{"id":1968,"date":"2016-11-02T23:01:50","date_gmt":"2016-11-02T23:01:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1968"},"modified":"2017-04-20T22:52:34","modified_gmt":"2017-04-20T22:52:34","slug":"evaluate-exponential-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/evaluate-exponential-functions\/","title":{"raw":"Evaluate Exponential Functions","rendered":"Evaluate Exponential Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Identify the base of an exponential function, and restrictions for it's value<\/li>\r\n \t<li>Evaluate an exponential growth function<\/li>\r\n \t<li>Use a compound interest Formula<\/li>\r\n \t<li>Evaluate exponential functions with base e<\/li>\r\n<\/ul>\r\n<\/div>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\nIndia is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year.[footnote]http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014.[\/footnote] If this rate continues, the population of India will exceed China\u2019s population by the year 2031. When populations grow rapidly, we often say that the growth is \"exponential,\" meaning that something is growing very rapidly. To a mathematician, however, the term <em data-effect=\"italics\">exponential growth <\/em>has a very specific meaning. In this section, we will take a look at <em data-effect=\"italics\">exponential functions<\/em>, which model this kind of rapid growth.\r\n\r\nRecall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base <em>b<\/em>\u00a0to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:\r\n<ul>\r\n \t<li>Let <em>b\u00a0<\/em>= \u20139 and [latex]x=\\frac{1}{2}[\/latex]. Then [latex]f\\left(x\\right)=f\\left(\\frac{1}{2}\\right)={\\left(-9\\right)}^{\\frac{1}{2}}=\\sqrt{-9}[\/latex], which is not a real number.<\/li>\r\n<\/ul>\r\nWhy do we limit the base to positive values other than 1? Because base 1\u00a0results in the constant function. Observe what happens if the base is\u00a01:\r\n<ul>\r\n \t<li>Let <em>b\u00a0<\/em>= 1. Then [latex]f\\left(x\\right)={1}^{x}=1[\/latex] for any value of <em>x<\/em>.<\/li>\r\n<\/ul>\r\nTo evaluate an exponential function with 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:\r\n\r\nLet [latex]f\\left(x\\right)={2}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?\r\n<p 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]<\/p>\r\nTo evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:\r\n\r\nLet [latex]f\\left(x\\right)=30{\\left(2\\right)}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?\r\n<p 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]<\/p>\r\nNote that if the order of operations were not followed, the result would be incorrect:\r\n<p style=\"text-align: center;\">[latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating Exponential Functions<\/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\r\n[reveal-answer q=\"454575\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"454575\"]\r\nFollow the order of operations. Be sure to pay attention to the parentheses.\r\n<p 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]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nLet [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.\r\n\r\n[reveal-answer q=\"860098\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"860098\"]5.5556[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=73212&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\">\r\n<\/iframe>\r\n\r\n<\/div>\r\nBecause the output of exponential functions increases very rapidly, the term \"exponential growth\" 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.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Exponential Growth<\/h3>\r\nA function that models <strong>exponential growth<\/strong> grows by a rate proportional to the amount present. 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\r\n\r\n[latex]\\text{ }f\\left(x\\right)=a{b}^{x}[\/latex]\r\n\r\nwhere\r\n<ul>\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\nIn 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].\r\n\r\nA few years of growth for these companies are illustrated below.\r\n<table summary=\"Six rows and three columns. The first column is labeled,\">\r\n<thead>\r\n<tr>\r\n<th>Year,\u00a0<em>x<\/em><\/th>\r\n<th>Stores, Company A<\/th>\r\n<th>Stores, Company B<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>100 + 50(0) = 100<\/td>\r\n<td>100(1 + 0.5)<sup>0<\/sup> = 100<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1<\/td>\r\n<td>100 + 50(1) = 150<\/td>\r\n<td>100(1 + 0.5)<sup>1<\/sup> = 150<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>100 + 50(2) = 200<\/td>\r\n<td>100(1 + 0.5)<sup>2<\/sup> = 225<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>100 + 50(3) = 250<\/td>\r\n<td>100(1 + 0.5)<sup>3<\/sup> =\u00a0337.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>x<\/em><\/td>\r\n<td><em>A<\/em>(<em>x<\/em>) = 100 + 50x<\/td>\r\n<td><em>B<\/em>(<em>x<\/em>) = 100(1 + 0.5)<sup><em>x<\/em><\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe graphs comparing the number of stores for each company over a five-year period are shown in below. 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=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225441\/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=\"487\" height=\"845\" data-media-type=\"image\/jpg\" \/> The graph shows the numbers of stores Companies A and B opened over a five-year period.[\/caption]\r\n\r\nNotice that the domain for both functions is [latex]\\left[0,\\infty \\right)[\/latex], and the range for both functions is [latex]\\left[100,\\infty \\right)[\/latex]. After year 1, Company B always has more stores than Company A.\r\n\r\nNow we will turn our attention to the function representing the number of stores for Company B, [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]. In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex], where 100 is the initial value, 1.5 is called the <em>base<\/em>, and <em>x<\/em>\u00a0is called the <em>exponent<\/em>.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating a Real-World Exponential Model<\/h3>\r\nAt the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with an annual growth rate of about 1.2%. 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 2013. To the nearest thousandth, what will the population of India be in 2031?\r\n\r\n[reveal-answer q=\"924755\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"924755\"]\r\nTo estimate the population in 2031, we evaluate the models for <em>t\u00a0<\/em>= 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth,\r\n<p style=\"text-align: center;\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/p>\r\nThere will be about 1.549 billion people in India in the year 2031.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe population of China was about 1.39 billion in the year 2013, with an annual growth rate of about 0.6%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.39{\\left(1.006\\right)}^{t}[\/latex], where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in the previous example?\r\n\r\n[reveal-answer q=\"891037\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"891037\"]About 1.548 billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129476&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<h2>Use a Compound Interest Formula<\/h2>\r\nSavings 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.\r\n\r\nThe <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.\r\n\r\nWe 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>:\r\n<p style=\"text-align: center;\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/p>\r\nFor example, observe the table below, 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.\r\n<table summary=\"Six rows and two columns. The first column is labeled,\">\r\n<thead>\r\n<tr>\r\n<th data-align=\"center\">Frequency<\/th>\r\n<th data-align=\"center\">Value after 1 year<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Annually<\/td>\r\n<td>$1100<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Semiannually<\/td>\r\n<td>$1102.50<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Quarterly<\/td>\r\n<td>$1103.81<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Monthly<\/td>\r\n<td>$1104.71<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Daily<\/td>\r\n<td>$1105.16<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Compound Interest Formula<\/h3>\r\n<strong>Compound interest<\/strong> can be calculated using the formula\r\n<p style=\"text-align: center;\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/p>\r\nwhere\r\n<ul>\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\n<div class=\"textbox exercises\">\r\n<h3>Example: Calculating Compound Interest<\/h3>\r\nIf we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years?\r\n\r\n[reveal-answer q=\"476919\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"476919\"]\r\nBecause we are starting with $3,000, <em>P\u00a0<\/em>= 3000. Our interest rate is 3%, so <em>r<\/em>\u00a0=\u00a00.03. Because we are compounding quarterly, we are compounding 4 times per year, so <em>n\u00a0<\/em>= 4. We want to know the value of the account in 10 years, so we are looking for <em>A<\/em>(10), the value when <em>t <\/em>= 10.\r\n<p 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}. \\\\ 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]<\/p>\r\nThe account will be worth about $4,045.05 in 10 years.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nAn 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?\r\n\r\n[reveal-answer q=\"939452\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"939452\"]about $3,644,675.88[\/hidden-answer]\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=2453&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"400\">\r\n<\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Compound Interest Formula to Solve for the Principal<\/h3>\r\nA 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?\r\n\r\n[reveal-answer q=\"550421\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"550421\"]\r\nThe nominal interest rate is 6%, so <em>r\u00a0<\/em>= 0.06. Interest is compounded twice a year, so <em>k\u00a0<\/em>= 2.\r\n\r\nWe want to find the initial investment, <em>P<\/em>, 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 <em>P<\/em>.\r\n<p 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]<\/p>\r\nLily will need to invest $13,801 to have $40,000 in 18 years.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nRefer to the previous\u00a0example. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?\r\n\r\n[reveal-answer q=\"895101\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"895101\"]$13,693[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Evaluate exponential functions with base e<\/h2>\r\nAs we saw earlier, the amount earned on an account increases as the compounding frequency increases. The table below\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.\r\n\r\nExamine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies.\r\n<table id=\"Table_04_01_04\" summary=\"Nine rows and three columns. The first column is labeled,\">\r\n<thead>\r\n<tr>\r\n<th data-align=\"center\">Frequency<\/th>\r\n<th data-align=\"center\">[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\r\n<th data-align=\"center\">Value<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Annually<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\r\n<td>$2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Semiannually<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\r\n<td>$2.25<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Quarterly<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\r\n<td>$2.441406<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Monthly<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\r\n<td>$2.613035<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Daily<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\r\n<td>$2.714567<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Hourly<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\r\n<td>$2.718127<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Once per minute<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\r\n<td>$2.718279<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Once per second<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\r\n<td>$2.718282<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThese values appear to be approaching a limit as <em>n<\/em>\u00a0increases without bound. In fact, as <em>n<\/em>\u00a0gets 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.\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Number <em>e<\/em><\/h3>\r\nThe letter <em>e<\/em> represents the irrational number\r\n<p style=\"text-align: center;\">[latex]{\\left(1+\\frac{1}{n}\\right)}^{n},\\text{as}n\\text{increases without bound}[\/latex]<\/p>\r\nThe letter <em>e <\/em>is used as a base for many real-world exponential models. To work with base <em>e<\/em>, we use the approximation, [latex]e\\approx 2.718282[\/latex]. The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using a Calculator to Find Powers of <em>e<\/em><\/h3>\r\nCalculate [latex]{e}^{3.14}[\/latex]. Round to five decimal places.\r\n\r\n[reveal-answer q=\"201253\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"201253\"]\r\nOn a calculator, press the button labeled [latex]\\left[{e}^{x}\\right][\/latex]. The window shows [<em>e<\/em>^(]. Type 3.14 and then close parenthesis, (]). Press [ENTER]. Rounding to 5 decimal places, [latex]{e}^{3.14}\\approx 23.10387[\/latex]. Caution: Many scientific calculators have an \"Exp\" button, which is used to enter numbers in scientific notation. It is not used to find powers of <em>e<\/em>.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse a calculator to find [latex]{e}^{-0.5}[\/latex]. Round to five decimal places.\r\n\r\n[reveal-answer q=\"168744\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"168744\"][latex]{e}^{-0.5}\\approx 0.60653[\/latex][\/hidden-answer]\r\n<iframe id=\"mom11\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1495&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"350\">\r\n<\/iframe>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Identify the base of an exponential function, and restrictions for it&#8217;s value<\/li>\n<li>Evaluate an exponential growth function<\/li>\n<li>Use a compound interest Formula<\/li>\n<li>Evaluate exponential functions with base e<\/li>\n<\/ul>\n<\/div>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><br \/>\nIndia is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year.<a class=\"footnote\" title=\"http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014.\" id=\"return-footnote-1968-1\" href=\"#footnote-1968-1\" 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. When populations grow rapidly, we often say that the growth is &#8220;exponential,&#8221; meaning that something is growing very rapidly. To a mathematician, however, the term <em data-effect=\"italics\">exponential growth <\/em>has a very specific meaning. In this section, we will take a look at <em data-effect=\"italics\">exponential functions<\/em>, which model this kind of rapid growth.<\/p>\n<p>Recall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base <em>b<\/em>\u00a0to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:<\/p>\n<ul>\n<li>Let <em>b\u00a0<\/em>= \u20139 and [latex]x=\\frac{1}{2}[\/latex]. Then [latex]f\\left(x\\right)=f\\left(\\frac{1}{2}\\right)={\\left(-9\\right)}^{\\frac{1}{2}}=\\sqrt{-9}[\/latex], which is not a real number.<\/li>\n<\/ul>\n<p>Why do we limit the base to positive values other than 1? Because base 1\u00a0results in the constant function. Observe what happens if the base is\u00a01:<\/p>\n<ul>\n<li>Let <em>b\u00a0<\/em>= 1. Then [latex]f\\left(x\\right)={1}^{x}=1[\/latex] for any value of <em>x<\/em>.<\/li>\n<\/ul>\n<p>To evaluate an exponential function with 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>Let [latex]f\\left(x\\right)={2}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\n<p 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]<\/p>\n<p>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>Let [latex]f\\left(x\\right)=30{\\left(2\\right)}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\n<p 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]<\/p>\n<p>Note that if the order of operations were not followed, the result would be incorrect:<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating Exponential Functions<\/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=\"q454575\">Solution<\/span><\/p>\n<div id=\"q454575\" class=\"hidden-answer\" style=\"display: none\">\nFollow the order of operations. Be sure to pay attention to the parentheses.<\/p>\n<p 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]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q860098\">Solution<\/span><\/p>\n<div id=\"q860098\" class=\"hidden-answer\" style=\"display: none\">5.5556<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=73212&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<p>Because the output of exponential functions increases very rapidly, the term &#8220;exponential growth&#8221; is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Exponential Growth<\/h3>\n<p>A function that models <strong>exponential growth<\/strong> grows by a rate proportional to the amount present. 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<p>[latex]\\text{ }f\\left(x\\right)=a{b}^{x}[\/latex]<\/p>\n<p>where<\/p>\n<ul>\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<p>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>A few years of growth for these companies are illustrated below.<\/p>\n<table summary=\"Six rows and three columns. The first column is labeled,\">\n<thead>\n<tr>\n<th>Year,\u00a0<em>x<\/em><\/th>\n<th>Stores, Company A<\/th>\n<th>Stores, Company B<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>100 + 50(0) = 100<\/td>\n<td>100(1 + 0.5)<sup>0<\/sup> = 100<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>100 + 50(1) = 150<\/td>\n<td>100(1 + 0.5)<sup>1<\/sup> = 150<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>100 + 50(2) = 200<\/td>\n<td>100(1 + 0.5)<sup>2<\/sup> = 225<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>100 + 50(3) = 250<\/td>\n<td>100(1 + 0.5)<sup>3<\/sup> =\u00a0337.5<\/td>\n<\/tr>\n<tr>\n<td><em>x<\/em><\/td>\n<td><em>A<\/em>(<em>x<\/em>) = 100 + 50x<\/td>\n<td><em>B<\/em>(<em>x<\/em>) = 100(1 + 0.5)<sup><em>x<\/em><\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The graphs comparing the number of stores for each company over a five-year period are shown in below. We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.<\/p>\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\/896\/2016\/11\/02225441\/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=\"487\" height=\"845\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">The graph shows the numbers of stores Companies A and B opened over a five-year period.<\/p>\n<\/div>\n<p>Notice that the domain for both functions is [latex]\\left[0,\\infty \\right)[\/latex], and the range for both functions is [latex]\\left[100,\\infty \\right)[\/latex]. After year 1, Company B always has more stores than Company A.<\/p>\n<p>Now we will turn our attention to the function representing the number of stores for Company B, [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]. In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex], where 100 is the initial value, 1.5 is called the <em>base<\/em>, and <em>x<\/em>\u00a0is called the <em>exponent<\/em>.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating a Real-World Exponential Model<\/h3>\n<p>At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with an annual growth rate of about 1.2%. 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 2013. To the nearest thousandth, what will the population of India be in 2031?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924755\">Solution<\/span><\/p>\n<div id=\"q924755\" class=\"hidden-answer\" style=\"display: none\">\nTo estimate the population in 2031, we evaluate the models for <em>t\u00a0<\/em>= 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth,<\/p>\n<p style=\"text-align: center;\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/p>\n<p>There will be about 1.549 billion people in India in the year 2031.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about 0.6%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.39{\\left(1.006\\right)}^{t}[\/latex], where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in the previous example?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q891037\">Solution<\/span><\/p>\n<div id=\"q891037\" class=\"hidden-answer\" style=\"display: none\">About 1.548 billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129476&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<h2>Use a Compound Interest Formula<\/h2>\n<p>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>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>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<p style=\"text-align: center;\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/p>\n<p>For example, observe the table below, which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases.<\/p>\n<table summary=\"Six rows and two columns. The first column is labeled,\">\n<thead>\n<tr>\n<th data-align=\"center\">Frequency<\/th>\n<th data-align=\"center\">Value after 1 year<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Annually<\/td>\n<td>$1100<\/td>\n<\/tr>\n<tr>\n<td>Semiannually<\/td>\n<td>$1102.50<\/td>\n<\/tr>\n<tr>\n<td>Quarterly<\/td>\n<td>$1103.81<\/td>\n<\/tr>\n<tr>\n<td>Monthly<\/td>\n<td>$1104.71<\/td>\n<\/tr>\n<tr>\n<td>Daily<\/td>\n<td>$1105.16<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox\">\n<h3>A General Note: The Compound Interest Formula<\/h3>\n<p><strong>Compound interest<\/strong> can be calculated using the formula<\/p>\n<p style=\"text-align: center;\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/p>\n<p>where<\/p>\n<ul>\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<div class=\"textbox exercises\">\n<h3>Example: Calculating Compound Interest<\/h3>\n<p>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=\"q476919\">Solution<\/span><\/p>\n<div id=\"q476919\" class=\"hidden-answer\" style=\"display: none\">\nBecause we are starting with $3,000, <em>P\u00a0<\/em>= 3000. Our interest rate is 3%, so <em>r<\/em>\u00a0=\u00a00.03. Because we are compounding quarterly, we are compounding 4 times per year, so <em>n\u00a0<\/em>= 4. We want to know the value of the account in 10 years, so we are looking for <em>A<\/em>(10), the value when <em>t <\/em>= 10.<\/p>\n<p 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}. \\\\ 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]<\/p>\n<p>The account will be worth about $4,045.05 in 10 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q939452\">Solution<\/span><\/p>\n<div id=\"q939452\" class=\"hidden-answer\" style=\"display: none\">about $3,644,675.88<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=2453&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"400\"><br \/>\n<\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Compound Interest Formula to Solve for the Principal<\/h3>\n<p>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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q550421\">Solution<\/span><\/p>\n<div id=\"q550421\" class=\"hidden-answer\" style=\"display: none\">\nThe nominal interest rate is 6%, so <em>r\u00a0<\/em>= 0.06. Interest is compounded twice a year, so <em>k\u00a0<\/em>= 2.<\/p>\n<p>We want to find the initial investment, <em>P<\/em>, 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 <em>P<\/em>.<\/p>\n<p 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]<\/p>\n<p>Lily will need to invest $13,801 to have $40,000 in 18 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Refer to the previous\u00a0example. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q895101\">Solution<\/span><\/p>\n<div id=\"q895101\" class=\"hidden-answer\" style=\"display: none\">$13,693<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Evaluate exponential functions with base e<\/h2>\n<p>As we saw earlier, the amount earned on an account increases as the compounding frequency increases. The table below\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>Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies.<\/p>\n<table id=\"Table_04_01_04\" summary=\"Nine rows and three columns. The first column is labeled,\">\n<thead>\n<tr>\n<th data-align=\"center\">Frequency<\/th>\n<th data-align=\"center\">[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\n<th data-align=\"center\">Value<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Annually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\n<td>$2<\/td>\n<\/tr>\n<tr>\n<td>Semiannually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\n<td>$2.25<\/td>\n<\/tr>\n<tr>\n<td>Quarterly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\n<td>$2.441406<\/td>\n<\/tr>\n<tr>\n<td>Monthly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\n<td>$2.613035<\/td>\n<\/tr>\n<tr>\n<td>Daily<\/td>\n<td>[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\n<td>$2.714567<\/td>\n<\/tr>\n<tr>\n<td>Hourly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\n<td>$2.718127<\/td>\n<\/tr>\n<tr>\n<td>Once per minute<\/td>\n<td>[latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\n<td>$2.718279<\/td>\n<\/tr>\n<tr>\n<td>Once per second<\/td>\n<td>[latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\n<td>$2.718282<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>These values appear to be approaching a limit as <em>n<\/em>\u00a0increases without bound. In fact, as <em>n<\/em>\u00a0gets 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 class=\"textbox\">\n<h3>A General Note: The Number <em>e<\/em><\/h3>\n<p>The letter <em>e<\/em> represents the irrational number<\/p>\n<p style=\"text-align: center;\">[latex]{\\left(1+\\frac{1}{n}\\right)}^{n},\\text{as}n\\text{increases without bound}[\/latex]<\/p>\n<p>The letter <em>e <\/em>is used as a base for many real-world exponential models. To work with base <em>e<\/em>, we use the approximation, [latex]e\\approx 2.718282[\/latex]. The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using a Calculator to Find Powers of <em>e<\/em><\/h3>\n<p>Calculate [latex]{e}^{3.14}[\/latex]. Round to five decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q201253\">Solution<\/span><\/p>\n<div id=\"q201253\" class=\"hidden-answer\" style=\"display: none\">\nOn a calculator, press the button labeled [latex]\\left[{e}^{x}\\right][\/latex]. The window shows [<em>e<\/em>^(]. Type 3.14 and then close parenthesis, (]). Press [ENTER]. Rounding to 5 decimal places, [latex]{e}^{3.14}\\approx 23.10387[\/latex]. Caution: Many scientific calculators have an &#8220;Exp&#8221; button, which is used to enter numbers in scientific notation. It is not used to find powers of <em>e<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use a calculator to find [latex]{e}^{-0.5}[\/latex]. Round to five decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q168744\">Solution<\/span><\/p>\n<div id=\"q168744\" class=\"hidden-answer\" style=\"display: none\">[latex]{e}^{-0.5}\\approx 0.60653[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom11\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1495&amp;theme=oea&amp;iframe_resize_id=mom11\" width=\"100%\" height=\"350\"><br \/>\n<\/iframe><\/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-1968\">\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>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>Question ID 1495. <strong>Authored by<\/strong>: WebWork-Rochester. <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>: IMathAS Community License CC-BY + GPL<\/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@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/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@5.2<\/li><li>Question ID 2453. <strong>Authored by<\/strong>: Anderson,Tophe, mb Sousa,James. <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>: IMathAS Community License CC-BY + GPL<\/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-1968-1\">http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014. <a href=\"#return-footnote-1968-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Question ID 1495\",\"author\":\"WebWork-Rochester\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"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@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 2453\",\"author\":\"Anderson,Tophe, mb Sousa,James\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"07251946-98ba-433c-862b-3c327756eda3","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1968","chapter","type-chapter","status-publish","hentry"],"part":1964,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1968","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1968\/revisions"}],"predecessor-version":[{"id":4258,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1968\/revisions\/4258"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1964"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1968\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/media?parent=1968"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1968"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1968"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/license?post=1968"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}