{"id":15985,"date":"2019-09-26T16:05:35","date_gmt":"2019-09-26T16:05:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/introduction-exponential-functions\/"},"modified":"2024-05-02T16:00:23","modified_gmt":"2024-05-02T16:00:23","slug":"introduction-exponential-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/introduction-exponential-functions\/","title":{"raw":"Evaluating Exponential Functions","rendered":"Evaluating Exponential Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify and evaluate exponential functions<\/li>\r\n \t<li>Use the compound interest formula<\/li>\r\n<\/ul>\r\n<\/div>\r\nLinear functions have a\u00a0constant rate of change \u2013 a constant number that the output increases for each increase in input. For example, in the equation [latex]f(x)=3x+4[\/latex]\u00a0, the slope tells us the output increases by three each time the input increases by one. Sometimes, on the other hand, quantities grow by a percent rate of change rather than by a fixed amount. In this lesson, we will define a function whose rate of change increases by a percent of the current value rather than a fixed quantity.\r\n\r\nTo illustrate\u00a0this difference consider two companies whose business is expanding: Company A has\u00a0[latex]100[\/latex] stores and expands by opening\u00a0[latex]50[\/latex] new stores a year, while Company B has\u00a0[latex]100[\/latex] stores and expands by increasing the number of stores by\u00a0[latex]50\\%[\/latex] of their total each year.\r\n\r\nThe table below compares\u00a0the growth of each company where company A increases the number of stores linearly, and company B increases the number of stores by a rate of\u00a0[latex]50\\%[\/latex] each year.\r\n<table style=\"width: 60%;\">\r\n<thead>\r\n<tr>\r\n<td style=\"width: 15.7534%;\">Year<\/td>\r\n<td style=\"width: 26.8836%;\">Stores, Company A<\/td>\r\n<td style=\"width: 27.0548%;\">\u00a0Description of Growth<\/td>\r\n<td style=\"width: 30.3082%;\">Stores, Company B<\/td>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 15.7534%;\">[latex]0[\/latex]<\/td>\r\n<td style=\"width: 26.8836%;\">[latex]100[\/latex]<\/td>\r\n<td style=\"width: 27.0548%;\">Starting with [latex]100[\/latex] each<\/td>\r\n<td style=\"width: 30.3082%;\">[latex]100[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.7534%;\">[latex]1[\/latex]<\/td>\r\n<td style=\"width: 26.8836%;\">[latex]100+50=150[\/latex]<\/td>\r\n<td style=\"width: 27.0548%;\">They both grow by\u00a0[latex]50[\/latex] stores in the first year.<\/td>\r\n<td style=\"width: 30.3082%;\">[latex]100[\/latex][latex]+50\\%[\/latex] of [latex]100[\/latex] [latex]100 + 0.50(100) = 150[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.7534%;\">[latex]2[\/latex]<\/td>\r\n<td style=\"width: 26.8836%;\">[latex]150+50=200[\/latex]<\/td>\r\n<td style=\"width: 27.0548%;\">Store A grows by\u00a0[latex]50[\/latex], Store B grows by\u00a0[latex]75[\/latex]<\/td>\r\n<td style=\"width: 30.3082%;\">[latex]150[\/latex][latex]+ 50\\%[\/latex] of\u00a0[latex]150[\/latex] [latex]150 + 0.50(150) = 225[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 15.7534%;\">[latex]3[\/latex]<\/td>\r\n<td style=\"width: 26.8836%;\">[latex]200+50=250[\/latex]<\/td>\r\n<td style=\"width: 27.0548%;\">Store A grows by\u00a0[latex]50[\/latex], Store B grows by\u00a0[latex]112.5[\/latex]<\/td>\r\n<td style=\"width: 30.3082%;\">[latex]225 + 50\\%[\/latex] of\u00a0[latex]225[\/latex] [latex]225 + 0.50(225) = 337.5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nCompany A has\u00a0[latex]100[\/latex] stores and expands by opening\u00a0[latex]50[\/latex] new stores a year, so its growth can be represented by the function [latex]A\\left(x\\right)=100+50x[\/latex]. Company B has\u00a0[latex]100[\/latex] stores and expands by increasing the number of stores by\u00a0[latex]50\\%[\/latex] each year, so its growth can be represented by the function [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex].\r\n\r\nThe graphs comparing the number of stores for each company over a five-year period are shown below<strong>.<\/strong> We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"338\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051913\/CNX_Precalc_Figure_04_01_0012.jpg\" alt=\"Graph of Companies A and B\u2019s functions, which values are found in the previous table.\" width=\"338\" height=\"586\" \/> The graph shows the number of stores Companies A and B opened over a five-year period.[\/caption]\r\n<p id=\"fs-id1165135209682\">Notice that the domain for both functions is [latex]\\left[0,\\infty \\right)[\/latex], and the range for both functions is [latex]\\left[100,\\infty \\right)[\/latex]. After year\u00a0[latex]1[\/latex], Company B always has more stores than Company A.<\/p>\r\n<p id=\"fs-id1165137836429\">Consider\u00a0the 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,\u00a0[latex]100[\/latex] represents the initial number of stores,\u00a0[latex]0.50[\/latex] represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex], where\u00a0[latex]100[\/latex] is the initial value,\u00a0[latex]1.5[\/latex] is called the <em>base<\/em>, and <em>x<\/em>\u00a0is called the <em>exponent<\/em>. This is an exponential function.\r\n<div id=\"fs-id1165137564690\" class=\"note textbox\">\r\n<h3 class=\"title\">Exponential Growth function<\/h3>\r\n<p id=\"fs-id1165137834019\">A function that models <strong>exponential growth<\/strong> grows by a rate proportional to the current amount. For any real number <em>x<\/em>\u00a0and any positive real numbers <em>a\u00a0<\/em>and <em>b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form<\/p>\r\n\r\n<div id=\"fs-id1165137851784\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\r\n<p id=\"eip-626\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137863819\">\r\n \t<li><em>a<\/em>\u00a0is the initial or starting value of the function.<\/li>\r\n \t<li><em>b<\/em>\u00a0is the growth factor or growth multiplier per unit <em>x<\/em>.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Evaluating Exponential Functions<\/h2>\r\n<p id=\"fs-id1165137644244\">To evaluate an exponential function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex], we simply substitute <em>x<\/em>\u00a0with 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=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill &amp; ={2}^{x}\\hfill &amp; \\hfill \\\\ f\\left(3\\right)\\hfill &amp; ={2}^{3}\\text{ }\\hfill &amp; \\text{Substitute }x=3.\\hfill \\\\ \\hfill &amp; =8\\text{ }\\hfill &amp; \\text{Evaluate the power}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\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=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill &amp; =30{\\left(2\\right)}^{x}\\hfill &amp; \\hfill \\\\ f\\left(3\\right)\\hfill &amp; =30{\\left(2\\right)}^{3}\\hfill &amp; \\text{Substitute }x=3.\\hfill \\\\ \\hfill &amp; =30\\left(8\\right)\\text{ }\\hfill &amp; \\text{Simplify the power first}\\text{.}\\hfill \\\\ \\hfill &amp; =240\\hfill &amp; \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\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=\"equation 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 class=\"equation unnumbered\" style=\"text-align: left;\"><\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\">In our first example, we will evaluate an exponential function without the aid of a calculator.<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLet [latex]f\\left(x\\right)=5{\\left(3\\right)}^{x+1}[\/latex]. Evaluate [latex]f\\left(2\\right)[\/latex] without using a calculator.\r\n[reveal-answer q=\"211228\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"211228\"]\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=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill &amp; =5{\\left(3\\right)}^{x+1}\\hfill &amp; \\hfill \\\\ f\\left(2\\right)\\hfill &amp; =5{\\left(3\\right)}^{2+1}\\hfill &amp; \\text{Substitute }x=2.\\hfill \\\\ \\hfill &amp; =5{\\left(3\\right)}^{3}\\hfill &amp; \\text{Add the exponents}.\\hfill \\\\ \\hfill &amp; =5\\left(27\\right)\\hfill &amp; \\text{Simplify the power}\\text{.}\\hfill \\\\ \\hfill &amp; =135\\hfill &amp; \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we present more examples of evaluating an exponential function at several different values.\r\n\r\nhttps:\/\/youtu.be\/QFFAoX0We34\r\n\r\nIn the next example, we will revisit the population of India.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nAt the beginning of this section, we learned that the population of India was about\u00a0[latex]1.25[\/latex] billion in the year\u00a0[latex]2013[\/latex], with an annual growth rate of about\u00a0[latex]1.2\\%[\/latex]. This situation is represented by the growth function [latex]P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t}[\/latex], where <em>t<\/em>\u00a0is the number of years since\u00a0[latex]2013[\/latex]. To the nearest thousandth, what will the population of India be in\u00a0[latex]2031[\/latex]?\r\n[reveal-answer q=\"385742\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"385742\"]\r\n<p id=\"fs-id1165137786635\">To estimate the population in\u00a0[latex]2031[\/latex], we evaluate the model for [latex]t=18[\/latex], because\u00a0[latex]2031[\/latex] is\u00a0[latex]18[\/latex] years after\u00a0[latex]2013[\/latex]. Rounding to the nearest thousandth,<\/p>\r\n\r\n<div id=\"eip-id1165135657117\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/div>\r\n<p id=\"fs-id1165135394343\">There will be about\u00a0[latex]1.549[\/latex] billion people in India in the year\u00a0[latex]2031[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of using an exponential function to predict the population of a small town.\r\n\r\nhttps:\/\/youtu.be\/SbIydBmJePE\r\n<h2>Compound Interest Formulas<\/h2>\r\nYou may have seen formulas that are used to calculate compound interest rates. \u00a0These formulas are another example of exponential growth.\u00a0The term <em>compounding<\/em> refers to interest earned not only on the original value but on the accumulated value of the account.\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\u00a0<em>nominal<\/em>\u00a0is 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 <em>t<\/em>, principal <em>P<\/em>, APR <em>r<\/em>, and number of compounding periods in a year\u00a0<em>n<\/em>:<\/p>\r\n\r\n<div id=\"eip-986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3 class=\"title\">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\" class=\"equation\" 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><em>A<\/em>(<em>t<\/em>) is the account value,<\/li>\r\n \t<li><i>t<\/i> is measured in years,<\/li>\r\n \t<li><em>P<\/em>\u00a0is the starting amount of the account, often called the principal, or more generally present value,<\/li>\r\n \t<li><em>r<\/em>\u00a0is the annual percentage rate (APR) expressed as a decimal, and<\/li>\r\n \t<li><em>n<\/em>\u00a0is the number of compounding periods in one year.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn our next example, we will calculate the value of an account after\u00a0[latex]10[\/latex] years of interest compounded quarterly.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf we invest\u00a0[latex]$3,000[\/latex] in an investment account paying\u00a0[latex]3\\%[\/latex] interest compounded quarterly, how much will the account be worth in\u00a0[latex]10[\/latex] years?\r\n[reveal-answer q=\"689928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"689928\"]\r\n<p id=\"fs-id1165137812832\">Because we are starting with\u00a0[latex]$3,000[\/latex], <em>P\u00a0<\/em>=\u00a0[latex]3000[\/latex]. Our interest rate is\u00a0[latex]3\\%[\/latex], so <em>r<\/em>\u00a0=\u00a0[latex]0.03[\/latex]. Because we are compounding quarterly, we are compounding\u00a0[latex]4[\/latex] times per year, so <em>n\u00a0<\/em>=\u00a0[latex]4[\/latex]. We want to know the value of the account in\u00a0[latex]10[\/latex] years, so we are looking for [latex]A(10)[\/latex] which is the value when <em>t <\/em>=\u00a0[latex]10[\/latex].<\/p>\r\n\r\n<div id=\"eip-id1402796\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{rll}A\\left(t\\right)\\hfill &amp; =P\\left(1+\\frac{r}{n}\\right)^{nt}\\hfill &amp; \\text{Use the compound interest formula}. \\\\ A\\left(10\\right)\\hfill &amp; =3000\\left(1+\\frac{0.03}{4}\\right)^{4\\cdot 10}\\hfill &amp; \\text{Substitute using given values}. \\\\ \\text{ }\\hfill &amp; \\approx 4045.05\\hfill &amp; \\text{Round to two decimal places}.\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137694040\">The account will be worth about\u00a0[latex]$4,045.05[\/latex] in\u00a0[latex]10[\/latex] years.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows an example of using exponential growth to calculate interest compounded quarterly.\r\n\r\nhttps:\/\/youtu.be\/3az4AKvUmmI\r\n\r\nIn our next example, we will use the compound interest formula to solve for the principal.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA\u00a0[latex]529[\/latex] 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\u00a0[latex]529[\/latex] account for her new granddaughter and wants the account to grow to\u00a0[latex]$40,000[\/latex] over\u00a0[latex]18[\/latex] years. She believes the account will earn\u00a0[latex]6\\%[\/latex] compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now?\r\n[reveal-answer q=\"254680\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"254680\"]\r\n<p id=\"fs-id1165137664627\">The nominal interest rate is\u00a0[latex]6\\%[\/latex], so <em>r\u00a0<\/em>=\u00a0[latex]0.06[\/latex]. Interest is compounded twice a year, so\u00a0\u00a0[latex]k=2[\/latex].<\/p>\r\n<p id=\"fs-id1165135209414\">We want to find the initial investment, <em>P<\/em>, needed so that the value of the account will be worth\u00a0[latex]$40,000[\/latex] in\u00a0[latex]18[\/latex] years. Substitute the given values into the compound interest formula, and solve for <em>P<\/em>.<\/p>\r\n\r\n<div id=\"eip-id1165131884554\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill &amp; =P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\hfill &amp; \\text{Use the compound interest formula}.\\hfill \\\\ 40,000\\hfill &amp; =P{\\left(1+\\frac{0.06}{2}\\right)}^{2\\left(18\\right)}\\hfill &amp; \\text{Substitute using given values }A\\text{, }r, n\\text{, and }t.\\hfill \\\\ 40,000\\hfill &amp; =P{\\left(1.03\\right)}^{36}\\hfill &amp; \\text{Simplify}.\\hfill \\\\ \\frac{40,000}{{\\left(1.03\\right)}^{36}}\\hfill &amp; =P\\hfill &amp; \\text{Isolate }P.\\hfill \\\\ P\\hfill &amp; \\approx 13,801\\hfill &amp; \\text{Divide and round to the nearest dollar}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137937589\">Lily will need to invest\u00a0[latex]$13,801[\/latex] to have\u00a0[latex]$40,000[\/latex] in\u00a0[latex]18[\/latex] years.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of finding the deposit amount necessary to obtain a future value from compounded interest.\r\n\r\nhttps:\/\/youtu.be\/saq9dF7a4r8\r\n<h2>Summary<\/h2>\r\nExponential growth grows by a rate proportional to the current amount.\u00a0For any real number <em>x<\/em>\u00a0and any positive real numbers <em>a\u00a0<\/em>and <em>b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form\u00a0[latex]f\\left(x\\right)=a{b}^{x}[\/latex]. \u00a0Evaluating exponential functions requires careful attention to the order of operations. Compound interest is an example of exponential growth.\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify and evaluate exponential functions<\/li>\n<li>Use the compound interest formula<\/li>\n<\/ul>\n<\/div>\n<p>Linear functions have a\u00a0constant rate of change \u2013 a constant number that the output increases for each increase in input. For example, in the equation [latex]f(x)=3x+4[\/latex]\u00a0, the slope tells us the output increases by three each time the input increases by one. Sometimes, on the other hand, quantities grow by a percent rate of change rather than by a fixed amount. In this lesson, we will define a function whose rate of change increases by a percent of the current value rather than a fixed quantity.<\/p>\n<p>To illustrate\u00a0this difference consider two companies whose business is expanding: Company A has\u00a0[latex]100[\/latex] stores and expands by opening\u00a0[latex]50[\/latex] new stores a year, while Company B has\u00a0[latex]100[\/latex] stores and expands by increasing the number of stores by\u00a0[latex]50\\%[\/latex] of their total each year.<\/p>\n<p>The table below compares\u00a0the growth of each company where company A increases the number of stores linearly, and company B increases the number of stores by a rate of\u00a0[latex]50\\%[\/latex] each year.<\/p>\n<table style=\"width: 60%;\">\n<thead>\n<tr>\n<td style=\"width: 15.7534%;\">Year<\/td>\n<td style=\"width: 26.8836%;\">Stores, Company A<\/td>\n<td style=\"width: 27.0548%;\">\u00a0Description of Growth<\/td>\n<td style=\"width: 30.3082%;\">Stores, Company B<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"width: 15.7534%;\">[latex]0[\/latex]<\/td>\n<td style=\"width: 26.8836%;\">[latex]100[\/latex]<\/td>\n<td style=\"width: 27.0548%;\">Starting with [latex]100[\/latex] each<\/td>\n<td style=\"width: 30.3082%;\">[latex]100[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.7534%;\">[latex]1[\/latex]<\/td>\n<td style=\"width: 26.8836%;\">[latex]100+50=150[\/latex]<\/td>\n<td style=\"width: 27.0548%;\">They both grow by\u00a0[latex]50[\/latex] stores in the first year.<\/td>\n<td style=\"width: 30.3082%;\">[latex]100[\/latex][latex]+50\\%[\/latex] of [latex]100[\/latex] [latex]100 + 0.50(100) = 150[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.7534%;\">[latex]2[\/latex]<\/td>\n<td style=\"width: 26.8836%;\">[latex]150+50=200[\/latex]<\/td>\n<td style=\"width: 27.0548%;\">Store A grows by\u00a0[latex]50[\/latex], Store B grows by\u00a0[latex]75[\/latex]<\/td>\n<td style=\"width: 30.3082%;\">[latex]150[\/latex][latex]+ 50\\%[\/latex] of\u00a0[latex]150[\/latex] [latex]150 + 0.50(150) = 225[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 15.7534%;\">[latex]3[\/latex]<\/td>\n<td style=\"width: 26.8836%;\">[latex]200+50=250[\/latex]<\/td>\n<td style=\"width: 27.0548%;\">Store A grows by\u00a0[latex]50[\/latex], Store B grows by\u00a0[latex]112.5[\/latex]<\/td>\n<td style=\"width: 30.3082%;\">[latex]225 + 50\\%[\/latex] of\u00a0[latex]225[\/latex] [latex]225 + 0.50(225) = 337.5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Company A has\u00a0[latex]100[\/latex] stores and expands by opening\u00a0[latex]50[\/latex] new stores a year, so its growth can be represented by the function [latex]A\\left(x\\right)=100+50x[\/latex]. Company B has\u00a0[latex]100[\/latex] stores and expands by increasing the number of stores by\u00a0[latex]50\\%[\/latex] 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>The graphs comparing the number of stores for each company over a five-year period are shown below<strong>.<\/strong> We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.<\/p>\n<div style=\"width: 348px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051913\/CNX_Precalc_Figure_04_01_0012.jpg\" alt=\"Graph of Companies A and B\u2019s functions, which values are found in the previous table.\" width=\"338\" height=\"586\" \/><\/p>\n<p class=\"wp-caption-text\">The graph shows the number of stores Companies A and B opened over a five-year period.<\/p>\n<\/div>\n<p id=\"fs-id1165135209682\">Notice that the domain for both functions is [latex]\\left[0,\\infty \\right)[\/latex], and the range for both functions is [latex]\\left[100,\\infty \\right)[\/latex]. After year\u00a0[latex]1[\/latex], Company B always has more stores than Company A.<\/p>\n<p id=\"fs-id1165137836429\">Consider\u00a0the 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,\u00a0[latex]100[\/latex] represents the initial number of stores,\u00a0[latex]0.50[\/latex] represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex], where\u00a0[latex]100[\/latex] is the initial value,\u00a0[latex]1.5[\/latex] is called the <em>base<\/em>, and <em>x<\/em>\u00a0is called the <em>exponent<\/em>. This is an exponential function.<\/p>\n<div id=\"fs-id1165137564690\" class=\"note textbox\">\n<h3 class=\"title\">Exponential Growth function<\/h3>\n<p id=\"fs-id1165137834019\">A function that models <strong>exponential growth<\/strong> grows by a rate proportional to the current amount. For any real number <em>x<\/em>\u00a0and any positive real numbers <em>a\u00a0<\/em>and <em>b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form<\/p>\n<div id=\"fs-id1165137851784\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\n<p id=\"eip-626\">where<\/p>\n<ul id=\"fs-id1165137863819\">\n<li><em>a<\/em>\u00a0is the initial or starting value of the function.<\/li>\n<li><em>b<\/em>\u00a0is the growth factor or growth multiplier per unit <em>x<\/em>.<\/li>\n<\/ul>\n<\/div>\n<h2>Evaluating Exponential Functions<\/h2>\n<p id=\"fs-id1165137644244\">To evaluate an exponential function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex], we simply substitute <em>x<\/em>\u00a0with 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=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill & ={2}^{x}\\hfill & \\hfill \\\\ f\\left(3\\right)\\hfill & ={2}^{3}\\text{ }\\hfill & \\text{Substitute }x=3.\\hfill \\\\ \\hfill & =8\\text{ }\\hfill & \\text{Evaluate the power}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137849020\">To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:<\/p>\n<p id=\"fs-id1165137849024\">Let [latex]f\\left(x\\right)=30{\\left(2\\right)}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\n<div id=\"eip-id1165134086025\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill & =30{\\left(2\\right)}^{x}\\hfill & \\hfill \\\\ f\\left(3\\right)\\hfill & =30{\\left(2\\right)}^{3}\\hfill & \\text{Substitute }x=3.\\hfill \\\\ \\hfill & =30\\left(8\\right)\\text{ }\\hfill & \\text{Simplify the power first}\\text{.}\\hfill \\\\ \\hfill & =240\\hfill & \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137841073\">Note that if the order of operations were not followed, the result would be incorrect:<\/p>\n<div id=\"eip-id1165135320147\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\"><\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\">In our first example, we will evaluate an exponential function without the aid of a calculator.<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q211228\">Show Solution<\/span><\/p>\n<div id=\"q211228\" 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=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill & =5{\\left(3\\right)}^{x+1}\\hfill & \\hfill \\\\ f\\left(2\\right)\\hfill & =5{\\left(3\\right)}^{2+1}\\hfill & \\text{Substitute }x=2.\\hfill \\\\ \\hfill & =5{\\left(3\\right)}^{3}\\hfill & \\text{Add the exponents}.\\hfill \\\\ \\hfill & =5\\left(27\\right)\\hfill & \\text{Simplify the power}\\text{.}\\hfill \\\\ \\hfill & =135\\hfill & \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we present more examples of evaluating an exponential function at several different values.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Determine Exponential Function Values and Graph the Function\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/QFFAoX0We34?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next example, we will revisit the population of India.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>At the beginning of this section, we learned that the population of India was about\u00a0[latex]1.25[\/latex] billion in the year\u00a0[latex]2013[\/latex], with an annual growth rate of about\u00a0[latex]1.2\\%[\/latex]. This situation is represented by the growth function [latex]P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t}[\/latex], where <em>t<\/em>\u00a0is the number of years since\u00a0[latex]2013[\/latex]. To the nearest thousandth, what will the population of India be in\u00a0[latex]2031[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q385742\">Show Solution<\/span><\/p>\n<div id=\"q385742\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137786635\">To estimate the population in\u00a0[latex]2031[\/latex], we evaluate the model for [latex]t=18[\/latex], because\u00a0[latex]2031[\/latex] is\u00a0[latex]18[\/latex] years after\u00a0[latex]2013[\/latex]. Rounding to the nearest thousandth,<\/p>\n<div id=\"eip-id1165135657117\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/div>\n<p id=\"fs-id1165135394343\">There will be about\u00a0[latex]1.549[\/latex] billion people in India in the year\u00a0[latex]2031[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show another example of using an exponential function to predict the population of a small town.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Evaluate a Given Exponential Function to Predict a Future Population\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/SbIydBmJePE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Compound Interest Formulas<\/h2>\n<p>You may have seen formulas that are used to calculate compound interest rates. \u00a0These formulas are another example of exponential growth.\u00a0The 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\u00a0<em>nominal<\/em>\u00a0is 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 <em>t<\/em>, principal <em>P<\/em>, APR <em>r<\/em>, and number of compounding periods in a year\u00a0<em>n<\/em>:<\/p>\n<div id=\"eip-986\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/div>\n<div class=\"textbox\">\n<h3 class=\"title\">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\" class=\"equation\" 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><em>A<\/em>(<em>t<\/em>) is the account value,<\/li>\n<li><i>t<\/i> is measured in years,<\/li>\n<li><em>P<\/em>\u00a0is the starting amount of the account, often called the principal, or more generally present value,<\/li>\n<li><em>r<\/em>\u00a0is the annual percentage rate (APR) expressed as a decimal, and<\/li>\n<li><em>n<\/em>\u00a0is the number of compounding periods in one year.<\/li>\n<\/ul>\n<\/div>\n<p>In our next example, we will calculate the value of an account after\u00a0[latex]10[\/latex] years of interest compounded quarterly.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If we invest\u00a0[latex]$3,000[\/latex] in an investment account paying\u00a0[latex]3\\%[\/latex] interest compounded quarterly, how much will the account be worth in\u00a0[latex]10[\/latex] years?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q689928\">Show Solution<\/span><\/p>\n<div id=\"q689928\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137812832\">Because we are starting with\u00a0[latex]$3,000[\/latex], <em>P\u00a0<\/em>=\u00a0[latex]3000[\/latex]. Our interest rate is\u00a0[latex]3\\%[\/latex], so <em>r<\/em>\u00a0=\u00a0[latex]0.03[\/latex]. Because we are compounding quarterly, we are compounding\u00a0[latex]4[\/latex] times per year, so <em>n\u00a0<\/em>=\u00a0[latex]4[\/latex]. We want to know the value of the account in\u00a0[latex]10[\/latex] years, so we are looking for [latex]A(10)[\/latex] which is the value when <em>t <\/em>=\u00a0[latex]10[\/latex].<\/p>\n<div id=\"eip-id1402796\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{rll}A\\left(t\\right)\\hfill & =P\\left(1+\\frac{r}{n}\\right)^{nt}\\hfill & \\text{Use the compound interest formula}. \\\\ A\\left(10\\right)\\hfill & =3000\\left(1+\\frac{0.03}{4}\\right)^{4\\cdot 10}\\hfill & \\text{Substitute using given values}. \\\\ \\text{ }\\hfill & \\approx 4045.05\\hfill & \\text{Round to two decimal places}.\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137694040\">The account will be worth about\u00a0[latex]$4,045.05[\/latex] in\u00a0[latex]10[\/latex] years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows an example of using exponential growth to calculate interest compounded quarterly.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 1:  Compounded Interest Formula - Quarterly\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/3az4AKvUmmI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In our next example, we will use the compound interest formula to solve for the principal.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A\u00a0[latex]529[\/latex] 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\u00a0[latex]529[\/latex] account for her new granddaughter and wants the account to grow to\u00a0[latex]$40,000[\/latex] over\u00a0[latex]18[\/latex] years. She believes the account will earn\u00a0[latex]6\\%[\/latex] compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q254680\">Show Solution<\/span><\/p>\n<div id=\"q254680\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137664627\">The nominal interest rate is\u00a0[latex]6\\%[\/latex], so <em>r\u00a0<\/em>=\u00a0[latex]0.06[\/latex]. Interest is compounded twice a year, so\u00a0\u00a0[latex]k=2[\/latex].<\/p>\n<p id=\"fs-id1165135209414\">We want to find the initial investment, <em>P<\/em>, needed so that the value of the account will be worth\u00a0[latex]$40,000[\/latex] in\u00a0[latex]18[\/latex] years. Substitute the given values into the compound interest formula, and solve for <em>P<\/em>.<\/p>\n<div id=\"eip-id1165131884554\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\hfill & \\text{Use the compound interest formula}.\\hfill \\\\ 40,000\\hfill & =P{\\left(1+\\frac{0.06}{2}\\right)}^{2\\left(18\\right)}\\hfill & \\text{Substitute using given values }A\\text{, }r, n\\text{, and }t.\\hfill \\\\ 40,000\\hfill & =P{\\left(1.03\\right)}^{36}\\hfill & \\text{Simplify}.\\hfill \\\\ \\frac{40,000}{{\\left(1.03\\right)}^{36}}\\hfill & =P\\hfill & \\text{Isolate }P.\\hfill \\\\ P\\hfill & \\approx 13,801\\hfill & \\text{Divide and round to the nearest dollar}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137937589\">Lily will need to invest\u00a0[latex]$13,801[\/latex] to have\u00a0[latex]$40,000[\/latex] in\u00a0[latex]18[\/latex] years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show another example of finding the deposit amount necessary to obtain a future value from compounded interest.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex: Compounded Interest Formula - Determine Deposit Needed (Present Value)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/saq9dF7a4r8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>Exponential growth grows by a rate proportional to the current amount.\u00a0For any real number <em>x<\/em>\u00a0and any positive real numbers <em>a\u00a0<\/em>and <em>b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form\u00a0[latex]f\\left(x\\right)=a{b}^{x}[\/latex]. \u00a0Evaluating exponential functions requires careful attention to the order of operations. Compound interest is an example of exponential growth.<\/p>\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-15985\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Determine Exponential Function Values and Graph the Function. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/QFFAoX0We34\">https:\/\/youtu.be\/QFFAoX0We34<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Evaluate a Given Exponential Function to Predict a Future Population. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/SbIydBmJePE\">https:\/\/youtu.be\/SbIydBmJePE<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 1: Compounded Interest Formula - Quarterly. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/3az4AKvUmmI\">https:\/\/youtu.be\/3az4AKvUmmI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Compounded Interest Formula - Determine Deposit Needed (Present Value). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/saq9dF7a4r8\">https:\/\/youtu.be\/saq9dF7a4r8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at : http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169554,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Determine Exponential Function Values and Graph the Function\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/QFFAoX0We34\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Evaluate a Given Exponential Function to Predict a Future Population\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/SbIydBmJePE\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Compounded Interest Formula - Quarterly\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/3az4AKvUmmI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Compounded Interest Formula - Determine Deposit Needed (Present Value)\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/saq9dF7a4r8\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay, et al.\",\"organization\":\"OpenStax\",\"url\":\" http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at : http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"f2a2ebc8661f4c09a4dd15b875789e9d, 705415bc38804807b93efe28fc81f62b, dee6d6bcb79040e2b8e9034e3468cc21","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-15985","chapter","type-chapter","status-publish","hentry"],"part":15981,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/15985","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/users\/169554"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/15985\/revisions"}],"predecessor-version":[{"id":19157,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/15985\/revisions\/19157"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/parts\/15981"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/15985\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/media?parent=15985"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapter-type?post=15985"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/contributor?post=15985"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/license?post=15985"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}