{"id":15986,"date":"2019-09-26T16:05:36","date_gmt":"2019-09-26T16:05:36","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/evaluate-exponential-functions-with-base-e\/"},"modified":"2024-05-02T16:00:40","modified_gmt":"2024-05-02T16:00:40","slug":"evaluate-exponential-functions-with-base-e","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/evaluate-exponential-functions-with-base-e\/","title":{"raw":"Exponential Functions with Base e","rendered":"Exponential Functions with Base e"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcome<\/h3>\r\n<ul>\r\n \t<li>Evaluate exponential functions with base\u00a0[latex]e[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\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<p id=\"fs-id1165135684377\">Examine the value of\u00a0[latex]$1[\/latex] invested at\u00a0[latex]100\\%[\/latex] interest for\u00a0[latex]1[\/latex] year, compounded at various frequencies.<\/p>\r\n\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>Frequency<\/th>\r\n<th>[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\r\n<th>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>[latex]$2[\/latex]<\/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>[latex]$2.25[\/latex]<\/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>[latex]$2.441406[\/latex]<\/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>[latex]$2.613035[\/latex]<\/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>[latex]$2.714567[\/latex]<\/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>[latex]$2.718127[\/latex]<\/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>[latex]$2.718279[\/latex]<\/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>[latex]$2.718282[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165137828146\">These values appear to be reaching a limit as <em>n<\/em>\u00a0increases. 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>\r\n\r\n<div id=\"fs-id1165135511324\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: The Number [latex]e[\/latex]<\/h3>\r\n<p id=\"fs-id1165135511335\">The letter <em>e<\/em> represents the irrational number<\/p>\r\n\r\n<div id=\"eip-id1165135378658\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">as n increases without bound<\/div>\r\n<p id=\"fs-id1165135369344\">The letter <em>e <\/em>is used as a base for many real-world exponential models. To work with base <em>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>\r\n\r\n<\/div>\r\nIn our first example, we will use a calculator to find powers of\u00a0<em>e.<\/em>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nCalculate [latex]{e}^{3.14}[\/latex].\u00a0 Round to five decimal places.\r\n[reveal-answer q=\"465847\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"465847\"]\r\n\r\nOn a calculator, press the button labeled [latex]\\left[{e}^{x}\\right][\/latex]. The window shows [<em>e<\/em>^(]. Type\u00a0[latex]3.14[\/latex] and then close parenthesis, (]). Press [ENTER]. Rounding to\u00a0[latex]5[\/latex] decimal places, [latex]{e}^{3.14}\\approx 23.10387[\/latex].\r\n\r\n[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>.][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137827923\">\r\n<h2>Investigating Continuous Growth<\/h2>\r\n<p id=\"fs-id1165137827929\">So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, <em>e <\/em>is used as the base for exponential functions. Exponential models that use <em>e<\/em>\u00a0as the base are called <em>continuous growth or decay models<\/em>. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.<\/p>\r\n\r\n<div id=\"fs-id1165137664673\" class=\"note textbox\">\r\n<h3 class=\"title\">The Continuous Growth\/Decay Formula<\/h3>\r\n<p id=\"fs-id1165135453868\">For all real numbers r,\u00a0<em>t<\/em>, and all positive numbers <em>a<\/em>, continuous growth or decay is represented by the formula<\/p>\r\n\r\n<div id=\"fs-id1165135536370\" class=\"equation\" style=\"text-align: center;\">[latex]A\\left(t\\right)=a{e}^{rt}[\/latex]<\/div>\r\n<p id=\"eip-101\">where<\/p>\r\n\r\n<ul id=\"fs-id1165135152052\">\r\n \t<li><em>a<\/em>\u00a0is the initial value,<\/li>\r\n \t<li><em>r<\/em>\u00a0is the continuous growth or decay rate per unit time,<\/li>\r\n \t<li>and <em>t<\/em>\u00a0is the elapsed time.<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165135560686\">If <em>r\u00a0<\/em>&gt;[latex]0[\/latex], then the formula represents continuous growth. If <em>r\u00a0<\/em>&lt;\u00a0[latex]0[\/latex], then the formula represents continuous decay.<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1165137812323\">For business applications, the continuous growth formula is called the continuous compounding formula and takes the form<\/p>\r\n\r\n<div id=\"eip-id1165134324899\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A\\left(t\\right)=P{e}^{rt}[\/latex]<\/div>\r\n<p id=\"eip-962\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137827330\">\r\n \t<li><em>P<\/em>\u00a0is the principal or the initial amount invested,<\/li>\r\n \t<li><em>r<\/em>\u00a0is the growth or interest rate per unit time,<\/li>\r\n \t<li>and <em>t<\/em>\u00a0is the period or term of the investment.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn our next example, we will calculate continuous growth of an investment. It is important to note the language that is used in the instructions for interest rate problems. \u00a0You will know to use the <em>continuous<\/em> growth or decay formula when you are asked to find an amount based on continuous compounding. \u00a0In previous examples we asked that you find an amount based on quarterly or monthly compounding where, in that case, you used the <em>compound<\/em> interest formula.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA person invested\u00a0[latex]$1,000[\/latex] in an account earning a nominal\u00a0[latex]10\\%[\/latex] per year compounded continuously. How much was in the account at the end of one year?\r\n[reveal-answer q=\"33008\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"33008\"]\r\n\r\nSince the account is growing in value, this is a continuous compounding problem with growth rate <em>r\u00a0<\/em>=[latex]0.10[\/latex]. The initial investment was\u00a0[latex]$1,000[\/latex], so <em>P\u00a0<\/em>=[latex]1000[\/latex]. We use the continuous compounding formula to find the value after <em>t\u00a0<\/em>=[latex]1[\/latex] year:\r\n<div id=\"eip-id1165133351794\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill &amp; =P{e}^{rt}\\hfill &amp; \\text{Use the continuous compounding formula}.\\hfill \\\\ \\hfill &amp; =1000{\\left(e\\right)}^{0.1} &amp; \\text{Substitute known values for }P, r,\\text{ and }t.\\hfill \\\\ \\hfill &amp; \\approx 1105.17\\hfill &amp; \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137895288\">The account is worth\u00a0[latex]$1,105.17[\/latex] after one year.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of interest compounded continuously.\r\n\r\nhttps:\/\/youtu.be\/fEjrYCog_8w\r\n<div id=\"fs-id1165135411368\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135411373\">How To: Given the initial value, rate of growth or decay, and time [latex]t[\/latex], solve a continuous growth or decay function<\/h3>\r\n<ol id=\"fs-id1165135511371\">\r\n \t<li>Use the information in the problem to determine <em>a<\/em>, the initial value of the function.<\/li>\r\n \t<li>Use the information in the problem to determine the growth rate <em>r<\/em>.\r\n<ol id=\"fs-id1165135188096\">\r\n \t<li>If the problem refers to continuous growth, then <em>r\u00a0<\/em>&gt; [latex]0[\/latex].<\/li>\r\n \t<li>If the problem refers to continuous decay, then <em>r\u00a0<\/em>&lt;\u00a0[latex]0[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use the information in the problem to determine the time <em>t<\/em>.<\/li>\r\n \t<li>Substitute the given information into the continuous growth formula and solve for <em>A<\/em>(<em>t<\/em>).<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn our next example, we will calculate continuous decay. Pay attention to the rate - it is negative which means we are considering a situation where an amount decreases or decays.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nRadon-222 decays at a continuous rate of\u00a0[latex]17.3\\%[\/latex] per day. How much will\u00a0[latex]100[\/latex] mg of Radon-[latex]222[\/latex] decay to in\u00a0[latex]3[\/latex] days?\r\n[reveal-answer q=\"995802\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"995802\"]\r\n\r\nSince the substance is decaying, the rate,\u00a0[latex]17.3\\%[\/latex], is negative. So, <em>r\u00a0<\/em>=\u00a0[latex]\u20130.173[\/latex]. The initial amount of radon-[latex]222[\/latex] was [latex]100[\/latex] mg, so <em>a\u00a0<\/em>=\u00a0[latex]100[\/latex]. We use the continuous decay formula to find the value after <em>t\u00a0<\/em>=[latex]3[\/latex] days:\r\n<div id=\"eip-id1165137779893\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill &amp; =a{e}^{rt}\\hfill &amp; \\text{Use the continuous growth formula}.\\hfill \\\\ \\hfill &amp; =100{e}^{-0.173\\left(3\\right)} &amp; \\text{Substitute known values for }a, r,\\text{ and }t.\\hfill \\\\ \\hfill &amp; \\approx 59.5115\\hfill &amp; \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137697132\">So\u00a0[latex]59.5115[\/latex] mg of radon-[latex]222[\/latex] will remain.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show an example of calculating the remaining amount of a radioactive substance after it decays for a length of time.\r\n\r\nhttps:\/\/youtu.be\/Vyl3NcTGRAo\r\n<h2>Summary<\/h2>\r\nContinuous growth or decay functions are of the form\u00a0[latex]A\\left(t\\right)=a{e}^{rt}[\/latex].\u00a0If <em>r\u00a0<\/em>&gt;\u00a0[latex]0[\/latex], then the formula represents continuous growth. If <em>r\u00a0<\/em>&lt;\u00a0[latex]0[\/latex], then the formula represents continuous decay.\u00a0For business applications, the continuous growth formula is called the continuous compounding formula and takes the form\u00a0[latex]A\\left(t\\right)=P{e}^{rt}[\/latex].\r\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcome<\/h3>\n<ul>\n<li>Evaluate exponential functions with base\u00a0[latex]e[\/latex]<\/li>\n<\/ul>\n<\/div>\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 id=\"fs-id1165135684377\">Examine the value of\u00a0[latex]$1[\/latex] invested at\u00a0[latex]100\\%[\/latex] interest for\u00a0[latex]1[\/latex] 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>Frequency<\/th>\n<th>[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\n<th>Value<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Annually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\n<td>[latex]$2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Semiannually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\n<td>[latex]$2.25[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Quarterly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\n<td>[latex]$2.441406[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Monthly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\n<td>[latex]$2.613035[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Daily<\/td>\n<td>[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\n<td>[latex]$2.714567[\/latex]<\/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>[latex]$2.718127[\/latex]<\/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>[latex]$2.718279[\/latex]<\/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>[latex]$2.718282[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137828146\">These values appear to be reaching a limit as <em>n<\/em>\u00a0increases. 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 id=\"fs-id1165135511324\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: The Number [latex]e[\/latex]<\/h3>\n<p id=\"fs-id1165135511335\">The letter <em>e<\/em> represents the irrational number<\/p>\n<div id=\"eip-id1165135378658\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">as n increases without bound<\/div>\n<p id=\"fs-id1165135369344\">The letter <em>e <\/em>is used as a base for many real-world exponential models. To work with base <em>e<\/em>, we use the approximation, [latex]e\\approx 2.718282[\/latex]. The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.<\/p>\n<\/div>\n<p>In our first example, we will use a calculator to find powers of\u00a0<em>e.<\/em><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Calculate [latex]{e}^{3.14}[\/latex].\u00a0 Round to five decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q465847\">Show Solution<\/span><\/p>\n<div id=\"q465847\" class=\"hidden-answer\" style=\"display: none\">\n<p>On a calculator, press the button labeled [latex]\\left[{e}^{x}\\right][\/latex]. The window shows [<em>e<\/em>^(]. Type\u00a0[latex]3.14[\/latex] and then close parenthesis, (]). Press [ENTER]. Rounding to\u00a0[latex]5[\/latex] decimal places, [latex]{e}^{3.14}\\approx 23.10387[\/latex].<\/p>\n<p>[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>.]<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1165137827923\">\n<h2>Investigating Continuous Growth<\/h2>\n<p id=\"fs-id1165137827929\">So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, <em>e <\/em>is used as the base for exponential functions. Exponential models that use <em>e<\/em>\u00a0as the base are called <em>continuous growth or decay models<\/em>. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.<\/p>\n<div id=\"fs-id1165137664673\" class=\"note textbox\">\n<h3 class=\"title\">The Continuous Growth\/Decay Formula<\/h3>\n<p id=\"fs-id1165135453868\">For all real numbers r,\u00a0<em>t<\/em>, and all positive numbers <em>a<\/em>, continuous growth or decay is represented by the formula<\/p>\n<div id=\"fs-id1165135536370\" class=\"equation\" style=\"text-align: center;\">[latex]A\\left(t\\right)=a{e}^{rt}[\/latex]<\/div>\n<p id=\"eip-101\">where<\/p>\n<ul id=\"fs-id1165135152052\">\n<li><em>a<\/em>\u00a0is the initial value,<\/li>\n<li><em>r<\/em>\u00a0is the continuous growth or decay rate per unit time,<\/li>\n<li>and <em>t<\/em>\u00a0is the elapsed time.<\/li>\n<\/ul>\n<p id=\"fs-id1165135560686\">If <em>r\u00a0<\/em>&gt;[latex]0[\/latex], then the formula represents continuous growth. If <em>r\u00a0<\/em>&lt;\u00a0[latex]0[\/latex], then the formula represents continuous decay.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137812323\">For business applications, the continuous growth formula is called the continuous compounding formula and takes the form<\/p>\n<div id=\"eip-id1165134324899\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A\\left(t\\right)=P{e}^{rt}[\/latex]<\/div>\n<p id=\"eip-962\">where<\/p>\n<ul id=\"fs-id1165137827330\">\n<li><em>P<\/em>\u00a0is the principal or the initial amount invested,<\/li>\n<li><em>r<\/em>\u00a0is the growth or interest rate per unit time,<\/li>\n<li>and <em>t<\/em>\u00a0is the period or term of the investment.<\/li>\n<\/ul>\n<\/div>\n<p>In our next example, we will calculate continuous growth of an investment. It is important to note the language that is used in the instructions for interest rate problems. \u00a0You will know to use the <em>continuous<\/em> growth or decay formula when you are asked to find an amount based on continuous compounding. \u00a0In previous examples we asked that you find an amount based on quarterly or monthly compounding where, in that case, you used the <em>compound<\/em> interest formula.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A person invested\u00a0[latex]$1,000[\/latex] in an account earning a nominal\u00a0[latex]10\\%[\/latex] per year compounded continuously. How much was in the account at the end of one year?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q33008\">Show Solution<\/span><\/p>\n<div id=\"q33008\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the account is growing in value, this is a continuous compounding problem with growth rate <em>r\u00a0<\/em>=[latex]0.10[\/latex]. The initial investment was\u00a0[latex]$1,000[\/latex], so <em>P\u00a0<\/em>=[latex]1000[\/latex]. We use the continuous compounding formula to find the value after <em>t\u00a0<\/em>=[latex]1[\/latex] year:<\/p>\n<div id=\"eip-id1165133351794\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =P{e}^{rt}\\hfill & \\text{Use the continuous compounding formula}.\\hfill \\\\ \\hfill & =1000{\\left(e\\right)}^{0.1} & \\text{Substitute known values for }P, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 1105.17\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137895288\">The account is worth\u00a0[latex]$1,105.17[\/latex] after one year.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show another example of interest compounded continuously.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Continuous Interest Formula\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/fEjrYCog_8w?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div id=\"fs-id1165135411368\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135411373\">How To: Given the initial value, rate of growth or decay, and time [latex]t[\/latex], solve a continuous growth or decay function<\/h3>\n<ol id=\"fs-id1165135511371\">\n<li>Use the information in the problem to determine <em>a<\/em>, the initial value of the function.<\/li>\n<li>Use the information in the problem to determine the growth rate <em>r<\/em>.\n<ol id=\"fs-id1165135188096\">\n<li>If the problem refers to continuous growth, then <em>r\u00a0<\/em>&gt; [latex]0[\/latex].<\/li>\n<li>If the problem refers to continuous decay, then <em>r\u00a0<\/em>&lt;\u00a0[latex]0[\/latex].<\/li>\n<\/ol>\n<\/li>\n<li>Use the information in the problem to determine the time <em>t<\/em>.<\/li>\n<li>Substitute the given information into the continuous growth formula and solve for <em>A<\/em>(<em>t<\/em>).<\/li>\n<\/ol>\n<\/div>\n<p>In our next example, we will calculate continuous decay. Pay attention to the rate &#8211; it is negative which means we are considering a situation where an amount decreases or decays.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Radon-222 decays at a continuous rate of\u00a0[latex]17.3\\%[\/latex] per day. How much will\u00a0[latex]100[\/latex] mg of Radon-[latex]222[\/latex] decay to in\u00a0[latex]3[\/latex] days?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q995802\">Show Solution<\/span><\/p>\n<div id=\"q995802\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the substance is decaying, the rate,\u00a0[latex]17.3\\%[\/latex], is negative. So, <em>r\u00a0<\/em>=\u00a0[latex]\u20130.173[\/latex]. The initial amount of radon-[latex]222[\/latex] was [latex]100[\/latex] mg, so <em>a\u00a0<\/em>=\u00a0[latex]100[\/latex]. We use the continuous decay formula to find the value after <em>t\u00a0<\/em>=[latex]3[\/latex] days:<\/p>\n<div id=\"eip-id1165137779893\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =a{e}^{rt}\\hfill & \\text{Use the continuous growth formula}.\\hfill \\\\ \\hfill & =100{e}^{-0.173\\left(3\\right)} & \\text{Substitute known values for }a, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 59.5115\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137697132\">So\u00a0[latex]59.5115[\/latex] mg of radon-[latex]222[\/latex] will remain.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we show an example of calculating the remaining amount of a radioactive substance after it decays for a length of time.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Determine a Continuous Exponential Decay Function and Make a Prediction\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Vyl3NcTGRAo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>Continuous growth or decay functions are of the form\u00a0[latex]A\\left(t\\right)=a{e}^{rt}[\/latex].\u00a0If <em>r\u00a0<\/em>&gt;\u00a0[latex]0[\/latex], then the formula represents continuous growth. If <em>r\u00a0<\/em>&lt;\u00a0[latex]0[\/latex], then the formula represents continuous decay.\u00a0For business applications, the continuous growth formula is called the continuous compounding formula and takes the form\u00a0[latex]A\\left(t\\right)=P{e}^{rt}[\/latex].<br \/>\n<\/section>\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-15986\">\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 a Continuous Exponential Decay Function and Make a Prediction. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Vyl3NcTGRAo\">https:\/\/youtu.be\/Vyl3NcTGRAo<\/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>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/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\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><li>Ex 1: Continuous Interest Formula. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/fEjrYCog_8w\">https:\/\/youtu.be\/fEjrYCog_8w<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/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><\/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":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"},{\"type\":\"cc\",\"description\":\"Ex 1: Continuous Interest Formula\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/fEjrYCog_8w\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Determine a Continuous Exponential Decay Function and Make a Prediction\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Vyl3NcTGRAo\",\"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\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"f2a2ebc8661f4c09a4dd15b875789e9d, 8bb96c39ee2d4a55952298fe737de654 ","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-15986","chapter","type-chapter","status-publish","hentry"],"part":15981,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/15986","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":6,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/15986\/revisions"}],"predecessor-version":[{"id":20526,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/15986\/revisions\/20526"}],"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\/15986\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/media?parent=15986"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapter-type?post=15986"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/contributor?post=15986"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/license?post=15986"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}