: 255.5px; text-align: center;\">Stores, Company B<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 12px;\">\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\">[latex]0[\/latex]<\/td>\n<td class=\"border\" style=\"height: 12px; width: 208.5px; text-align: center;\">[latex]100+50\\left(0\\right)=100[\/latex]<\/td>\n<td class=\"border\" style=\"width: 155.5px;\">Starting with 100 each<\/td>\n<td class=\"border\" style=\"height: 12px; width: 255.5px; text-align: center;\">[latex]100{\\left(1+0.5\\right)}^{0}=100[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\">[latex]1[\/latex]<\/td>\n<td class=\"border\" style=\"height: 12px; width: 208.5px; text-align: center;\">[latex]100+50\\left(1\\right)=150[\/latex]<\/td>\n<td class=\"border\" style=\"width: 155.5px;\">Both grow by 50 stores in the first year.<\/td>\n<td class=\"border\" style=\"height: 12px; width: 255.5px; text-align: center;\">[latex]100{\\left(1+0.5\\right)}^{1}=150[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\">[latex]2[\/latex]<\/td>\n<td class=\"border\" style=\"height: 12px; width: 208.5px; text-align: center;\">[latex]100+50\\left(2\\right)=200[\/latex]<\/td>\n<td class=\"border\" style=\"width: 155.5px;\">Company A grows by 50 stores and company B by 75 stores.<\/td>\n<td class=\"border\" style=\"height: 12px; width: 255.5px; text-align: center;\">[latex]100{\\left(1+0.5\\right)}^{2}=225[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\">[latex]3[\/latex]<\/td>\n<td class=\"border\" style=\"height: 12px; width: 208.5px; text-align: center;\">[latex]100+50\\left(3\\right)=250[\/latex]<\/td>\n<td class=\"border\" style=\"width: 155.5px;\">Company A grows by 50 stores and company B by 112.5 stores.<\/td>\n<td class=\"border\" style=\"height: 12px; width: 255.5px; text-align: center;\">[latex]100{\\left(1+0.5\\right)}^{3}=337.5[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 26px;\">\n<td class=\"border\" style=\"height: 26px; width: 86.5px; text-align: center;\">[latex]x[\/latex]<\/td>\n<td class=\"border\" style=\"height: 26px; width: 208.5px; text-align: center;\">[latex]A\\left(x\\right)=100+50x[\/latex]<\/td>\n<td class=\"border\" style=\"width: 155.5px;\"><\/td>\n<td class=\"border\" style=\"height: 26px; width: 255.5px; text-align: center;\">[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 2<\/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\">\n<div style=\"width: 237px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210233\/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=\"227\" height=\"394\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.\u00a0<\/strong><\/p>\n<\/div>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<p id=\"fs-id1165135209682\">Notice that the contextual 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,<\/p>\n<p style=\"text-align: center;\">[latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}.[\/latex]<\/p>\n<p>In this exponential function, 100 represents the initial number of stores, 0.50 represents the <strong>growth rate<\/strong>, [latex]r,[\/latex] and [latex]b=1+0.5=1.5[\/latex] represents the <strong>growth factor<\/strong>. Generalizing further, we can write this function as<\/p>\n<p style=\"text-align: center;\">[latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}.[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div id=\"Example_04_01_03\" class=\"textbox examples\">\n<div id=\"fs-id1165137535640\">\n<div id=\"fs-id1165137535642\">\n<h3>Example 3:\u00a0 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 1.37 billion in the year 2019, with an annual growth rate of about 1.08%.\u00a0 Let [latex]t[\/latex] be the number of years since 2019. Find an exponential function to model this growth.\u00a0 Then, to the nearest thousandth, determine what will the population of India be in 2031.<\/p>\n<\/div>\n<div id=\"fs-id1165137786632\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137786632\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137786632\" class=\"hidden-answer\" style=\"display: none\">The growth rate is given as 1.08% so [latex]r=0.0108.[\/latex]\u00a0 Since the growth factor is [latex]b=1+r[\/latex], we have that [latex]b=1+0.0108=1.0108.[\/latex]\u00a0 There are approximately 1.37 billion people at our initial time [latex]t=0[\/latex] which represents 2019 so [latex]a=1.37.[\/latex]\u00a0 Note that we do not add a bunch of zeros to the equation but rather keep the billions unit in mind when we interpret the function&#8217;s output.\u00a0 Finally our function is [latex]P\\left(t\\right)=1.37{\\left(1.0108\\right)}^{t}.[\/latex]<\/p>\n<div><\/div>\n<div>\n<p>To\u00a0estimate the population in 2031, we evaluate the model for [latex]t=12[\/latex] because 2031 is 12 years after 2019. Rounding to the nearest thousandth,<\/p>\n<div id=\"eip-id1165135657117\" class=\"unnumbered\" style=\"text-align: center;\">[latex]P\\left(12\\right)=1.37{\\left(1.0108\\right)}^{12}\\approx 1.558[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165135394343\">There will be about 1.558 billion people in India in the year 2031.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 4:\u00a0 Real World Exponential Models<\/h3>\n<p>Bismuth-210 is an isotope that radioactively decays by about 13% each day, meaning 13% of the remaining Bismuth-210 transforms into another atom (polonium-210 in this case) each day. If you begin with 100 mg of Bismuth-210, how much remains after one week?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q685223\">Show Solution<\/span><\/p>\n<div id=\"q685223\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">With radioactive decay, instead of the quantity increasing at a percent rate, the quantity is decreasing at a percent rate. Our initial quantity is [latex]a=100[\/latex] mg, and our growth rate will be negative 13%, since we are decreasing: [latex]r=-0.13.[\/latex] This gives the equation:<\/p>\n<p style=\"text-align: center;\">[latex]Q\\left(d\\right)=100\\left(1\u2212 0.13\\right)^{d}=100\\left(0.87\\right)^{d}[\/latex]<\/p>\n<p style=\"text-align: left;\">This can also be explained by recognizing that if 13% decays, then 87% remains.<\/p>\n<p>Next, we are asked to find how much remains after one week so we evaluate the function when [latex]d=7.[\/latex] Therefore,\u00a0[latex]Q\\left(7\\right)=100\\left(0.87\\right)^{7}=37.73.[\/latex]\u00a0 After one week, 37.73 mg of Bismuth-210 remains.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135536569\" class=\"precalculus tryit\">\n<h3>Try it #3<\/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 0.6%. Find an exponential function that models this situation where [latex]t[\/latex] represents the number of years since 2013.\u00a0 To the nearest thousandth, what will the population of China be for the year 2031?<\/p>\n<\/div>\n<div id=\"fs-id1165134200184\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134200184\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134200184\" class=\"hidden-answer\" style=\"display: none\">\nThis 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 2013.<\/p>\n<p id=\"fs-id1165134200186\">China&#8217;s population will be about 1.548 billion people in 2031.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135526980\" class=\"bc-section section\">\n<div class=\"textbox tryit\">\n<h3>Try It #4<\/h3>\n<p>A population of 1000 animals is decreasing 3% each year. Find the population in 30 years.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q972350\">Show Solution<\/span><\/p>\n<div id=\"q972350\" class=\"hidden-answer\" style=\"display: none\">\n<p>The function modeling this situation is [latex]P\\left(t\\right)=1000(0.97)^{t}[\/latex] and in 30 years the population will be 401 animals.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Finding Equations of Exponential Functions From Data<\/h3>\n<p id=\"fs-id1165135526985\">In the previous examples, we were given an exponential growth or decay rate, which we used to find a formula to model the situation and then evaluated for a given input. Sometimes we are given data points for an exponential function and we must use them to first find the growth factor and then the formula for the function before we can answer questions about the situation.<\/p>\n<div id=\"fs-id1165135369632\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165135180102\"><strong>Given two data points with the initial value known, write an exponential model.<\/strong><\/p>\n<ol id=\"fs-id1165135180107\" type=\"1\">\n<li>Identify the initial value from the data point of the form [latex]\\left(0,a\\right).[\/latex] Then [latex]a[\/latex] is the initial value.<\/li>\n<li>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]\u00a0 To do this, divide both sides by [latex]a[\/latex] and then raise both sides to the appropriate power to solve the equation.<\/li>\n<li>Write the equation using the values you found for [latex]a[\/latex] and [latex]b[\/latex] in the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_01_04\" class=\"textbox examples\">\n<div id=\"fs-id1165137580876\">\n<div id=\"fs-id1165137580878\">\n<h3>Example 5:\u00a0 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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135188416\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135188416\" class=\"hidden-answer\" style=\"display: none\">\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\" style=\"text-align: center;\">[latex]\\begin{align*}N\\left(t\\right)&=80{b}^{t}&&\\hfill \\\\180\\hfill &=80{b}^{6}&& \\text{Substitute using point }\\left(6, 180\\right).\\hfill \\\\\\frac{9}{4}&={b}^{6}&& \\text{Divide and write in lowest terms}.\\hfill \\\\b&={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}}&& \\text{Isolate }b\\text{ using properties of exponents}.\\hfill \\\\b&\\approx 1.1447&& \\text{Round to 4 decimal places}. \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1165135193230\"><strong>NOTE:\u00a0<\/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 id=\"fs-id1165137724117\">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 3<\/a>\u00a0passes 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<div id=\"CNX_Precalc_Figure_04_01_002\" class=\"small\">\n<div style=\"width: 254px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210236\/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=\"244\" height=\"350\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 3.\u00a0<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137761908\" class=\"precalculus tryit\">\n<h3>Try it #5<\/h3>\n<div id=\"ti_04_01_04\">\n<div id=\"fs-id1165135496544\">\n<p id=\"fs-id1165135496547\">A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013, the population had reached 236 wolves. Let [latex]t=0[\/latex] represent the year 2011.\u00a0 \u00a0What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population, [latex]N\\text{, }[\/latex]of wolves over time [latex]t.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q189152\">Show Solution<\/span><\/p>\n<div id=\"q189152\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left(0,129\\right)[\/latex] and [latex]\\left(2,236\\right);\\text{ }\\text{ }\\text{ }N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165135180102\"><strong>Given two data points, write an exponential model with the initial value unknown.<\/strong><\/p>\n<p><strong><em>Method 1: Substitution<\/em><\/strong><\/p>\n<ol id=\"fs-id1165135180107\" type=\"1\">\n<li>Substitute both points into two equations with the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}.[\/latex]<\/li>\n<li>Solve the first equation for [latex]a[\/latex] in terms of [latex]b.[\/latex]<\/li>\n<li>Substitute this expression for [latex]a[\/latex] into the second equation and then solve for [latex]b.[\/latex]<\/li>\n<li>Once you have found [latex]b,[\/latex] substitute\u00a0a data point and [latex]b[\/latex] into the general equation and solve for [latex]a[\/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<p><em><strong>Method 2: Ratios<\/strong><\/em><\/p>\n<ol id=\"fs-id1165135180107\" type=\"1\">\n<li>Choose two points to work with:\u00a0 [latex]\\left(c, f\\left(c\\right)\\right)[\/latex] and [latex]\\left(d, f\\left(d\\right)\\right).[\/latex] Order your points so [latex]c < d.[\/latex]<\/li>\n<li>Set up your ratio: [latex]\\frac{f\\left(d\\right)}{f\\left(c\\right)}=\\frac{ab^d}{ab^c}.[\/latex] Note that the values on the left are gotten from your output values of the data points and the values on the right come from the general equation with [latex]c[\/latex] and [latex]d[\/latex] filled in.<\/li>\n<li>Simplify. Note that [latex]\\frac{a}{a}=1[\/latex] so essentially they always cancel and you will always find [latex]b[\/latex] first.\u00a0 Remember that [latex]\\frac{b^d}{b^c}=b^{d-c}.[\/latex] You will then need to find the (d-c)<sup>th<\/sup> root of both sides.<\/li>\n<li>Once you have found [latex]b,[\/latex] substitute\u00a0a data point and [latex]b[\/latex] into the general equation and solve for [latex]a[\/latex].<\/li>\n<li>Write your equation using the values of [latex]a[\/latex] and [latex]b[\/latex]\u00a0in the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_01_05\" class=\"textbox examples\">\n<div id=\"fs-id1165135411399\">\n<div id=\"fs-id1165135411402\">\n<h3>Example 6:\u00a0 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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135646190\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135646190\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135646192\"><em><strong>Method 1:<\/strong> <\/em>Because we don\u2019t have the initial value, we substitute both points into the 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]\u00a0 \u00a0(Equation 1)<\/li>\n<li>Substituting [latex]\\left(2,1\\right)[\/latex] gives [latex]1=a{b}^{2}[\/latex]\u00a0 \u00a0(Equation 2)<\/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 style=\"text-align: center;\">[latex]\\begin{align*}6&=ab^{-2}\\\\\\frac{6}{b^{-2}}&=a&&\\text{Divide by }b^{-2}.\\\\a&=6b^2&&\\text{Use properties of exponents to rewrite the denominator.}\\end{align*}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165135180085\">Substitute [latex]a[\/latex] in the second equation, and solve for [latex]b:[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align*}1&=ab^2\\\\1&=\\left(6b^2\\right)b^2=6b^4&&\\text{Substitute }a.\\\\b&=\\left(\\frac{1}{6}\\right)^{\\frac{1}{4}}&&\\text{Use properties of exponents to isolate }b.\\\\b&\\approx0.6389&&\\text{Round 4 decimal places.}\\end{align*}[\/latex]<\/p>\n<p>&nbsp;<\/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]<\/p>\n<div style=\"text-align: center;\">[latex]a=6b^2\\approx6\\left(0.6389\\right)^2\\approx2.4492[\/latex]<\/div>\n<p>&nbsp;<\/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><em><strong>Method 2:<\/strong><\/em>\u00a0 The two points\u00a0[latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(2,1\\right)[\/latex] were given so we start with step 2; setting up the ratio.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{6}=\\frac{ab^2}{ab^{-2}}.[\/latex][latex]\\\\[\/latex]<\/p>\n<p>Next we simplify using the properties of exponents.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align*}\\frac{1}{6}&=b^{2-(-2)}\\\\\\frac{1}{6}&=b^{4}\\\\\\left(\\frac{1}{6}\\right)^{\\frac{1}{4}}&=\\left(b^4\\right)^{\\frac{1}{4}}\\\\b&\\approx0.6389\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\n<p>Then we find [latex]a[\/latex] by substituting in a point [latex]\\left(-2, 6\\right)[\/latex] and the known value of [latex]b\\approx0.6389[\/latex] into [latex]f\\left(x\\right)=ab^x.[\/latex]\u00a0 We get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align*}6&=a\\left(0.6389\\right)^{-2}\\\\6&=2.44982027a\\\\a&\\approx2.4492.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\n<p>Finally, the equation is [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}.[\/latex]<\/p>\n<p id=\"fs-id1165135482016\"><em><strong>Check:<\/strong><\/em> 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 4<\/a>\u00a0passes 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 <strong><span class=\"no-emphasis\">exponential decay<\/span><\/strong> function.<\/p>\n<div id=\"CNX_Precalc_Figure_04_01_003\" class=\"small\">\n<div style=\"width: 301px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210248\/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=\"291\" height=\"266\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 4.\u00a0<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"ti_11_04_01\" class=\"precalculus tryit\">\n<h3>Try it #6<\/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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137732255\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137732255\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137732257\">[latex]f\\left(x\\right)=2{\\left(1.5\\right)}^{x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137737999\" class=\"precalculus qa key-takeaways\">\n<h3>Q&amp;A<\/h3>\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 examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137771578\"><strong>Given a graph or a table 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 or table. Choose the vertical 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>Use one of the methods for finding an equation given two points above.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_01_06\" class=\"textbox examples\">\n<div id=\"fs-id1165137806415\">\n<div id=\"fs-id1165137806417\">\n<h3>\u00a0Example 7:\u00a0 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 5<\/a>.<\/p>\n<div id=\"CNX_Precalc_Figure_04_01_004\" class=\"medium\">\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210251\/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\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 5.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135415647\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135415647\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135415647\" class=\"hidden-answer\" style=\"display: none\">\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\" style=\"text-align: center;\">[latex]\\begin{align*}f\\left(x\\right)&=a{b}^{x}&& \\text{Write the general form of an exponential equation}.\\hfill \\\\ f\\left(x\\right)&=3{b}^{x}&& \\text{Substitute the initial value 3 for }a.\\hfill \\\\ 12&=3{b}^{2}&& \\text{Substitute in 12 for }y\\text{ and 2 for }x.\\hfill \\\\4&={b}^{2}&& \\text{Divide by 3}.\\hfill \\\\ b&=\u00b12&& \\text{Take the square root}.\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/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]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135169326\" class=\"precalculus tryit\">\n<h3>Try it #7<\/h3>\n<div id=\"ti_04_01_06\">\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 6<\/a>.<\/p>\n<div id=\"CNX_Precalc_Figure_04_01_005\" class=\"small\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/02\/08210254\/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\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 6.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191210\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135191210\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135191210\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135191213\">[latex]f\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2}2]{2}{\\left(\\sqrt[\\leftroot{1}\\uproot{2}2]{2}\\right)}^{x}.[\/latex] Answers may vary due to round-off error. The answer should be very close to [latex]f\\left(x\\right)=1.4142{\\left(1.4142\\right)}^{x}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137871708\" class=\"bc-section section\">\n<div class=\"textbox examples\">\n<h3>Example 8: Writing an Exponential Function Given a Table of Values<\/h3>\n<p>Find an equation for the exponential function given by Table 4.<\/p>\n<table style=\"border-collapse: collapse;\">\n<caption>Table 4<\/caption>\n<tbody>\n<tr style=\"height: 11px;\">\n<td class=\"border\" style=\"width: 117.656px; height: 11px; text-align: center;\">[latex]x[\/latex]<\/td>\n<td class=\"border\" style=\"width: 100.656px; height: 11px; text-align: center;\">1<\/td>\n<td class=\"border\" style=\"width: 100.656px; height: 11px; text-align: center;\">3<\/td>\n<td class=\"border\" style=\"width: 100.656px; height: 11px; text-align: center;\">6<\/td>\n<td class=\"border\" style=\"width: 101.656px; height: 11px; text-align: center;\">8<\/td>\n<td class=\"border\" style=\"width: 101.656px; height: 11px; text-align: center;\">12<\/td>\n<\/tr>\n<tr style=\"height: 23px;\">\n<td class=\"border\" style=\"width: 117.656px; height: 23px; text-align: center;\">[latex]f\\left(x\\right)[\/latex]<\/td>\n<td class=\"border\" style=\"width: 100.656px; height: 23px; text-align: center;\">4.8<\/td>\n<td class=\"border\" style=\"width: 100.656px; height: 23px; text-align: center;\">6.912<\/td>\n<td class=\"border\" style=\"width: 100.656px; height: 23px; text-align: center;\">11.943936<\/td>\n<td class=\"border\" style=\"width: 101.656px; height: 23px; text-align: center;\">17.19926784<\/td>\n<td class=\"border\" style=\"width: 101.656px; height: 23px; text-align: center;\">35.664401793<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q841908\">Show Solution<\/span><\/p>\n<div id=\"q841908\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we choose any two points to use in finding our equation.\u00a0 Let&#8217;s use (1, 4.8) and (3, 6.912).<\/p>\n<p>For this example, we will use ratios to find the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{6.912}{4.8}=\\frac{ab^3}{ab^1}[\/latex]<\/p>\n<p>Next, we simplify and solve.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align*}1.44&=b^{3-1}\\\\1.44&=b^2\\\\1.44^{\\frac{1}{2}}&=\\left(b^2\\right)^{\\frac{1}{2}}\\\\b&=1.2\\end{align*}[\/latex]<\/p>\n<p>Finally, we find [latex]a[\/latex] by plugging in 1.2 for [latex]b[\/latex] and 1 for [latex]x[\/latex] and 4.8 for [latex]f\\left(x\\right)[\/latex] in [latex]f\\left(x\\right)=ab^x[\/latex] to get<\/p>\n<p style=\"text-align: center;\">[latex]4.8=a\\left(1.2\\right)^1[\/latex] or [latex]a=4.[\/latex]<\/p>\n<p>The final equation is [latex]f\\left(x\\right)=4\\left(1.2\\right)^x.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\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 <strong>compound interest<\/strong>. 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 <strong>annual percentage rate (APR)<\/strong> of an account, also called the <strong>nominal rate<\/strong>, 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\" style=\"text-align: center;\">[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\">Table 5<\/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 id=\"Table_04_01_03\" 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<caption>Table 5<\/caption>\n<thead>\n<tr>\n<th class=\"border\">Frequency<\/th>\n<th class=\"border\">Value after 1 year<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td class=\"border\">Annually<\/td>\n<td class=\"border\">$1100<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">Semiannually<\/td>\n<td class=\"border\">$1102.50<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">Quarterly<\/td>\n<td class=\"border\">$1103.81<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">Monthly<\/td>\n<td class=\"border\">$1104.71<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">Daily<\/td>\n<td class=\"border\">$1105.16<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1165137793679\">\n<div class=\"textbox shaded\">\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\" style=\"text-align: center;\">[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>\n<div id=\"Example_04_01_08\" class=\"textbox examples\">\n<div id=\"fs-id1165137812820\">\n<h3>Example 9: 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q318094\">Show Solution<\/span><\/p>\n<div id=\"q318094\" class=\"hidden-answer\" style=\"display: none\">\n<p>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=10.[\/latex]<\/p>\n<div id=\"eip-id1402796\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}A\\left(t\\right)&=P{\\left(1+\\frac{r}{n}\\right)}^{nt}&&\\text{Use the compound interest formula}.\\\\A\\left(10\\right)&=3000{\\left(1+\\frac{0.03}{4}\\right)}^{4\\cdot10}&&\\text{Substitute using given values}.\\\\&\\approx4045.05.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<div class=\"unnumbered\" style=\"text-align: center;\"><\/div>\n<p id=\"fs-id1165137694040\">The account will be worth about $4,045.05 in 10 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137694046\" class=\"precalculus tryit\">\n<h3>Try it #8<\/h3>\n<div id=\"ti_04_01_08\">\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<\/div>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q89519\">Show Solution<\/span><\/p>\n<div id=\"q89519\" class=\"hidden-answer\" style=\"display: none\">\n<p>about $3,644,675.88<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_01_09\" class=\"textbox examples\">\n<div id=\"fs-id1165135180446\">\n<div>\n<h3>Example 10:\u00a0 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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135175338\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135175338\" class=\"hidden-answer\" style=\"display: none\">\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]n=2.[\/latex]<\/p>\n<p id=\"fs-id1165135209414\">We want to find the initial investment, [latex]P,[\/latex] needed so that the value of the account will be worth $40,000 in 18 years. Substitute the given values into the compound interest formula, and solve for [latex]P.[\/latex]<\/p>\n<div id=\"eip-id1165131884554\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}A\\left(t\\right)&=P{\\left(1+\\frac{r}{n}\\right)}^{nt}&&\\text{Use the compound interest formula}.\\\\40,000&=P{\\left(1+\\frac{0.06}{2}\\right)}^{2\\left(18\\right)}&&\\text{Substitute using given values }A\\text{, }r, n\\text{, and }t.\\\\40,000&=P{\\left(1.03\\right)}^{36}&&\\text{Simplify}.\\\\\\frac{40,000}{{\\left(1.03\\right)}^{36}}&=P&&\\text{Isolate }P.\\\\P&\\approx13,801&&\\text{Divide and round to the nearest dollar}.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1165137937589\">Lily will need to invest $13,801 to have $40,000 in 18 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"precalculus tryit\">\n<h3>Try it #9<\/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\">Example 10<\/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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135176736\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135176736\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135176738\">$13,693<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137724961\" class=\"bc-section section\">\n<h3>Evaluating Functions with Base <em><span class=\"e2\">e<\/span><\/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\">Table 6<\/a>\u00a0shows 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\">Table 6<\/a>.<\/p>\n<table id=\"Table_04_01_04\" 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 2, and Semiannually has the input value of (1+1\/2)^2 which equals to 2.25, Quarterly has the input value of (1+1\/4)^4 which equals to 2.441406, Monthly has the input value of (1+1\/12)^12 which equals to 2.613035, Daily has the input value of (1+1\/365)^365 which equals to 2.714567, Hourly has the input value of (1+1\/8766)^8766 which equals to 2.718127, One per minute has the input value of (1+1\/525960)^525960 which equals to 2.718279, and Once per second has the input value of (1+1\/31557600)^31557600 which equals to 2.718282.\">\n<caption>Table 6<\/caption>\n<thead>\n<tr>\n<th class=\"border\">Frequency<\/th>\n<th class=\"border\">[latex]A\\left(n\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\n<th class=\"border\">Value<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td class=\"border\">Annually<\/td>\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\n<td class=\"border\">$2<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">Semiannually<\/td>\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\n<td class=\"border\">$2.25<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">Quarterly<\/td>\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\n<td class=\"border\">$2.441406<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">Monthly<\/td>\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\n<td class=\"border\">$2.613035<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">Daily<\/td>\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\n<td class=\"border\">$2.714567<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">Hourly<\/td>\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{\\text{8760}}\\right)}^{\\text{8760}}[\/latex]<\/td>\n<td class=\"border\">$2.718127<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">Once per minute<\/td>\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{\\text{525600}}\\right)}^{\\text{525600}}[\/latex]<\/td>\n<td class=\"border\">$2.718279<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">Once per second<\/td>\n<td class=\"border\">[latex]{\\left(1+\\frac{1}{31536000}\\right)}^{31536000}[\/latex]<\/td>\n<td class=\"border\">$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\">\n<div class=\"textbox shaded\">\n<h3>Euler&#8217;s Number<\/h3>\n<p id=\"fs-id1165135511335\">The letter <em><span class=\"e2\">e<\/span><\/em> represents the irrational number [latex]{\\left(1+\\frac{1}{n}\\right)}^{n},[\/latex] as [latex]n[\/latex] increases without bound.<\/p>\n<p>&nbsp;<\/p>\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><span class=\"e2\">e<\/span><\/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>\n<div id=\"Example_04_01_10\" class=\"textbox examples\">\n<div id=\"fs-id1165135344893\">\n<div id=\"fs-id1165135344895\">\n<h3>Example 11:\u00a0 Using a Calculator to Find Powers of <em><span class=\"e2\">e<\/span><\/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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135192743\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135192743\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135192745\">On a calculator, press the button labeled [latex]\\left[{e}^{x}\\right].[\/latex] The window shows [latex]\\left[e^{\\wedge}\\text{ ( }\\right].[\/latex]Type [latex]3.14[\/latex] and then close parenthesis, [latex]\\left[\\text{ ) }\\right].[\/latex] Press [ENTER]. Rounding to\u00a0 5 decimal places, [latex]{e}^{3.14}\\approx 23.10387.[\/latex]<\/p>\n<p>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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137447188\" class=\"precalculus tryit\">\n<h3>Try it #10<\/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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135176286\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135176286\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135176288\">[latex]{e}^{-0.5}\\approx 0.60653[\/latex]<\/p>\n<\/div>\n<\/div>\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 many 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<strong> <em>continuous growth or decay models<\/em><\/strong>. 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\">\n<div class=\"textbox shaded\">\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]k,[\/latex] continuous growth or decay is represented by the formula<\/p>\n<div id=\"fs-id1165135536370\" style=\"text-align: center;\">[latex]A\\left(t\\right)=a{e}^{kt}[\/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]k[\/latex] is the <strong>continuous rate<\/strong> per unit time, and<\/li>\n<li>[latex]t[\/latex] is the elapsed time.<\/li>\n<\/ul>\n<p id=\"fs-id1165135560686\">If [latex]k>0[\/latex], then the formula represents continuous growth. If [latex]k<0[\/latex], then the formula represents continuous decay.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135411368\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165135411373\"><strong>Given the initial value, continuous rate of growth or decay, and time, 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]k.[\/latex]\n<ul id=\"fs-id1165135188096\" type=\"a\">\n<li>If the problem refers to continuous growth, then [latex]k>0.[\/latex]<\/li>\n<li>If the problem refers to continuous decay, then [latex]k<0.[\/latex]<\/li>\n<\/ul>\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 simplify.<\/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>Example 12: 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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137694203\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137694203\" class=\"hidden-answer\" style=\"display: none\">The rate given is continuous so we use the formula [latex]A\\left(t\\right)=a{e}^{kt}.[\/latex] Since the account is growing in value, this is a continuous compounding problem with growth rate [latex]k=0.10.[\/latex] The initial investment was $1,000, so [latex]a=1000.[\/latex] To find the value after [latex]t=1[\/latex] year:<\/p>\n<div id=\"eip-id1165133351794\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{lll}A\\left(t\\right)\\hfill & =a{e}^{kt}\\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 }a, k,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 1105.17\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137895288\">The account is worth $1,105.17 after one year.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137895295\" class=\"precalculus tryit\">\n<h3>Try it #11<\/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<p id=\"fs-id1165134389978\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q105348\">Show Solution<\/span><\/p>\n<div id=\"q105348\" class=\"hidden-answer\" style=\"display: none\">\n<p>$3,659,823.44<\/p><\/div>\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>Example 13: 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\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137803706\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137803706\" class=\"hidden-answer\" style=\"display: none\">Since the substance is decaying, the rate, 17.3%, is negative. So, [latex]k=\\text{ }-0.173.[\/latex] The initial amount of radon-222 was [latex]100\\text{ }[\/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\" style=\"text-align: center;\">[latex]\\begin{array}{lll}A\\left(t\\right)\\hfill & =a{e}^{kt}\\hfill & \\text{Use the continuous decay formula}.\\hfill \\\\ \\hfill & =100{e}^{-0.173\\left(3\\right)}\\begin{array}{cccc}& & & \\end{array}\\hfill & \\text{Substitute known values for }a, k,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 59.5115\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165137697132\">So 59.5115 mg of radon-222 will remain after 3 days.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135348462\" class=\"precalculus tryit\">\n<h3>Try it #12<\/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\">Example 13<\/a>, how much radon-222 will remain after one year?<\/p>\n<\/div>\n<div id=\"fs-id1165135245732\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135245732\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135245732\" class=\"hidden-answer\" style=\"display: none\">\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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135191186\" class=\"precalculus media\">\n<div class=\"textbox shaded\">\n<div id=\"fs-id1165137827923\" class=\"bc-section section\">\n<div id=\"fs-id1165135191186\" class=\"precalculus media\">\n<h3>Media<\/h3>\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:\/\/openstax.org\/l\/expgrowth\">Exponential Growth Function<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/compoundint\">Compound Interest<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165135191191\"><span style=\"color: #6c64ad; font-size: 1em; font-weight: 600;\">Key Equations<\/span><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135264762\" class=\"key-equations\">\n<table id=\"fs-id2306479\" summary=\"...\">\n<tbody>\n<tr>\n<td class=\"border\">definition of exponential growth or decay<\/td>\n<td class=\"border\">[latex]f\\left(x\\right)=a{b}^{x},\\text{ where }a>0,b>0,b\\ne 1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">compound interest formula<\/td>\n<td class=\"border\">[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>\n<td class=\"border\">continuous growth formula<\/td>\n<td class=\"border\">[latex]A\\left(t\\right)=a{e}^{kt},\\text{ where}[\/latex]<\/p>\n<div><\/div>\n<p>[latex]t[\/latex] is the number of unit time periods of growth<\/p>\n<div><\/div>\n<p>[latex]a[\/latex] is the starting amount (in the continuous compounding formula [latex]a[\/latex] may be replaced with P, the principal)<\/p>\n<div><\/div>\n<p>[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.<\/li>\n<li>A function is evaluated by solving at a specific value.<\/li>\n<li>An exponential model can be found when the growth rate and initial value are known.<\/li>\n<li>An exponential model can be found when the two data points from the model are known.<\/li>\n<li>An exponential model can be found using two data points from a graph or a table of the model.<\/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.<\/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.<\/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>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 continuous growth or decay rate are known.<\/li>\n<\/ul>\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>\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<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-238\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li> Exponential Functions. <strong>Authored by<\/strong>: Douglas Hoffman. <strong>Provided by<\/strong>: Openstax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:aU0u6kE1@14\/Exponential-Functions\">https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:aU0u6kE1@14\/Exponential-Functions<\/a>. <strong>Project<\/strong>: Essential Precalcus, Part 1. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-238-1\">http:\/\/www.worldometers.info\/world-population\/. Accessed April 22, 2019. <a href=\"#return-footnote-238-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":311,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\" Exponential Functions\",\"author\":\"Douglas Hoffman\",\"organization\":\"Openstax\",\"url\":\"https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:aU0u6kE1@14\/Exponential-Functions\",\"project\":\"Essential Precalcus, Part 1\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-238","chapter","type-chapter","status-publish","hentry"],"part":223,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/238","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":34,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/238\/revisions"}],"predecessor-version":[{"id":3272,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/238\/revisions\/3272"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/parts\/223"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/238\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/media?parent=238"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=238"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/contributor?post=238"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/license?post=238"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}