{"id":5380,"date":"2021-10-13T21:10:33","date_gmt":"2021-10-13T21:10:33","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/?post_type=chapter&#038;p=5380"},"modified":"2021-11-23T04:13:47","modified_gmt":"2021-11-23T04:13:47","slug":"empty-6","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/empty-6\/","title":{"raw":"Growth and Decay Applications","rendered":"Growth and Decay Applications"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Graph exponential growth and decay functions.<\/li>\r\n \t<li>Use the compound interest formula.<\/li>\r\n \t<li>Solve problems involving radioactive decay, carbon dating, and half life.<\/li>\r\n \t<li>Use Newton's Law of Cooling.<\/li>\r\n \t<li>Use a logistic growth model.<\/li>\r\n \t<li>Choose an appropriate model for data.<\/li>\r\n<\/ul>\r\n<\/div>\r\nWe have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth including radioactive isotopes and Newton\u2019s Law of Cooling.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181335\/CNX_Precalc_Figure_04_07_001F2.jpg\" alt=\"Inside a nuclear research reactor.\" width=\"325\" height=\"409\" \/> A nuclear research reactor inside the Neely Nuclear Research Center on the Georgia Institute of Technology campus. (credit: Georgia Tech Research Institute)[\/caption]\r\n<h2><\/h2>\r\n<h2>Exponential Growth and Decay<\/h2>\r\nIn real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:\r\n<p style=\"text-align: center;\">[latex]y={A}_{0}{e}^{kt}[\/latex]<\/p>\r\nwhere [latex]{A}_{0}[\/latex] is equal to the value at time zero, <em>e<\/em>\u00a0is Euler\u2019s constant, and <em>k<\/em>\u00a0is a positive constant that determines the rate (percentage) of growth. We may use the <strong>exponential growth<\/strong> function in applications involving <strong>doubling time<\/strong>, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based odn a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.\r\n\r\nOn the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the <strong>exponential decay<\/strong> model. Again, we have the form [latex]y={A}_{0}{e}^{-kt}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, and <em>e<\/em>\u00a0is Euler\u2019s constant. Now <em>k<\/em>\u00a0is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating <strong>half-life<\/strong>, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.\r\n\r\nIn our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in the graphs below. It is important to remember that, although parts of each of the two graphs seem to lie on the <em>x<\/em>-axis, they are really a tiny distance above the <em>x<\/em>-axis.\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\/03181337\/CNX_Precalc_Figure_04_07_0022.jpg\" alt=\"Graph of y=2e^(3x) with the labeled points (-1\/3, 2\/e), (0, 2), and (1\/3, 2e) and with the asymptote at y=0.\" width=\"487\" height=\"326\" \/> A graph showing exponential growth. The equation is [latex]y=2{e}^{3x}[\/latex].[\/caption][caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181339\/CNX_Precalc_Figure_04_07_0032.jpg\" alt=\"Graph of y=3e^(-2x) with the labeled points (-1\/2, 3e), (0, 3), and (1\/2, 3\/e) and with the asymptote at y=0.\" width=\"487\" height=\"438\" \/> A graph showing exponential decay. The equation is [latex]y=3{e}^{-2x}[\/latex].[\/caption]Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The <strong>order of magnitude<\/strong> is the power of ten when the number is expressed in scientific notation with one digit to the left of the decimal. For example, the distance to the nearest star, <strong>Proxima Centauri<\/strong>, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is [latex]4.01134972\\times {10}^{13}[\/latex]. We could describe this number as having order of magnitude [latex]{10}^{13}[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Characteristics of the Exponential Function [latex]y=A_{0}e^{kt}[\/latex]<\/h3>\r\nAn exponential function of the form [latex]y={A}_{0}{e}^{kt}[\/latex] has the following characteristics:\r\n<ul>\r\n \t<li>one-to-one function<\/li>\r\n \t<li>horizontal asymptote: <em>y\u00a0<\/em>= 0<\/li>\r\n \t<li>domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\r\n \t<li>range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\r\n \t<li>x intercept: none<\/li>\r\n \t<li>y-intercept: [latex]\\left(0,{A}_{0}\\right)[\/latex]<\/li>\r\n \t<li>increasing if <em>k\u00a0<\/em>&gt; 0<\/li>\r\n \t<li>decreasing if <em>k\u00a0<\/em>&lt; 0<\/li>\r\n<\/ul>\r\nAn exponential function models exponential growth when k\u00a0&gt; 0 and exponential decay when k\u00a0&lt; 0.\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<h3>Using the Compound Interest Formula<\/h3>\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\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>Frequency<\/th>\r\n<th>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&nbsp;\r\n\r\nWe can calculate 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 times compounded in a year\u00a0<em>n.<\/em>\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 accumulated value of the account<\/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<\/li>\r\n \t<li><em>n<\/em>\u00a0is the number of times compounded in a 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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"476919\"]\r\n\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}{llllll}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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"939452\"]\r\n\r\nabout $3,644,675.88[\/hidden-answer]\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"550421\"]\r\n\r\nThe nominal interest rate is 6%, so <em>r\u00a0<\/em>= 0.06. Interest is compounded twice a year, so <em>n<\/em><em>\u00a0<\/em>= 2.\r\n\r\nWe want to find the initial investment\u00a0<em>P\u00a0<\/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\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"895101\"]\r\n\r\n$13,693[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<h3><\/h3>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing Exponential Growth<\/h3>\r\nA population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.\r\n\r\n[reveal-answer q=\"151444\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"151444\"]\r\n\r\nWhen an amount grows at a fixed percent per unit time, the growth is exponential. To find [latex]{A}_{0}[\/latex] we use the fact that [latex]{A}_{0}[\/latex] is the amount at time zero, so [latex]{A}_{0}=10[\/latex]. To find <em>k<\/em>, use the fact that after one hour [latex]\\left(t=1\\right)[\/latex] the population doubles from 10 to 20. The formula is derived as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }20=10{e}^{k\\cdot 1}\\hfill &amp; \\hfill \\\\ \\text{ }2={e}^{k}\\hfill &amp; \\text{Divide both sides by 10}\\hfill \\\\ \\mathrm{ln}2=k\\hfill &amp; \\text{Take the natural logarithm of both sides}\\hfill \\end{array}[\/latex]<\/p>\r\nso [latex]k=\\mathrm{ln}\\left(2\\right)[\/latex]. Thus the equation we want to graph is [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}=10{\\left({e}^{\\mathrm{ln}2}\\right)}^{t}=10\\cdot {2}^{t}[\/latex]. The graph is shown below.\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\/03181344\/CNX_Precalc_Figure_04_07_0052.jpg\" alt=\"A graph starting at ten on the y-axis and rising rapidly to the right.\" width=\"487\" height=\"438\" \/> The graph of [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}[\/latex]<b>.<\/b>[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nThe population of bacteria after ten hours is 10,240. We could describe this amount as being of the order of magnitude [latex]{10}^{4}[\/latex]. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude [latex]{10}^{7}[\/latex], so we could say that the population has increased by three orders of magnitude in ten hours.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Calculating Doubling Time<\/h2>\r\nFor growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the <strong>doubling time<\/strong>.\r\n\r\nGiven the basic <strong>exponential growth<\/strong> equation [latex]A={A}_{0}{e}^{kt}[\/latex], doubling time can be found by solving for when the original quantity has doubled, that is, by solving [latex]2{A}_{0}={A}_{0}{e}^{kt}[\/latex].\r\n\r\nThe formula is derived as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2{A}_{0}={A}_{0}{e}^{kt}\\hfill &amp; \\hfill \\\\ 2={e}^{kt}\\hfill &amp; \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}2=kt\\hfill &amp; \\text{Take the natural logarithm of both sides}.\\hfill \\\\ t=\\frac{\\mathrm{ln}2}{k}\\hfill &amp; \\text{Divide by the coefficient of }t.\\hfill \\end{array}[\/latex]<\/p>\r\nThus the doubling time is\r\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}2}{k}[\/latex]<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Function That Describes Exponential Growth<\/h3>\r\nAccording to Moore\u2019s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.\r\n\r\n[reveal-answer q=\"567900\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"567900\"]\r\n\r\nThe formula is derived as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=\\frac{\\mathrm{ln}2}{k}\\hfill &amp; \\text{The doubling time formula}.\\hfill \\\\ 2=\\frac{\\mathrm{ln}2}{k}\\hfill &amp; \\text{Use a doubling time of two years}.\\hfill \\\\ k=\\frac{\\mathrm{ln}2}{2}\\hfill &amp; \\text{Multiply by }k\\text{ and divide by 2}.\\hfill \\\\ A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}\\hfill &amp; \\text{Substitute }k\\text{ into the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\r\nThe function is [latex]A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}[\/latex].\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\nRecent data suggests that, as of 2013, the rate of growth predicted by Moore\u2019s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.\r\n\r\n[reveal-answer q=\"271465\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"271465\"]\r\n\r\n[latex]f\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{3}t}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129930&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Half-Life<\/h2>\r\nWe now turn to <strong>exponential decay<\/strong>. One of the common terms associated with exponential decay, as stated above, is <strong>half-life<\/strong>, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.\r\n\r\nTo find the half-life of a function describing exponential decay, solve the following equation:\r\n<p style=\"text-align: center;\">[latex]\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}[\/latex]<\/p>\r\nWe find that the half-life depends only on the constant <em>k<\/em>\u00a0and not on the starting quantity [latex]{A}_{0}[\/latex].\r\n\r\nThe formula is derived as follows\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}\\hfill &amp; \\hfill \\\\ \\frac{1}{2}={e}^{kt}\\hfill &amp; \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{1}{2}\\right)=kt\\hfill &amp; \\text{Take the natural log of both sides}.\\hfill \\\\ -\\mathrm{ln}\\left(2\\right)=kt\\hfill &amp; \\text{Apply properties of logarithms}.\\hfill \\\\ -\\frac{\\mathrm{ln}\\left(2\\right)}{k}=t\\hfill &amp; \\text{Divide by }k.\\hfill \\end{array}[\/latex]<\/p>\r\nSince <em>t<\/em>, the time, is positive, <em>k<\/em>\u00a0must, as expected, be negative. This gives us the half-life formula\r\n<p style=\"text-align: center;\">[latex]t=-\\frac{\\mathrm{ln}\\left(2\\right)}{k}[\/latex]<\/p>\r\nIn previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.\r\n\r\nOne such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its <strong>half-life<\/strong>. The table below\u00a0lists the half-life for several of the more common radioactive substances.\r\n<table summary=\"Seven rows and three columns. The first column is labeled,\">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center;\">Substance<\/th>\r\n<th style=\"text-align: center;\">Use<\/th>\r\n<th style=\"text-align: center;\">Half-life<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>gallium-67<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>80 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>cobalt-60<\/td>\r\n<td>manufacturing<\/td>\r\n<td>5.3 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>technetium-99m<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>6 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>americium-241<\/td>\r\n<td>construction<\/td>\r\n<td>432 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>carbon-14<\/td>\r\n<td>archeological dating<\/td>\r\n<td>5,715 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>uranium-235<\/td>\r\n<td>atomic power<\/td>\r\n<td>703,800,000 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}A\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{T}t}\\hfill \\\\ A\\left(t\\right)={A}_{0}{e}^{\\mathrm{ln}\\left(0.5\\right)\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left({e}^{\\mathrm{ln}\\left(0.5\\right)}\\right)}^{\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left(\\frac{1}{2}\\right)}^{\\frac{t}{T}}\\hfill \\end{array}[\/latex]<\/p>\r\nwhere\r\n<ul>\r\n \t<li>[latex]{A}_{0}[\/latex] is the amount initially present<\/li>\r\n \t<li><em>T<\/em>\u00a0is the half-life of the substance<\/li>\r\n \t<li><em>t<\/em>\u00a0is the time period over which the substance is studied<\/li>\r\n \t<li><em>y<\/em>\u00a0is the amount of the substance present after time\u00a0<em>t<\/em><\/li>\r\n<\/ul>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the half-life, find the decay rate<\/h3>\r\n<ol>\r\n \t<li>Write [latex]A={A}_{o}{e}^{kt}[\/latex].<\/li>\r\n \t<li>Replace <em>A<\/em>\u00a0by [latex]\\frac{1}{2}{A}_{0}[\/latex] and replace <em>t<\/em>\u00a0by the given half-life.<\/li>\r\n \t<li>Solve to find <em>k<\/em>. Express <em>k<\/em>\u00a0as an exact value (do not round).<\/li>\r\n<\/ol>\r\nNote: <em>It is also possible to find the decay rate using<\/em> [latex]k=-\\frac{\\mathrm{ln}\\left(2\\right)}{t}[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Formula for Radioactive Decay to Find the Quantity of a Substance<\/h3>\r\nHow long will it take for 10% of a 1000-gram sample of uranium-235 to decay?\r\n\r\n[reveal-answer q=\"536683\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"536683\"]\r\n\r\n[latex]\\begin{array}{l}\\text{ }y=\\text{1000}e\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill &amp; \\hfill \\\\ \\text{ }900=1000{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill &amp; \\text{After 10% decays, 900 grams are left}.\\hfill \\\\ \\text{ }0.9={e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill &amp; \\text{Divide by 1000}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\mathrm{ln}\\left({e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\right)\\hfill &amp; \\text{Take ln of both sides}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill &amp; \\text{ln}\\left({e}^{M}\\right)=M\\hfill \\\\ \\text{}\\text{}t=\\text{703,800,000}\\times \\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{years}\\hfill &amp; \\text{Solve for }t.\\hfill \\\\ \\text{}\\text{}t\\approx \\text{106,979,777 years}\\hfill &amp; \\hfill \\end{array}[\/latex]\r\n<h4>Analysis of the Solution<\/h4>\r\nTen percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.\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\nHow long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?\r\n\r\n[reveal-answer q=\"923702\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"923702\"]\r\n\r\n[latex]t=703,800,000\\times \\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Function that Describes Radioactive Decay<\/h3>\r\nThe half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, <em>t<\/em>.\r\n\r\n[reveal-answer q=\"594419\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"594419\"]\r\n\r\nThis formula is derived as follows.\r\n\r\n[latex]\\begin{array}{l}\\text{}A={A}_{0}{e}^{kt}\\hfill &amp; \\text{The continuous growth formula}.\\hfill \\\\ 0.5{A}_{0}={A}_{0}{e}^{k\\cdot 5730}\\hfill &amp; \\text{Substitute the half-life for }t\\text{ and }0.5{A}_{0}\\text{ for }f\\left(t\\right).\\hfill \\\\ \\text{}0.5={e}^{5730k}\\hfill &amp; \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(0.5\\right)=5730k\\hfill &amp; \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{}k=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\hfill &amp; \\text{Divide by the coefficient of }k.\\hfill \\\\ \\text{}A={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}\\hfill &amp; \\text{Substitute for }r\\text{ in the continuous growth formula}.\\hfill \\end{array}[\/latex]\r\n\r\nThe function that describes this continuous decay is [latex]f\\left(t\\right)={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]. We observe that the coefficient of <em>t<\/em>, [latex]\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\approx -1.2097[\/latex] is negative, as expected in the case of exponential decay.\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 half-life of plutonium-244 is 80,000,000 years. Find a function that gives the amount of carbon-14 remaining as a function of time measured in years.\r\n\r\n[reveal-answer q=\"267261\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"267261\"]\r\n\r\n[latex]f\\left(t\\right)={A}_{0}{e}^{-0.0000000087t}[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom150\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=100026&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Radiocarbon Dating<\/h2>\r\nThe formula for radioactive decay is important in <strong>radiocarbon dating\u00a0<\/strong>which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.\r\n\r\nCarbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12 which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years using tree rings and other organic samples of known dates\u2014although the ratio has changed slightly over the centuries.\r\n\r\nAs long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.\r\n\r\nSince the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after <em>t<\/em>\u00a0years is\r\n<p style=\"text-align: center;\">[latex]A\\approx {A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]<\/p>\r\nwhere\r\n<ul>\r\n \t<li><em>A<\/em>\u00a0is the amount of carbon-14 remaining<\/li>\r\n \t<li>[latex]{A}_{0}[\/latex] is the amount of carbon-14 when the plant or animal began decaying.<\/li>\r\n<\/ul>\r\nThis formula is derived as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }A={A}_{0}{e}^{kt}\\hfill &amp; \\text{The continuous growth formula}.\\hfill \\\\ \\text{ }0.5{A}_{0}={A}_{0}{e}^{k\\cdot 5730}\\hfill &amp; \\text{Substitute the half-life for }t\\text{ and }0.5{A}_{0}\\text{ for }f\\left(t\\right).\\hfill \\\\ \\text{ }0.5={e}^{5730k}\\hfill &amp; \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(0.5\\right)=5730k\\hfill &amp; \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\hfill &amp; \\text{Divide both sides by the coefficient of }k.\\hfill \\\\ \\text{ }A={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}\\hfill &amp; \\text{Substitute for }r\\text{ in the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\r\nTo find the age of an object we solve this equation for <em>t<\/em>:\r\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}\\left(\\frac{A}{{A}_{0}}\\right)}{-0.000121}[\/latex]<\/p>\r\nOut of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let <em>r<\/em>\u00a0be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated determined by a method called liquid scintillation. From the equation [latex]A\\approx {A}_{0}{e}^{-0.000121t}[\/latex] we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is [latex]r=\\frac{A}{{A}_{0}}\\approx {e}^{-0.000121t}[\/latex]. We solve this equation for <em>t<\/em>, to get\r\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>How To: Given the percentage of carbon-14 in an object, determine its age<\/h3>\r\n<ol>\r\n \t<li>Express the given percentage of carbon-14 as an equivalent decimal <i>r<\/i>.<\/li>\r\n \t<li>Substitute for <i>r\u00a0<\/i>in the equation [latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex] and solve for the age, <em>t<\/em>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Age of a Bone<\/h3>\r\nA bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?\r\n\r\n[reveal-answer q=\"954630\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"954630\"]\r\n\r\nWe substitute 20% = 0.20 for <i>r\u00a0<\/i>in the equation and solve for <em>t<\/em>:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}\\hfill &amp; \\text{Use the general form of the equation}.\\hfill \\\\ =\\frac{\\mathrm{ln}\\left(0.20\\right)}{-0.000121}\\hfill &amp; \\text{Substitute for }r.\\hfill \\\\ \\approx 13301\\hfill &amp; \\text{Round to the nearest year}.\\hfill \\end{array}[\/latex]<\/p>\r\nThe bone fragment is about 13,301 years old.\r\n<h4>Analysis of the Solution<\/h4>\r\nThe instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as [latex]\\text{13,301 years}\\pm \\text{1% or 13,301 years}\\pm \\text{133 years}[\/latex].\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\nCesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?\r\n\r\n[reveal-answer q=\"758746\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"758746\"]\r\n\r\nless than 230 years; 229.3157 to be exact[\/hidden-answer]\r\n<iframe id=\"mom15\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29686&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Bounded Growth and Decay<\/h2>\r\n<h3>Using Newton\u2019s Law of Cooling<\/h3>\r\nExponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object\u2019s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a <strong>vertical shift<\/strong> of the generic <strong>exponential decay function<\/strong>. This translation leads to <strong>Newton\u2019s Law of Cooling<\/strong>, the scientific formula for temperature as a function of time as an object\u2019s temperature is equalized with the ambient temperature.\r\n\r\nThe formula is derived as follows:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}T\\left(t\\right)=A{b}^{ct}+{T}_{s}\\hfill &amp; \\hfill \\\\ T\\left(t\\right)=A{e}^{\\mathrm{ln}\\left({b}^{ct}\\right)}+{T}_{s}\\hfill &amp; \\text{Properties of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{ct\\mathrm{ln}b}+{T}_{s}\\hfill &amp; \\text{Properties of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{kt}+{T}_{s}\\hfill &amp; \\text{Rename the constant }c \\mathrm{ln} b,\\text{ calling it }k.\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Newton\u2019s Law of Cooling<\/h3>\r\nThe temperature of an object, <em>T<\/em>, in surrounding air with temperature [latex]{T}_{s}[\/latex] will behave according to the formula\r\n\r\n[latex]T\\left(t\\right)=A{e}^{kt}+{T}_{s}[\/latex]\r\nwhere\r\n<ul>\r\n \t<li><em>t<\/em>\u00a0is time<\/li>\r\n \t<li><em>A<\/em>\u00a0is the difference between the initial temperature of the object and the surroundings<\/li>\r\n \t<li><em>k<\/em>\u00a0is a constant, the continuous rate of cooling of the object<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a set of conditions, apply Newton\u2019s Law of Cooling<\/h3>\r\n<ol>\r\n \t<li>Set [latex]{T}_{s}[\/latex] equal to the <em>y<\/em>-coordinate of the horizontal asymptote (usually the ambient temperature).<\/li>\r\n \t<li>Substitute the given values into the continuous growth formula [latex]T\\left(t\\right)=A{e}^{k}{}^{t}+{T}_{s}[\/latex] to find the parameters <em>A<\/em>\u00a0and <em>k<\/em>.<\/li>\r\n \t<li>Substitute in the desired time to find the temperature or the desired temperature to find the time.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Newton\u2019s Law of Cooling<\/h3>\r\nA cheesecake is taken out of the oven with an ideal internal temperature of [latex]165^\\circ\\text{F}[\/latex] and is placed into a [latex]35^\\circ\\text{F}[\/latex] refrigerator. After 10 minutes, the cheesecake has cooled to [latex]150^\\circ\\text{F}[\/latex]. If we must wait until the cheesecake has cooled to [latex]70^\\circ\\text{F}[\/latex] before we eat it, how long will we have to wait?\r\n\r\n[reveal-answer q=\"705928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"705928\"]\r\n\r\nBecause the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake\u2019s temperature will decay exponentially toward 35, following the equation\r\n<p style=\"text-align: center;\">[latex]T\\left(t\\right)=A{e}^{kt}+35[\/latex]<\/p>\r\nWe know the initial temperature was 165, so [latex]T\\left(0\\right)=165[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}165=A{e}^{k0}+35\\hfill &amp; \\text{Substitute }\\left(0,165\\right).\\hfill \\\\ A=130\\hfill &amp; \\text{Solve for }A.\\hfill \\end{array}[\/latex]<\/p>\r\nWe were given another data point, [latex]T\\left(10\\right)=150[\/latex], which we can use to solve for <em>k<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }150=130{e}^{k10}+35\\hfill &amp; \\text{Substitute (10, 150)}.\\hfill \\\\ \\text{ }115=130{e}^{k10}\\hfill &amp; \\text{Subtract 35 from both sides}.\\hfill \\\\ \\text{ }\\frac{115}{130}={e}^{10k}\\hfill &amp; \\text{Divide both sides by 130}.\\hfill \\\\ \\text{ }\\mathrm{ln}\\left(\\frac{115}{130}\\right)=10k\\hfill &amp; \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(\\frac{115}{130}\\right)}{10}=-0.0123\\hfill &amp; \\text{Divide both sides by the coefficient of }k.\\hfill \\end{array}[\/latex]<\/p>\r\nThis gives us the equation for the cooling of the cheesecake: [latex]T\\left(t\\right)=130{e}^{-0.0123t}+35[\/latex].\r\n\r\nNow we can solve for the time it will take for the temperature to cool to 70 degrees.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}70=130{e}^{-0.0123t}+35\\hfill &amp; \\text{Substitute in 70 for }T\\left(t\\right).\\hfill \\\\ 35=130{e}^{-0.0123t}\\hfill &amp; \\text{Subtract 35 from both sides}.\\hfill \\\\ \\frac{35}{130}={e}^{-0.0123t}\\hfill &amp; \\text{Divide both sides by 130}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{35}{130}\\right)=-0.0123t\\hfill &amp; \\text{Take the natural log of both sides}.\\hfill \\\\ t=\\frac{\\mathrm{ln}\\left(\\frac{35}{130}\\right)}{-0.0123}\\approx 106.68\\hfill &amp; \\text{Divide both sides by the coefficient of }t.\\hfill \\end{array}[\/latex]<\/p>\r\nIt will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to [latex]70^\\circ\\text{F}[\/latex].\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\nA pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?\r\n\r\n[reveal-answer q=\"846066\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"846066\"]6.026 hours[\/hidden-answer]\r\n<iframe id=\"mom150\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114392&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Logistic-Growth Model<\/h3>\r\nAn influenza epidemic spreads through a population rapidly at a rate that depends on two factors. The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model good for studying the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.\r\n\r\nFor example, at time <em>t\u00a0<\/em>= 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is <em>b\u00a0<\/em>= 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.\r\n\r\n[reveal-answer q=\"346660\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"346660\"]\r\n\r\nWe substitute the given data into the logistic growth model\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\r\nBecause at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is <em>c\u00a0<\/em>= 1000. To find <em>a<\/em>, we use the formula that the number of cases at time <em>t\u00a0<\/em>= 0 is [latex]\\frac{c}{1+a}=1[\/latex], from which it follows that <em>a\u00a0<\/em>= 999. This model predicts that, after ten days, the number of people who have had the flu is [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}\\approx 293.8[\/latex]. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, <em>c\u00a0<\/em>= 1000.\r\n<h4>Analysis of the Solution<\/h4>\r\nRemember that because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.\r\n\r\nThe graph below gives a good picture of how this model fits the data.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181349\/CNX_Precalc_Figure_04_07_0072.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as\" width=\"731\" height=\"492\" \/> The graph of [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUsing the model in the previous example, estimate the number of cases of flu on day 15.\r\n\r\n[reveal-answer q=\"421768\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"421768\"]\r\n\r\n895 cases on day 15[\/hidden-answer]\r\n<iframe id=\"mom150\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5801&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"300\"><\/iframe>\r\n\r\n<\/div>\r\n<h2><\/h2>\r\n<h2><span style=\"font-size: 1.15em;\">Key Concepts<\/span><\/h2>\r\n<ul>\r\n \t<li>The basic exponential function is [latex]f\\left(x\\right)=a{b}^{x}[\/latex]. If <em>b\u00a0<\/em>&gt; 1, we have exponential growth; if 0 &lt; <em>b\u00a0<\/em>&lt; 1, we have exponential decay.<\/li>\r\n \t<li>We can also write [latex]f\\left(x\\right)=a{b}^{x}[\/latex] in terms of continuous growth as [latex]A={A}_{0}{e}^{kx}[\/latex], where [latex]{A}_{0}[\/latex] is the starting value. If [latex]{A}_{0}[\/latex] is positive, then we have exponential growth when <em>k\u00a0<\/em>&gt; 0 and exponential decay when <em>k\u00a0<\/em>&lt; 0.<\/li>\r\n \t<li>We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors.<\/li>\r\n \t<li>We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data.<\/li>\r\n \t<li>Exponential growth functions are used to model situations where growth begins slowly and then accelerates rapidly without bound or where decay begins rapidly and then slows down to get closer and closer to zero.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>carrying capacity<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">in a logistic model, the limiting value of the output<\/dd>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>doubling time<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">the time it takes for a quantity to double<\/dd>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>half-life<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">the length of time it takes for a substance to exponentially decay to half of its original quantity<\/dd>\r\n<\/dl>\r\n<dl class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><strong>logistic growth model<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137838640\">a function of the form [latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex] where [latex]\\frac{c}{1+a}[\/latex] is the initial value, <em>c<\/em>\u00a0is the carrying capacity, or limiting value, and <em>b<\/em>\u00a0is a constant determined by the rate of growth<\/dd>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Graph exponential growth and decay functions.<\/li>\n<li>Use the compound interest formula.<\/li>\n<li>Solve problems involving radioactive decay, carbon dating, and half life.<\/li>\n<li>Use Newton&#8217;s Law of Cooling.<\/li>\n<li>Use a logistic growth model.<\/li>\n<li>Choose an appropriate model for data.<\/li>\n<\/ul>\n<\/div>\n<p>We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth including radioactive isotopes and Newton\u2019s Law of Cooling.<\/p>\n<div style=\"width: 335px\" 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\/03181335\/CNX_Precalc_Figure_04_07_001F2.jpg\" alt=\"Inside a nuclear research reactor.\" width=\"325\" height=\"409\" \/><\/p>\n<p class=\"wp-caption-text\">A nuclear research reactor inside the Neely Nuclear Research Center on the Georgia Institute of Technology campus. (credit: Georgia Tech Research Institute)<\/p>\n<\/div>\n<h2><\/h2>\n<h2>Exponential Growth and Decay<\/h2>\n<p>In real-world applications, we need to model the behavior of a function. In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:<\/p>\n<p style=\"text-align: center;\">[latex]y={A}_{0}{e}^{kt}[\/latex]<\/p>\n<p>where [latex]{A}_{0}[\/latex] is equal to the value at time zero, <em>e<\/em>\u00a0is Euler\u2019s constant, and <em>k<\/em>\u00a0is a positive constant that determines the rate (percentage) of growth. We may use the <strong>exponential growth<\/strong> function in applications involving <strong>doubling time<\/strong>, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based odn a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.<\/p>\n<p>On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the <strong>exponential decay<\/strong> model. Again, we have the form [latex]y={A}_{0}{e}^{-kt}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, and <em>e<\/em>\u00a0is Euler\u2019s constant. Now <em>k<\/em>\u00a0is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating <strong>half-life<\/strong>, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.<\/p>\n<p>In our choice of a function to serve as a mathematical model, we often use data points gathered by careful observation and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and decay graphs have a distinctive shape, as we can see in the graphs below. It is important to remember that, although parts of each of the two graphs seem to lie on the <em>x<\/em>-axis, they are really a tiny distance above the <em>x<\/em>-axis.<\/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\/03181337\/CNX_Precalc_Figure_04_07_0022.jpg\" alt=\"Graph of y=2e^(3x) with the labeled points (-1\/3, 2\/e), (0, 2), and (1\/3, 2e) and with the asymptote at y=0.\" width=\"487\" height=\"326\" \/><\/p>\n<p class=\"wp-caption-text\">A graph showing exponential growth. The equation is [latex]y=2{e}^{3x}[\/latex].<\/p>\n<\/div>\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\/03181339\/CNX_Precalc_Figure_04_07_0032.jpg\" alt=\"Graph of y=3e^(-2x) with the labeled points (-1\/2, 3e), (0, 3), and (1\/2, 3\/e) and with the asymptote at y=0.\" width=\"487\" height=\"438\" \/><\/p>\n<p class=\"wp-caption-text\">A graph showing exponential decay. The equation is [latex]y=3{e}^{-2x}[\/latex].<\/p>\n<\/div>\n<p>Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The <strong>order of magnitude<\/strong> is the power of ten when the number is expressed in scientific notation with one digit to the left of the decimal. For example, the distance to the nearest star, <strong>Proxima Centauri<\/strong>, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is [latex]4.01134972\\times {10}^{13}[\/latex]. We could describe this number as having order of magnitude [latex]{10}^{13}[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Characteristics of the Exponential Function [latex]y=A_{0}e^{kt}[\/latex]<\/h3>\n<p>An exponential function of the form [latex]y={A}_{0}{e}^{kt}[\/latex] has the following characteristics:<\/p>\n<ul>\n<li>one-to-one function<\/li>\n<li>horizontal asymptote: <em>y\u00a0<\/em>= 0<\/li>\n<li>domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\n<li>range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\n<li>x intercept: none<\/li>\n<li>y-intercept: [latex]\\left(0,{A}_{0}\\right)[\/latex]<\/li>\n<li>increasing if <em>k\u00a0<\/em>&gt; 0<\/li>\n<li>decreasing if <em>k\u00a0<\/em>&lt; 0<\/li>\n<\/ul>\n<p>An exponential function models exponential growth when k\u00a0&gt; 0 and exponential decay when k\u00a0&lt; 0.<\/p>\n<\/div>\n<h3><\/h3>\n<h3>Using the Compound Interest Formula<\/h3>\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>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>Frequency<\/th>\n<th>Value after 1 year<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Annually<\/td>\n<td>$1100<\/td>\n<\/tr>\n<tr>\n<td>Semiannually<\/td>\n<td>$1102.50<\/td>\n<\/tr>\n<tr>\n<td>Quarterly<\/td>\n<td>$1103.81<\/td>\n<\/tr>\n<tr>\n<td>Monthly<\/td>\n<td>$1104.71<\/td>\n<\/tr>\n<tr>\n<td>Daily<\/td>\n<td>$1105.16<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>&nbsp;<\/p>\n<p>We can calculate 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 times compounded in a year\u00a0<em>n.<\/em><\/p>\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 accumulated value of the account<\/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<\/li>\n<li><em>n<\/em>\u00a0is the number of times compounded in a 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\">Show Solution<\/span><\/p>\n<div id=\"q476919\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because 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}{llllll}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\">Show Solution<\/span><\/p>\n<div id=\"q939452\" class=\"hidden-answer\" style=\"display: none\">\n<p>about $3,644,675.88<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.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\">Show Solution<\/span><\/p>\n<div id=\"q550421\" class=\"hidden-answer\" style=\"display: none\">\n<p>The nominal interest rate is 6%, so <em>r\u00a0<\/em>= 0.06. Interest is compounded twice a year, so <em>n<\/em><em>\u00a0<\/em>= 2.<\/p>\n<p>We want to find the initial investment\u00a0<em>P\u00a0<\/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\">Show Solution<\/span><\/p>\n<div id=\"q895101\" class=\"hidden-answer\" style=\"display: none\">\n<p>$13,693<\/p><\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<h3><\/h3>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing Exponential Growth<\/h3>\n<p>A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q151444\">Show Solution<\/span><\/p>\n<div id=\"q151444\" class=\"hidden-answer\" style=\"display: none\">\n<p>When an amount grows at a fixed percent per unit time, the growth is exponential. To find [latex]{A}_{0}[\/latex] we use the fact that [latex]{A}_{0}[\/latex] is the amount at time zero, so [latex]{A}_{0}=10[\/latex]. To find <em>k<\/em>, use the fact that after one hour [latex]\\left(t=1\\right)[\/latex] the population doubles from 10 to 20. The formula is derived as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }20=10{e}^{k\\cdot 1}\\hfill & \\hfill \\\\ \\text{ }2={e}^{k}\\hfill & \\text{Divide both sides by 10}\\hfill \\\\ \\mathrm{ln}2=k\\hfill & \\text{Take the natural logarithm of both sides}\\hfill \\end{array}[\/latex]<\/p>\n<p>so [latex]k=\\mathrm{ln}\\left(2\\right)[\/latex]. Thus the equation we want to graph is [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}=10{\\left({e}^{\\mathrm{ln}2}\\right)}^{t}=10\\cdot {2}^{t}[\/latex]. The graph is shown below.<\/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\/03181344\/CNX_Precalc_Figure_04_07_0052.jpg\" alt=\"A graph starting at ten on the y-axis and rising rapidly to the right.\" width=\"487\" height=\"438\" \/><\/p>\n<p class=\"wp-caption-text\">The graph of [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}[\/latex]<b>.<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>The population of bacteria after ten hours is 10,240. We could describe this amount as being of the order of magnitude [latex]{10}^{4}[\/latex]. The population of bacteria after twenty hours is 10,485,760 which is of the order of magnitude [latex]{10}^{7}[\/latex], so we could say that the population has increased by three orders of magnitude in ten hours.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Calculating Doubling Time<\/h2>\n<p>For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the <strong>doubling time<\/strong>.<\/p>\n<p>Given the basic <strong>exponential growth<\/strong> equation [latex]A={A}_{0}{e}^{kt}[\/latex], doubling time can be found by solving for when the original quantity has doubled, that is, by solving [latex]2{A}_{0}={A}_{0}{e}^{kt}[\/latex].<\/p>\n<p>The formula is derived as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2{A}_{0}={A}_{0}{e}^{kt}\\hfill & \\hfill \\\\ 2={e}^{kt}\\hfill & \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}2=kt\\hfill & \\text{Take the natural logarithm of both sides}.\\hfill \\\\ t=\\frac{\\mathrm{ln}2}{k}\\hfill & \\text{Divide by the coefficient of }t.\\hfill \\end{array}[\/latex]<\/p>\n<p>Thus the doubling time is<\/p>\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}2}{k}[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Function That Describes Exponential Growth<\/h3>\n<p>According to Moore\u2019s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q567900\">Show Solution<\/span><\/p>\n<div id=\"q567900\" class=\"hidden-answer\" style=\"display: none\">\n<p>The formula is derived as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=\\frac{\\mathrm{ln}2}{k}\\hfill & \\text{The doubling time formula}.\\hfill \\\\ 2=\\frac{\\mathrm{ln}2}{k}\\hfill & \\text{Use a doubling time of two years}.\\hfill \\\\ k=\\frac{\\mathrm{ln}2}{2}\\hfill & \\text{Multiply by }k\\text{ and divide by 2}.\\hfill \\\\ A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}\\hfill & \\text{Substitute }k\\text{ into the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\n<p>The function is [latex]A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Recent data suggests that, as of 2013, the rate of growth predicted by Moore\u2019s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q271465\">Show Solution<\/span><\/p>\n<div id=\"q271465\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{3}t}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129930&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Half-Life<\/h2>\n<p>We now turn to <strong>exponential decay<\/strong>. One of the common terms associated with exponential decay, as stated above, is <strong>half-life<\/strong>, the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.<\/p>\n<p>To find the half-life of a function describing exponential decay, solve the following equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}[\/latex]<\/p>\n<p>We find that the half-life depends only on the constant <em>k<\/em>\u00a0and not on the starting quantity [latex]{A}_{0}[\/latex].<\/p>\n<p>The formula is derived as follows<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}\\hfill & \\hfill \\\\ \\frac{1}{2}={e}^{kt}\\hfill & \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{1}{2}\\right)=kt\\hfill & \\text{Take the natural log of both sides}.\\hfill \\\\ -\\mathrm{ln}\\left(2\\right)=kt\\hfill & \\text{Apply properties of logarithms}.\\hfill \\\\ -\\frac{\\mathrm{ln}\\left(2\\right)}{k}=t\\hfill & \\text{Divide by }k.\\hfill \\end{array}[\/latex]<\/p>\n<p>Since <em>t<\/em>, the time, is positive, <em>k<\/em>\u00a0must, as expected, be negative. This gives us the half-life formula<\/p>\n<p style=\"text-align: center;\">[latex]t=-\\frac{\\mathrm{ln}\\left(2\\right)}{k}[\/latex]<\/p>\n<p>In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/p>\n<p>One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its <strong>half-life<\/strong>. The table below\u00a0lists the half-life for several of the more common radioactive substances.<\/p>\n<table summary=\"Seven rows and three columns. The first column is labeled,\">\n<thead>\n<tr>\n<th style=\"text-align: center;\">Substance<\/th>\n<th style=\"text-align: center;\">Use<\/th>\n<th style=\"text-align: center;\">Half-life<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>gallium-67<\/td>\n<td>nuclear medicine<\/td>\n<td>80 hours<\/td>\n<\/tr>\n<tr>\n<td>cobalt-60<\/td>\n<td>manufacturing<\/td>\n<td>5.3 years<\/td>\n<\/tr>\n<tr>\n<td>technetium-99m<\/td>\n<td>nuclear medicine<\/td>\n<td>6 hours<\/td>\n<\/tr>\n<tr>\n<td>americium-241<\/td>\n<td>construction<\/td>\n<td>432 years<\/td>\n<\/tr>\n<tr>\n<td>carbon-14<\/td>\n<td>archeological dating<\/td>\n<td>5,715 years<\/td>\n<\/tr>\n<tr>\n<td>uranium-235<\/td>\n<td>atomic power<\/td>\n<td>703,800,000 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}A\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{T}t}\\hfill \\\\ A\\left(t\\right)={A}_{0}{e}^{\\mathrm{ln}\\left(0.5\\right)\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left({e}^{\\mathrm{ln}\\left(0.5\\right)}\\right)}^{\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left(\\frac{1}{2}\\right)}^{\\frac{t}{T}}\\hfill \\end{array}[\/latex]<\/p>\n<p>where<\/p>\n<ul>\n<li>[latex]{A}_{0}[\/latex] is the amount initially present<\/li>\n<li><em>T<\/em>\u00a0is the half-life of the substance<\/li>\n<li><em>t<\/em>\u00a0is the time period over which the substance is studied<\/li>\n<li><em>y<\/em>\u00a0is the amount of the substance present after time\u00a0<em>t<\/em><\/li>\n<\/ul>\n<div class=\"textbox\">\n<h3>How To: Given the half-life, find the decay rate<\/h3>\n<ol>\n<li>Write [latex]A={A}_{o}{e}^{kt}[\/latex].<\/li>\n<li>Replace <em>A<\/em>\u00a0by [latex]\\frac{1}{2}{A}_{0}[\/latex] and replace <em>t<\/em>\u00a0by the given half-life.<\/li>\n<li>Solve to find <em>k<\/em>. Express <em>k<\/em>\u00a0as an exact value (do not round).<\/li>\n<\/ol>\n<p>Note: <em>It is also possible to find the decay rate using<\/em> [latex]k=-\\frac{\\mathrm{ln}\\left(2\\right)}{t}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Formula for Radioactive Decay to Find the Quantity of a Substance<\/h3>\n<p>How long will it take for 10% of a 1000-gram sample of uranium-235 to decay?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q536683\">Show Solution<\/span><\/p>\n<div id=\"q536683\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}\\text{ }y=\\text{1000}e\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill & \\hfill \\\\ \\text{ }900=1000{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill & \\text{After 10% decays, 900 grams are left}.\\hfill \\\\ \\text{ }0.9={e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill & \\text{Divide by 1000}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\mathrm{ln}\\left({e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\right)\\hfill & \\text{Take ln of both sides}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill & \\text{ln}\\left({e}^{M}\\right)=M\\hfill \\\\ \\text{}\\text{}t=\\text{703,800,000}\\times \\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{years}\\hfill & \\text{Solve for }t.\\hfill \\\\ \\text{}\\text{}t\\approx \\text{106,979,777 years}\\hfill & \\hfill \\end{array}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q923702\">Show Solution<\/span><\/p>\n<div id=\"q923702\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]t=703,800,000\\times \\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Function that Describes Radioactive Decay<\/h3>\n<p>The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, <em>t<\/em>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q594419\">Show Solution<\/span><\/p>\n<div id=\"q594419\" class=\"hidden-answer\" style=\"display: none\">\n<p>This formula is derived as follows.<\/p>\n<p>[latex]\\begin{array}{l}\\text{}A={A}_{0}{e}^{kt}\\hfill & \\text{The continuous growth formula}.\\hfill \\\\ 0.5{A}_{0}={A}_{0}{e}^{k\\cdot 5730}\\hfill & \\text{Substitute the half-life for }t\\text{ and }0.5{A}_{0}\\text{ for }f\\left(t\\right).\\hfill \\\\ \\text{}0.5={e}^{5730k}\\hfill & \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(0.5\\right)=5730k\\hfill & \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{}k=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\hfill & \\text{Divide by the coefficient of }k.\\hfill \\\\ \\text{}A={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}\\hfill & \\text{Substitute for }r\\text{ in the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\n<p>The function that describes this continuous decay is [latex]f\\left(t\\right)={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]. We observe that the coefficient of <em>t<\/em>, [latex]\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\approx -1.2097[\/latex] is negative, as expected in the case of exponential decay.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The half-life of plutonium-244 is 80,000,000 years. Find a function that gives the amount of carbon-14 remaining as a function of time measured in years.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q267261\">Show Solution<\/span><\/p>\n<div id=\"q267261\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(t\\right)={A}_{0}{e}^{-0.0000000087t}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom150\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=100026&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Radiocarbon Dating<\/h2>\n<p>The formula for radioactive decay is important in <strong>radiocarbon dating\u00a0<\/strong>which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.<\/p>\n<p>Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12 which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years using tree rings and other organic samples of known dates\u2014although the ratio has changed slightly over the centuries.<\/p>\n<p>As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.<\/p>\n<p>Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after <em>t<\/em>\u00a0years is<\/p>\n<p style=\"text-align: center;\">[latex]A\\approx {A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]<\/p>\n<p>where<\/p>\n<ul>\n<li><em>A<\/em>\u00a0is the amount of carbon-14 remaining<\/li>\n<li>[latex]{A}_{0}[\/latex] is the amount of carbon-14 when the plant or animal began decaying.<\/li>\n<\/ul>\n<p>This formula is derived as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }A={A}_{0}{e}^{kt}\\hfill & \\text{The continuous growth formula}.\\hfill \\\\ \\text{ }0.5{A}_{0}={A}_{0}{e}^{k\\cdot 5730}\\hfill & \\text{Substitute the half-life for }t\\text{ and }0.5{A}_{0}\\text{ for }f\\left(t\\right).\\hfill \\\\ \\text{ }0.5={e}^{5730k}\\hfill & \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(0.5\\right)=5730k\\hfill & \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\hfill & \\text{Divide both sides by the coefficient of }k.\\hfill \\\\ \\text{ }A={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}\\hfill & \\text{Substitute for }r\\text{ in the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\n<p>To find the age of an object we solve this equation for <em>t<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}\\left(\\frac{A}{{A}_{0}}\\right)}{-0.000121}[\/latex]<\/p>\n<p>Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let <em>r<\/em>\u00a0be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated determined by a method called liquid scintillation. From the equation [latex]A\\approx {A}_{0}{e}^{-0.000121t}[\/latex] we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is [latex]r=\\frac{A}{{A}_{0}}\\approx {e}^{-0.000121t}[\/latex]. We solve this equation for <em>t<\/em>, to get<\/p>\n<p style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the percentage of carbon-14 in an object, determine its age<\/h3>\n<ol>\n<li>Express the given percentage of carbon-14 as an equivalent decimal <i>r<\/i>.<\/li>\n<li>Substitute for <i>r\u00a0<\/i>in the equation [latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex] and solve for the age, <em>t<\/em>.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Age of a Bone<\/h3>\n<p>A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q954630\">Show Solution<\/span><\/p>\n<div id=\"q954630\" class=\"hidden-answer\" style=\"display: none\">\n<p>We substitute 20% = 0.20 for <i>r\u00a0<\/i>in the equation and solve for <em>t<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}\\hfill & \\text{Use the general form of the equation}.\\hfill \\\\ =\\frac{\\mathrm{ln}\\left(0.20\\right)}{-0.000121}\\hfill & \\text{Substitute for }r.\\hfill \\\\ \\approx 13301\\hfill & \\text{Round to the nearest year}.\\hfill \\end{array}[\/latex]<\/p>\n<p>The bone fragment is about 13,301 years old.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as [latex]\\text{13,301 years}\\pm \\text{1% or 13,301 years}\\pm \\text{133 years}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q758746\">Show Solution<\/span><\/p>\n<div id=\"q758746\" class=\"hidden-answer\" style=\"display: none\">\n<p>less than 230 years; 229.3157 to be exact<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom15\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29686&amp;theme=oea&amp;iframe_resize_id=mom15\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Bounded Growth and Decay<\/h2>\n<h3>Using Newton\u2019s Law of Cooling<\/h3>\n<p>Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object\u2019s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a <strong>vertical shift<\/strong> of the generic <strong>exponential decay function<\/strong>. This translation leads to <strong>Newton\u2019s Law of Cooling<\/strong>, the scientific formula for temperature as a function of time as an object\u2019s temperature is equalized with the ambient temperature.<\/p>\n<p>The formula is derived as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}T\\left(t\\right)=A{b}^{ct}+{T}_{s}\\hfill & \\hfill \\\\ T\\left(t\\right)=A{e}^{\\mathrm{ln}\\left({b}^{ct}\\right)}+{T}_{s}\\hfill & \\text{Properties of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{ct\\mathrm{ln}b}+{T}_{s}\\hfill & \\text{Properties of logarithms}.\\hfill \\\\ T\\left(t\\right)=A{e}^{kt}+{T}_{s}\\hfill & \\text{Rename the constant }c \\mathrm{ln} b,\\text{ calling it }k.\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Newton\u2019s Law of Cooling<\/h3>\n<p>The temperature of an object, <em>T<\/em>, in surrounding air with temperature [latex]{T}_{s}[\/latex] will behave according to the formula<\/p>\n<p>[latex]T\\left(t\\right)=A{e}^{kt}+{T}_{s}[\/latex]<br \/>\nwhere<\/p>\n<ul>\n<li><em>t<\/em>\u00a0is time<\/li>\n<li><em>A<\/em>\u00a0is the difference between the initial temperature of the object and the surroundings<\/li>\n<li><em>k<\/em>\u00a0is a constant, the continuous rate of cooling of the object<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a set of conditions, apply Newton\u2019s Law of Cooling<\/h3>\n<ol>\n<li>Set [latex]{T}_{s}[\/latex] equal to the <em>y<\/em>-coordinate of the horizontal asymptote (usually the ambient temperature).<\/li>\n<li>Substitute the given values into the continuous growth formula [latex]T\\left(t\\right)=A{e}^{k}{}^{t}+{T}_{s}[\/latex] to find the parameters <em>A<\/em>\u00a0and <em>k<\/em>.<\/li>\n<li>Substitute in the desired time to find the temperature or the desired temperature to find the time.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Newton\u2019s Law of Cooling<\/h3>\n<p>A cheesecake is taken out of the oven with an ideal internal temperature of [latex]165^\\circ\\text{F}[\/latex] and is placed into a [latex]35^\\circ\\text{F}[\/latex] refrigerator. After 10 minutes, the cheesecake has cooled to [latex]150^\\circ\\text{F}[\/latex]. If we must wait until the cheesecake has cooled to [latex]70^\\circ\\text{F}[\/latex] before we eat it, how long will we have to wait?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q705928\">Show Solution<\/span><\/p>\n<div id=\"q705928\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake\u2019s temperature will decay exponentially toward 35, following the equation<\/p>\n<p style=\"text-align: center;\">[latex]T\\left(t\\right)=A{e}^{kt}+35[\/latex]<\/p>\n<p>We know the initial temperature was 165, so [latex]T\\left(0\\right)=165[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}165=A{e}^{k0}+35\\hfill & \\text{Substitute }\\left(0,165\\right).\\hfill \\\\ A=130\\hfill & \\text{Solve for }A.\\hfill \\end{array}[\/latex]<\/p>\n<p>We were given another data point, [latex]T\\left(10\\right)=150[\/latex], which we can use to solve for <em>k<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }150=130{e}^{k10}+35\\hfill & \\text{Substitute (10, 150)}.\\hfill \\\\ \\text{ }115=130{e}^{k10}\\hfill & \\text{Subtract 35 from both sides}.\\hfill \\\\ \\text{ }\\frac{115}{130}={e}^{10k}\\hfill & \\text{Divide both sides by 130}.\\hfill \\\\ \\text{ }\\mathrm{ln}\\left(\\frac{115}{130}\\right)=10k\\hfill & \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{ }k=\\frac{\\mathrm{ln}\\left(\\frac{115}{130}\\right)}{10}=-0.0123\\hfill & \\text{Divide both sides by the coefficient of }k.\\hfill \\end{array}[\/latex]<\/p>\n<p>This gives us the equation for the cooling of the cheesecake: [latex]T\\left(t\\right)=130{e}^{-0.0123t}+35[\/latex].<\/p>\n<p>Now we can solve for the time it will take for the temperature to cool to 70 degrees.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}70=130{e}^{-0.0123t}+35\\hfill & \\text{Substitute in 70 for }T\\left(t\\right).\\hfill \\\\ 35=130{e}^{-0.0123t}\\hfill & \\text{Subtract 35 from both sides}.\\hfill \\\\ \\frac{35}{130}={e}^{-0.0123t}\\hfill & \\text{Divide both sides by 130}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{35}{130}\\right)=-0.0123t\\hfill & \\text{Take the natural log of both sides}.\\hfill \\\\ t=\\frac{\\mathrm{ln}\\left(\\frac{35}{130}\\right)}{-0.0123}\\approx 106.68\\hfill & \\text{Divide both sides by the coefficient of }t.\\hfill \\end{array}[\/latex]<\/p>\n<p>It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to [latex]70^\\circ\\text{F}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q846066\">Show Solution<\/span><\/p>\n<div id=\"q846066\" class=\"hidden-answer\" style=\"display: none\">6.026 hours<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom150\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114392&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Logistic-Growth Model<\/h3>\n<p>An influenza epidemic spreads through a population rapidly at a rate that depends on two factors. The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model good for studying the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.<\/p>\n<p>For example, at time <em>t\u00a0<\/em>= 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is <em>b\u00a0<\/em>= 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q346660\">Show Solution<\/span><\/p>\n<div id=\"q346660\" class=\"hidden-answer\" style=\"display: none\">\n<p>We substitute the given data into the logistic growth model<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\n<p>Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is <em>c\u00a0<\/em>= 1000. To find <em>a<\/em>, we use the formula that the number of cases at time <em>t\u00a0<\/em>= 0 is [latex]\\frac{c}{1+a}=1[\/latex], from which it follows that <em>a\u00a0<\/em>= 999. This model predicts that, after ten days, the number of people who have had the flu is [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}\\approx 293.8[\/latex]. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, <em>c\u00a0<\/em>= 1000.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Remember that because we are dealing with a virus, we cannot predict with certainty the number of people infected. The model only approximates the number of people infected and will not give us exact or actual values.<\/p>\n<p>The graph below gives a good picture of how this model fits the data.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03181349\/CNX_Precalc_Figure_04_07_0072.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as\" width=\"731\" height=\"492\" \/><\/p>\n<p class=\"wp-caption-text\">The graph of [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Using the model in the previous example, estimate the number of cases of flu on day 15.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q421768\">Show Solution<\/span><\/p>\n<div id=\"q421768\" class=\"hidden-answer\" style=\"display: none\">\n<p>895 cases on day 15<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom150\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5801&amp;theme=oea&amp;iframe_resize_id=mom150\" width=\"100%\" height=\"300\"><\/iframe><\/p>\n<\/div>\n<h2><\/h2>\n<h2><span style=\"font-size: 1.15em;\">Key Concepts<\/span><\/h2>\n<ul>\n<li>The basic exponential function is [latex]f\\left(x\\right)=a{b}^{x}[\/latex]. If <em>b\u00a0<\/em>&gt; 1, we have exponential growth; if 0 &lt; <em>b\u00a0<\/em>&lt; 1, we have exponential decay.<\/li>\n<li>We can also write [latex]f\\left(x\\right)=a{b}^{x}[\/latex] in terms of continuous growth as [latex]A={A}_{0}{e}^{kx}[\/latex], where [latex]{A}_{0}[\/latex] is the starting value. If [latex]{A}_{0}[\/latex] is positive, then we have exponential growth when <em>k\u00a0<\/em>&gt; 0 and exponential decay when <em>k\u00a0<\/em>&lt; 0.<\/li>\n<li>We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors.<\/li>\n<li>We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data.<\/li>\n<li>Exponential growth functions are used to model situations where growth begins slowly and then accelerates rapidly without bound or where decay begins rapidly and then slows down to get closer and closer to zero.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt><strong>carrying capacity<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">in a logistic model, the limiting value of the output<\/dd>\n<\/dl>\n<dl class=\"definition\">\n<dt>\n<\/dt>\n<dt><strong>doubling time<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">the time it takes for a quantity to double<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt><strong>half-life<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">the length of time it takes for a substance to exponentially decay to half of its original quantity<\/dd>\n<\/dl>\n<dl class=\"definition\">\n<dt>\n<\/dt>\n<dt><strong>logistic growth model<\/strong><\/dt>\n<dd id=\"fs-id1165137838640\">a function of the form [latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex] where [latex]\\frac{c}{1+a}[\/latex] is the initial value, <em>c<\/em>\u00a0is the carrying capacity, or limiting value, and <em>b<\/em>\u00a0is a constant determined by the rate of growth<\/dd>\n<\/dl>\n","protected":false},"author":167848,"menu_order":10,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5380","chapter","type-chapter","status-publish","hentry"],"part":5352,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5380","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/users\/167848"}],"version-history":[{"count":16,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5380\/revisions"}],"predecessor-version":[{"id":5755,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5380\/revisions\/5755"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/5352"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5380\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/media?parent=5380"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=5380"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/contributor?post=5380"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/license?post=5380"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}