{"id":476,"date":"2019-03-07T13:15:02","date_gmt":"2019-03-07T13:15:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/exponential-and-logarithmic-models\/"},"modified":"2025-03-31T20:22:32","modified_gmt":"2025-03-31T20:22:32","slug":"exponential-and-logarithmic-models","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/exponential-and-logarithmic-models\/","title":{"raw":"2.7 Exponential and Logarithmic Models","rendered":"2.7 Exponential and Logarithmic Models"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nIn this section, you will:\r\n<ul>\r\n \t<li>Model exponential growth and decay.<\/li>\r\n \t<li>Doubling time and half-life.<\/li>\r\n<\/ul>\r\nOptional:\r\n<ul>\r\n \t<li>Use Newton\u2019s Law of Cooling.<\/li>\r\n \t<li>Use logistic-growth models.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"CNX_Precalc_Figure_04_07_001\" class=\"small\">\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\/3896\/2019\/03\/07131409\/CNX_Precalc_Figure_04_07_001F.jpg\" alt=\"Inside a nuclear research reactor.\" width=\"325\" height=\"409\" \/> <strong>Figure 1.\u00a0<\/strong>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\r\n<\/div>\r\n<p id=\"fs-id1165134081045\">We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth.<\/p>\r\n\r\n<div id=\"fs-id1165135190498\" class=\"bc-section section\">\r\n<h3>Modeling Exponential Growth and Decay<\/h3>\r\n<p id=\"fs-id1165135169375\">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>\r\n\r\n<div class=\"unnumbered\" style=\"text-align: center;\">[latex]y={A}_{0}{e}^{kt}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"eip-292\">where [latex]{A}_{0}[\/latex] is equal to the value at time zero, [latex]e[\/latex] is the natural base (Euler\u2019s constant), and [latex]k[\/latex] is a positive constant that determines the rate (percentage) of continuous growth.\u00a0 We also may use<\/p>\r\n<p style=\"text-align: center;\">[latex]y={A}_{0}{b}^{t}[\/latex][latex]\\\\[\/latex]<\/p>\r\nwhere [latex]{A}_{0}[\/latex] is equal to the value at time zero, and [latex]b[\/latex] is the growth factor which is greater than 1.\u00a0 We may use the <strong><span class=\"no-emphasis\">exponential growth<\/span><\/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 on a doubling time.\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><span class=\"no-emphasis\">exponential decay<\/span><\/strong> model. Again, we can use the form [latex]y={A}_{0}{e}^{kt}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, [latex]e[\/latex] is Euler\u2019s constant, and [latex]k[\/latex] is the (negative) continuous decay rate or we can use the form [latex]y={A}_{0}{b}^{t}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, and [latex]b[\/latex] is the decay factor between zero and one.\u00a0 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<p id=\"fs-id1165137824748\">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>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_07_002\" class=\"small\"><\/div>\r\n<div id=\"CNX_Precalc_Figure_04_07_003\" class=\"small\"><\/div>\r\n<div>\r\n<div class=\"textbox shaded\">\r\n<h3>Characteristics of the Exponential Function, [latex]y={A}_{0}{e}^{kt}[\/latex] and [latex]y={A}_{0}{b}^{t}[\/latex]<\/h3>\r\n<p id=\"fs-id1165137551809\">An exponential function with the form [latex]y={A}_{0}{e}^{kt}[\/latex] or [latex]y={A}_{0}{b}^{t}[\/latex] has the following characteristics:<\/p>\r\n\r\n<ul id=\"fs-id1165137530021\">\r\n \t<li>one-to-one function<\/li>\r\n \t<li>horizontal asymptote: [latex]y=0[\/latex]<\/li>\r\n \t<li>domain: [latex]\\left(\u2013\\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 [latex]k&gt;0[\/latex] or [latex]b&gt;1[\/latex] and decreasing if [latex]k \\lt 0[\/latex] or [latex]0 \\lt b \\lt 1[\/latex]<\/li>\r\n<\/ul>\r\n<div id=\"CNX_Precalc_Figure_04_07_004\" class=\"medium\">[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131419\/CNX_Precalc_Figure_04_07_004new.jpg\" alt=\"\" width=\"731\" height=\"337\" \/> <strong>Figure 2.\u00a0<\/strong>An exponential function models exponential growth when [latex]k&gt;0[\/latex] and exponential decay when [latex]k \\lt 0.[\/latex][\/caption]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135190498\" class=\"bc-section section\">\r\n<div id=\"fs-id1165134040514\" class=\"bc-section section\">\r\n<div id=\"fs-id1165137749158\" class=\"bc-section section\">\r\n<h3>Expressing an Exponential Model in Base <em><span class=\"e2\">e<\/span><\/em><\/h3>\r\n<p id=\"fs-id1165137852030\">While powers and logarithms of any base can be used in modeling, the two most common bases are [latex]10[\/latex] and [latex]e.[\/latex] In science and mathematics, the base [latex]e[\/latex] is often preferred. We can use laws of exponents and laws of logarithms to change any base to base [latex]e.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165134342616\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137452983\"><strong>Given a model with the form [latex]y=a{b}^{x},[\/latex] change it to the form [latex]y={A}_{0}{e}^{kx}.[\/latex]<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165134148496\" type=\"1\">\r\n \t<li>Since [latex]{b}^{x}={e}^{\\mathrm{ln}\\left({b}^{x}\\right)}[\/latex], rewrite [latex]y=a{b}^{x}[\/latex] as [latex]y=a{e}^{\\mathrm{ln}\\left({b}^{x}\\right)}.[\/latex]<\/li>\r\n \t<li>Use the power rule of logarithms to rewrite [latex]y[\/latex] as [latex]y=a{e}^{x\\mathrm{ln}\\left(b\\right)}=a{e}^{\\mathrm{ln}\\left(b\\right)x}.[\/latex]<\/li>\r\n \t<li>Note that [latex]a={A}_{0}[\/latex] and [latex]k=\\mathrm{ln}\\left(b\\right)[\/latex] in the equation [latex]y={A}_{0}{e}^{kx}.[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_07_08\" class=\"textbox examples\">\r\n<div id=\"fs-id1165134261257\">\r\n<div id=\"fs-id1165132957903\">\r\n<h3>Example 1:\u00a0 Changing to The Natural Base<\/h3>\r\n<p id=\"fs-id1165132957912\">Change the function [latex]y=2.5{\\left(3.1\\right)}^{x}[\/latex] so that this same function is written in the form [latex]y={A}_{0}{e}^{kx}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137432377\">[reveal-answer q=\"fs-id1165137432377\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137432377\"]\r\n<p id=\"fs-id1165137432379\">The formula is derived as follows<\/p>\r\n\r\n<div id=\"eip-id1165134193507\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}y&amp;=2.5{\\left(3.1\\right)}^{x}&amp;&amp; \\hfill \\\\ &amp;=2.5{e}^{\\mathrm{ln}\\left({3.1}^{x}\\right)}&amp;&amp; \\text{Insert exponential and its inverse}\\text{.}\\hfill \\\\ &amp;=2.5{e}^{x\\mathrm{ln}\\left(3.1\\right)}&amp;&amp; \\text{Laws of logs}\\text{.}\\hfill \\\\ &amp;=2.5{e}^{\\left(\\mathrm{ln}\\left(3.1\\right)\\right)}{}^{x}&amp;&amp; \\text{Commutative law of multiplication.}\\hfill \\end{align*}[\/latex]<\/div>\r\n<div><\/div>\r\n<div class=\"unnumbered\" style=\"text-align: left;\">In the above equation, we see that [latex]{A}_{0}=2.5, [\/latex] and [latex]k=\\mathrm{ln}\\left(3.1\\right)\\approx 1.1314.[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135664951\" class=\"precalculus tryit\">\r\n<h3>Try it #1<\/h3>\r\n<div id=\"ti_04_07_07\">\r\n<div id=\"fs-id1165135664961\">\r\n<p id=\"fs-id1165135664963\">Change the function [latex]y=3{\\left(0.5\\right)}^{x}[\/latex] to one having [latex]e[\/latex] as the base.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135333684\">[reveal-answer q=\"fs-id1165135333684\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135333684\"]\r\n<p id=\"fs-id1165135333685\">[latex]y=3{e}^{\\left(\\mathrm{ln}\\left(0.5\\right)\\right)x}\\approx3{e}^{-.6931x}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h4>Calculating Doubling Time<\/h4>\r\n<p id=\"fs-id1165137897897\">For growing quantities, we might want to find out how long it takes for a quantity to double. The time it takes for a quantity to double is called the <strong>doubling time<\/strong>.\u00a0 Let [latex]{A}_{0}[\/latex] represent the initial quantity at time zero.\u00a0 We then want to find when the quantity is doubled or equal to [latex]{2A}_{0}.[\/latex]\u00a0 Using the exponential growth equation [latex]f\\left(t\\right)={A}_{0}{b}^{t} \\text{ or } f\\left(t\\right)={A}_{0}{e}^{kt},[\/latex] we solve the equation [latex]{2A}_{0}={A}_{0}{b}^{t}[\/latex] or the equation [latex]{2A}_{0}={A}_{0}{e}^{kt}[\/latex] for [latex]t[\/latex] to find the doubling time [latex]t,[\/latex]\u00a0 if the growth rate or growth factor is known.\u00a0 Similarly, if we know the doubling time but not the growth rate or factor, we can use\u00a0[latex]{2A}_{0}={A}_{0}{e}^{kt}[\/latex] to solve for the continuous growth rate [latex]k[\/latex] or we can use\u00a0[latex]{2A}_{0}={A}_{0}{b}^{t}[\/latex] to solve for the growth factor [latex]b.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0 Graphing Exponential Growth<\/h3>\r\n<div id=\"fs-id1165135190498\" class=\"bc-section section\">\r\n<div id=\"fs-id1165135190498\" class=\"bc-section section\">\r\n<div>\r\n<p id=\"fs-id1165134381668\">A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135388513\">[reveal-answer q=\"65549\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"65549\"]\r\n<p id=\"fs-id1165135261481\">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] and we can use the model [latex]A=10e^{kt}.[\/latex]\u00a0 To find [latex]k,[\/latex] use the fact that after one hour [latex]\\left(t=1\\right)[\/latex] the population doubles from [latex]10[\/latex] to [latex]20,[\/latex] and plug in the point [latex]\\left(1,\\text{ } 20\\right)[\/latex] to the formula. The formula is derived as follows:<\/p>\r\n\r\n<div id=\"fs-id1165137530969\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}20&amp;=10{e}^{k\\cdot 1}&amp;&amp;\\text{ }\\\\2&amp;={e}^{k}&amp;&amp;\\text{Divide by 10.}\\\\ \\mathrm{ln}\\left(2\\right)&amp;=k&amp;&amp;\\text{Take the natural logarithm of both sides.}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id3079167\">Therefore, [latex]k=\\mathrm{ln}\\left(2\\right)\\approx0.6931.[\/latex] Thus the equation we want to graph is [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}[\/latex] or [latex]y\\approx10e^{0.6931t}.[\/latex]\u00a0 We can also write this as, [latex]y=10{\\left({e}^{\\mathrm{ln}2}\\right)}^{t}=10{\\left(2\\right)}^{t}.[\/latex]<\/p>\r\nThe graph is shown in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_005\">Figure 3<\/a>.\r\n\r\n&nbsp;\r\n<div id=\"CNX_Precalc_Figure_04_07_005\" class=\"small\">[caption id=\"\" align=\"aligncenter\" width=\"293\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131422\/CNX_Precalc_Figure_04_07_005_fixed.jpg\" alt=\"A graph starting at ten on the y-axis and rising rapidly to the right.\" width=\"293\" height=\"263\" \/> <strong>Figure 3.\u00a0<\/strong>The graph of [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}[\/latex][\/caption]\r\n<h3>Analysis<\/h3>\r\nThe population of bacteria after 10 hours is 10,240.\u00a0 We could describe this amount as being of the order of magnitude [latex]{10}^4.[\/latex]\u00a0 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 order of magnitude in 10 hours.\r\n\r\n<\/div>\r\n<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134040514\" class=\"bc-section section\">\r\n<div id=\"Example_04_07_04\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137657275\">\r\n<div id=\"fs-id1165137464856\">\r\n<h3>Example 3:\u00a0 Finding a Function That Describes Exponential Growth<\/h3>\r\n<p id=\"fs-id1165135306923\">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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134433337\">[reveal-answer q=\"fs-id1165134433337\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134433337\"]\r\n<p id=\"fs-id1165134433339\">Let [latex]{A}_{0}[\/latex] be the quantity of transistors that can currently be put on the computer chip.\u00a0 We will then work with the model [latex]y={A}_{0}{b}^{t}.[\/latex]\u00a0 We know that in two years, the number of transistors on the computer chip will double to be [latex]{2A}_{0}.[\/latex] We have the second point [latex]\\left(2, 2A_{0}\\right)[\/latex] that can be plugged into our model to get:<\/p>\r\n\r\n<div><\/div>\r\n<div id=\"eip-id1165135333736\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}{2A}_{0}&amp;={A}_{0}{b}^{2} &amp;&amp;\\text{ }\\\\ 2&amp;={b}^{2}&amp;&amp; \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ b&amp;=\\sqrt[\\leftroot{1}\\uproot{2} ]{2}\\approx1.414\\hfill &amp;&amp; \\text{Take the square root of both sides }\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165135365796\">Substitute [latex]b[\/latex] into the\u00a0 formula to get the function [latex]y={A}_{0}{1.414}^{t}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135190645\" class=\"precalculus tryit\">\r\n<h3>Try it #2<\/h3>\r\n<div id=\"ti_04_07_03\">\r\n<div id=\"fs-id1165137834355\">\r\n<p id=\"fs-id1165137758818\">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>\r\n<span style=\"font-size: 1em;\">[reveal-answer q=\"fs-id1165135525924\"]Show Solution[\/reveal-answer]<\/span>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135525924\">\r\n\r\n[hidden-answer a=\"fs-id1165135525924\"]\r\n<p id=\"fs-id1165135525926\">[latex]f\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{3}t}\\approx{A}_{0}{e}^{0.2310t}[\/latex] or [latex]f\\left(t\\right)={A}_{0}{1.26}^{t}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137803583\" class=\"bc-section section\">\r\n<h4>Half-Life<\/h4>\r\n<div id=\"fs-id1165137803583\" class=\"bc-section section\">\r\n\r\nWe now turn to <strong><span class=\"no-emphasis\">exponential decay<\/span><\/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<p id=\"eip-346\">To find the half-life of a function describing exponential decay, solve one of the following equations:<\/p>\r\n\r\n<div id=\"eip-942\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}[\/latex]\u00a0 or\u00a0\u00a0[latex]\\frac{1}{2}{A}_{0}={A}_{o}{b}^{t}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1165135177747\">The half-life depends only on the constant [latex]k[\/latex] or [latex]b[\/latex] and not on the starting quantity [latex]{A}_{0}.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165137418076\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137528941\"><strong>Given the half-life, find the continuous decay rate.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165135186559\" type=\"1\">\r\n \t<li>Write [latex]A={A}_{o}{e}^{kt}.[\/latex]<\/li>\r\n \t<li>Replace [latex]A[\/latex] by [latex]\\frac{1}{2}{A}_{0}[\/latex] and replace [latex]t[\/latex] by the given half-life.<\/li>\r\n \t<li>Solve to find [latex]k.[\/latex] Express [latex]k[\/latex] as an exact value (do not round).<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1165135613329\">Recall from Section 2.5, you can change the continuous rate to a noncontinuous rate if that is the desired rate.<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_04_07_02\" class=\"textbox examples\">\r\n<div>\r\n<div id=\"fs-id1165137732189\">\r\n<h3>Example 4:\u00a0 Finding the Function that Describes Radioactive Decay<\/h3>\r\n<p id=\"fs-id1165137501905\">The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, [latex]t.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137812032\">[reveal-answer q=\"fs-id1165137812032\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137812032\"]\r\n<p id=\"fs-id1165137805508\">Let [latex]{A}_{0}[\/latex] be the initial amount of carbon-14. We can work with the continuous or noncontinuous growth rate but for this example choose to work with [latex]A={A}_{0}{e}^{kt}.[\/latex] When [latex]t=5730,[\/latex] there will be [latex]\\frac{1}{2}{A}_{0}[\/latex] of the carbon-14 remaining.\u00a0 Substitute [latex]\\left(5730, \\frac{1}{2}{A}_{0}\\right)[\/latex] into the equation\u00a0 and solve as follows.<\/p>\r\n\r\n<div id=\"eip-id1165134317970\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}A&amp;={A}_{0}{e}^{kt}&amp;&amp; \\text{The continuous growth formula}.\\hfill \\\\ 0.5{A}_{0}&amp;={A}_{0}{e}^{k\\cdot 5730}&amp;&amp; \\text{Substitute the half-life for }t\\text{ and }0.5{A}_{0}\\text{ for }A.\\hfill \\\\ 0.5&amp;={e}^{5730k}&amp;&amp; \\text{Divide by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(0.5\\right)&amp;=5730k&amp;&amp; \\text{Take the natural logarithm of both sides}.\\hfill \\\\ k&amp;=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}&amp;&amp; \\text{Divide by the coefficient of }k.\\hfill \\\\ A&amp;={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}&amp;&amp; \\text{Substitute for }k\\text{ in the continuous growth formula}.\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1165137423070\">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 [latex]t,[\/latex] [latex]\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\approx -1.2097\u00d7{10}^{-4}[\/latex] is negative, as expected in the case of exponential decay.<\/p>\r\nEvaluating [latex]{e}^\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730},[\/latex] we have a growth factor of [latex]b\\approx0.999879[\/latex] and an equivalent equation of\u00a0[latex]f\\left(t\\right)\\approx{A}_{0}{0.999879}^{t}.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137639710\" class=\"precalculus tryit\">\r\n<h3>Try it #3<\/h3>\r\n<div id=\"ti_04_07_01\">\r\n<div id=\"fs-id1165135527002\">\r\n<p id=\"fs-id1165135527004\">The half-life of plutonium-244 is 80,000,000 years. Find function gives the amount of plutonium-244 remaining as a function of time, measured in years.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165134431212\">[reveal-answer q=\"fs-id1165134431212\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165134431212\"]\r\n<p id=\"fs-id1165135484584\">[latex]f\\left(t\\right)={A}_{0}{e}^{-0.0000000087t}[\/latex]\u00a0 or\u00a0\u00a0[latex]f\\left(t\\right)={A}_{0}{0.9999999913}^{t}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137828123\" class=\"bc-section section\">\r\n<h4><a class=\"autogenerated-content\" style=\"font-size: 1rem; font-weight: normal; text-align: initial;\" href=\"#Table_04_06_001\">Table 1<\/a><span style=\"font-size: 1rem; font-weight: normal; text-align: initial; color: #373d3f;\">\u00a0lists the half-life for several of the more common radioactive substances.<\/span><\/h4>\r\n<\/div>\r\n<table id=\"Table_04_06_001\" summary=\"Seven rows and three columns. The first column is labeled, \u201csubstance\u201d, the second column is labeled, \u201cuse\u201d, and the third column is labeled, \u201chalf-life\u201d. Gallium-67 is used for nuclear medicine and has a half-life of 80 hours. Cobalt-60 is used for manufacturing and has a half-life of 5.3 years. Technetium-99m is used for nuclear medicine and has a half-life of 6 hours. Americium-241 is used for construction and has a half-life of 432 years. Carbon-14 is used for archeological dating and has a half-life of 5,715 years. Uranium-235 is used for atomic power and has a half-life of 703,800,000 years.\"><caption>Table 1<\/caption>\r\n<thead>\r\n<tr>\r\n<th class=\"border\">Substance<\/th>\r\n<th class=\"border\">Use<\/th>\r\n<th class=\"border\">Half-life<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\">gallium-67<\/td>\r\n<td class=\"border\">nuclear medicine<\/td>\r\n<td class=\"border\">80 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">cobalt-60<\/td>\r\n<td class=\"border\">manufacturing<\/td>\r\n<td class=\"border\">5.3 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">technetium-99m<\/td>\r\n<td class=\"border\">nuclear medicine<\/td>\r\n<td class=\"border\">6 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">americium-241<\/td>\r\n<td class=\"border\">construction<\/td>\r\n<td class=\"border\">432 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">carbon-14<\/td>\r\n<td class=\"border\">archeological dating<\/td>\r\n<td class=\"border\">5,715 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\">uranium-235<\/td>\r\n<td class=\"border\">atomic power<\/td>\r\n<td class=\"border\">703,800,000 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165135209669\">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.<\/p>\r\n\r\n<div id=\"Example_04_06_13\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137628651\">\r\n<div id=\"fs-id1165137628653\">\r\n<h3>Example 5:\u00a0 Using the Formula for Radioactive Decay to Find the Quantity of a Substance<\/h3>\r\n<p id=\"fs-id1165137628659\">How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137628663\">[reveal-answer q=\"fs-id1165137628663\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137628663\"]\r\nWe start with 1000 grams so we let [latex]{A}_{0}=1000.[\/latex]\u00a0 We will use the formula [latex]f\\left(t\\right)={A}_{0} {b}^{t}[\/latex] so [latex]f\\left(t\\right)=1000{b}^{t}.[\/latex]\u00a0<span style=\"font-size: 0.9em;\">The half life of uranium-235 is 703,800,000. Therefore, we will plug in the point (703800000, 500) to the equation to find the growth factor b.<\/span>\r\n<div><\/div>\r\n<div>The initial equation is [latex]500=1000{b}^{703800000}[\/latex] and we divide both sides by 1000 to get [latex]0.5={b}^{703800000}.[\/latex] Take the 703,800,000th root of both sides and we have [latex]b\\approx0.999999999015036[\/latex] and the equation<\/div>\r\n<div><\/div>\r\n<div style=\"text-align: center;\">[latex]f\\left(t\\right)=1000\\left(0.999999999015036\\right)^t.[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div>Now that we have found the model to use, we can turn our attention to the question asked.\u00a0 Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.[latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1165137628651\">\r\n<div>\r\n<div><\/div>\r\n<div>\r\n<div>Solve [latex]900=\\text{1000}\\left(0.999999999015036\\right)^{t}[\/latex] for [latex]t.[\/latex]\u00a0 We begin by dividing both sides by 1000 to get<\/div>\r\n<div style=\"text-align: center;\">[latex]0.9=0.999999999015036^t.[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div>Next take the natural logarithm of both sides to get<\/div>\r\n<div style=\"text-align: center;\">[latex]\\mathrm{ln}\\left(0.9\\right) = \\mathrm{ln}\\left(0.999999999015036^t\\right).[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div>Using the power property of logarithms the t comes out front to give<\/div>\r\n<div style=\"text-align: center;\">[latex]\\mathrm{ln}\\left(0.9\\right) = t\\mathrm{ln}\\left(0.999999999015036\\right).[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div>Finally, dividing both sides by [latex]\\mathrm{ln}\\left(0.999999999015036\\right)[\/latex] gives<\/div>\r\n<div style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.999999999015036\\right)}\\approx\\text{106,968,248}.[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div class=\"unnumbered\" style=\"text-align: center;\"><\/div>\r\n<\/div>\r\n<div>There will be 900 grams of uranium-235 after approximately 107 million years have passed.<\/div>\r\n<div>\r\n<div><\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"unnumbered\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137453455\">\r\n<p id=\"fs-id1165137453460\"><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135250678\" class=\"precalculus tryit\">\r\n<h3>Try it #4<\/h3>\r\n<div id=\"ti_04_06_13\">\r\n<div id=\"fs-id1165137426963\">\r\n<p id=\"fs-id1165137426966\">How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137426971\">[reveal-answer q=\"fs-id1165137426971\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137426971\"]\r\n<p id=\"fs-id1165137426973\" style=\"text-align: center;\">[latex]t=703,800,000\u00d7\\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137828123\" class=\"bc-section section\"><\/div>\r\n<div class=\"bc-section section\">\r\n<div id=\"fs-id1165135511570\" class=\"bc-section section\">\r\n<h3>Choosing an Appropriate Model for Data<\/h3>\r\n<p id=\"fs-id1165135511575\">Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.<\/p>\r\n<p id=\"fs-id1165135532198\">Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered.<\/p>\r\n<p id=\"fs-id1165134550686\">In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down.<\/p>\r\n<p id=\"fs-id1165135511577\">After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.<\/p>\r\n\r\n<div id=\"Example_04_07_07\" class=\"textbox examples\">\r\n<div id=\"fs-id1165134550694\">\r\n<div id=\"fs-id1165135196760\">\r\n<h3>Example 6:\u00a0 Choosing a Mathematical Model<\/h3>\r\n<p id=\"fs-id1165135196765\">Does a linear, exponential, logarithmic, or quadratic model best fit the values listed in <a class=\"autogenerated-content\" href=\"#Table_04_07_001\">Table 2<\/a>? Find the model, and use a graph to check your choice.<\/p>\r\n\r\n<table id=\"Table_04_07_001\" summary=\"..\"><caption>Table 2<\/caption>\r\n<tbody>\r\n<tr>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\"><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">1<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">2<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">3<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">4<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">5<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">6<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">7<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">8<\/td>\r\n<td class=\"border\" style=\"width: 43.6563px; text-align: center;\">9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\"><strong>[latex]y[\/latex]<\/strong><\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">0<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">1.386<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">2.197<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">2.773<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">3.219<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">3.584<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">3.892<\/td>\r\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">4.159<\/td>\r\n<td class=\"border\" style=\"width: 43.6563px; text-align: center;\">4.394<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"fs-id1165135416447\">[reveal-answer q=\"fs-id1165135416447\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135416447\"]\r\n<p id=\"fs-id1165135416449\">First, plot the data on a graph as in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_008\">Figure 4<\/a>. For the purpose of graphing, round the data to two significant digits.<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_07_008\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"309\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131432\/CNX_Precalc_Figure_04_07_008.jpg\" alt=\"Graph of the previous table\u2019s values.\" width=\"309\" height=\"302\" \/> <strong>Figure 4.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nClearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try [latex]y=a\\mathrm{ln}\\left(bx\\right).[\/latex] Plugging in the first point, [latex]\\left(\\text{1,0}\\right)\\text{,}[\/latex] gives [latex]0=a\\mathrm{ln}b.[\/latex] We reject the case that [latex]a=0[\/latex] (if it were, all outputs would be 0), so we know [latex]\\mathrm{ln}\\left(b\\right)=0.[\/latex] Thus, [latex]b=1[\/latex] and [latex]y=a\\mathrm{ln}\\left(\\text{x}\\right).[\/latex] Next, we can use the point [latex]\\left(\\text{9,4}\\text{.394}\\right)[\/latex] to solve for [latex]a:[\/latex]\r\n\r\n&nbsp;\r\n<div id=\"eip-id1165134187109\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}y&amp;=a\\mathrm{ln}\\left(x\\right)\\hfill \\\\ 4.394&amp;=a\\mathrm{ln}\\left(9\\right)\\hfill \\\\a&amp;=\\frac{4.394}{\\mathrm{ln}\\left(9\\right)}\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1165134550694\">\r\n<div id=\"fs-id1165135416447\">\r\n<p id=\"fs-id1165135536301\">Because [latex]a=\\frac{4.394}{\\mathrm{ln}\\left(9\\right)}\\approx 2,[\/latex] an appropriate model for the data is [latex]y=2\\mathrm{ln}\\left(x\\right).[\/latex]<\/p>\r\nTo check the accuracy of the model, we graph the function together with the given points as in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_009\">Figure 5<\/a>.\r\n<div id=\"CNX_Precalc_Figure_04_07_009\" class=\"small\">[caption id=\"\" align=\"aligncenter\" width=\"305\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131436\/CNX_Precalc_Figure_04_07_009a.jpg\" alt=\"Graph of previous table\u2019s values showing that it fits the function y=2ln(x) with an asymptote at x=0.\" width=\"305\" height=\"298\" \/> <strong>Figure 5.\u00a0<\/strong>The graph of [latex]y=2\\mathrm{ln}\\left(x\\right).[\/latex][\/caption]<\/div>\r\n<div><\/div>\r\n<div>\r\n<p id=\"fs-id1165135369595\">We can conclude that the model is a good fit to the data.<\/p>\r\n<p id=\"fs-id1165135369598\">Compare <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_009\">Figure 5<\/a>\u00a0to the graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] shown in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_009a\">Figure 6<\/a>.<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_07_009a\" class=\"small\">[caption id=\"\" align=\"aligncenter\" width=\"310\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131439\/CNX_Precalc_Figure_04_07_009b.jpg\" alt=\"Graph of previous table\u2019s values showing that it fits the function y=2ln(x) with an asymptote at x=0.\" width=\"310\" height=\"303\" \/> <strong>Figure 6.\u00a0<\/strong>The graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right).[\/latex][\/caption]<\/div>\r\n<div class=\"wp-caption-text\"><\/div>\r\n<p id=\"fs-id1165137836798\">The graphs appear to be identical when [latex]x&gt;0.[\/latex] A quick check confirms this conclusion:\u00a0 [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)=2\\mathrm{ln}\\left(x\\right)[\/latex] for [latex]x&gt;0.[\/latex]<\/p>\r\n<p id=\"fs-id1165137771398\">However, if [latex]x&lt;0,[\/latex] the graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] includes a \u201cextra\u201d branch, as shown in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_010\">Figure 9<\/a>. This occurs because, while [latex]y=2\\mathrm{ln}\\left(x\\right)[\/latex] cannot have negative values in the domain (as such values would force the argument to be negative), the function [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] can have negative domain values.<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_07_010\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131442\/CNX_Precalc_Figure_04_07_010.jpg\" alt=\"Graph of y=ln(x^2).\" width=\"487\" height=\"216\" \/> <strong>Figure 7.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"small\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165134054938\" class=\"precalculus tryit\">\r\n<h3>Try it #5<\/h3>\r\n<div id=\"ti_04_07_06\">\r\n<div id=\"fs-id1165137611513\">\r\n<p id=\"fs-id1165137611515\">Does a linear, exponential, or logarithmic model best fit the data in <a class=\"autogenerated-content\" href=\"#Table_04_07_02\">Table 3<\/a>? Find the model.<\/p>\r\n\r\n<table id=\"Table_04_07_02\" style=\"height: 24px;\" summary=\"..\"><caption>Table 3<\/caption>\r\n<tbody>\r\n<tr style=\"height: 12px;\">\r\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\"><strong>[latex]x[\/latex]<\/strong><\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">1<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">2<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">3<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">4<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">5<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">6<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">7<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 46.5px; text-align: center;\">8<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 46.5px; text-align: center;\">9<\/td>\r\n<\/tr>\r\n<tr style=\"height: 12px;\">\r\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\"><strong>[latex]y[\/latex]<\/strong><\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">3.297<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">5.437<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">8.963<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">14.778<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">24.365<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">40.172<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">66.231<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 46.5px; text-align: center;\">109.196<\/td>\r\n<td class=\"border\" style=\"height: 12px; width: 46.5px; text-align: center;\">180.034<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div>\r\n<p id=\"fs-id1165135173377\">[reveal-answer q=\"165360\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"165360\"]<\/p>\r\nExponential.\u00a0 [latex]y=2{e}^{0.5x}.[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137749158\" class=\"bc-section section\"><\/div>\r\n<h3>Using Newton\u2019s Law of Cooling (Optional)<\/h3>\r\n<p id=\"fs-id1165137854986\">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><span class=\"no-emphasis\">vertical shift<\/span><\/strong> of the generic <strong><span class=\"no-emphasis\">exponential decay<\/span><\/strong> function. 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>\r\n\r\n<div id=\"eip-809\" class=\"unnumbered\" style=\"text-align: center;\">[latex]T\\left(t\\right)=a{e}^{kt}+{T}_{s}[\/latex].<\/div>\r\n<div id=\"eip-741\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1165135629984\">\r\n<div class=\"textbox shaded\">\r\n<h3>Newton\u2019s Law of Cooling<\/h3>\r\n<p id=\"fs-id1165137644683\">The temperature of an object, [latex]T,[\/latex] in surrounding air with temperature [latex]{T}_{s}[\/latex] will behave according to the formula<\/p>\r\n\r\n<div id=\"fs-id1165137660934\" style=\"text-align: center;\">[latex]T\\left(t\\right)=A{e}^{kt}+{T}_{s}[\/latex]<\/div>\r\nwhere\r\n<ul id=\"fs-id1165137415924\">\r\n \t<li>[latex]t[\/latex] is time,<\/li>\r\n \t<li>[latex]A[\/latex] is the difference between the initial temperature of the object and the surroundings, and<\/li>\r\n \t<li>[latex]k[\/latex] is a constant, the continuous rate of cooling of the object.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137667467\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"eip-id2434562\"><strong>Given a set of conditions, apply Newton\u2019s Law of Cooling.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137563116\" type=\"1\">\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 [latex]A[\/latex] and [latex]k.[\/latex]<\/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 id=\"Example_04_07_05\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135422904\">\r\n<div id=\"fs-id1165135422906\">\r\n<h3>\u00a0Example 7:\u00a0 Using Newton\u2019s Law of Cooling<\/h3>\r\n<p id=\"fs-id1165135439846\">A cheesecake is taken out of the oven with an ideal internal temperature of [latex]\\text{165\u00b0F,}[\/latex] and is placed into a [latex]35\u00b0F[\/latex] refrigerator. After 10 minutes, the cheesecake has cooled to [latex]\\text{150\u00b0F}.[\/latex] If we must wait until the cheesecake has cooled to [latex]\\text{70\u00b0F}[\/latex] before we eat it, how long will we have to wait?<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137705784\">[reveal-answer q=\"fs-id1165137705784\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137705784\"]\r\n<p id=\"fs-id1165137705786\">Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake\u2019s temperature will decay exponentially toward 35, we have the following equation<\/p>\r\n\r\n<div id=\"eip-id1165134081701\" class=\"unnumbered\" style=\"text-align: center;\">[latex]T\\left(t\\right)=A{e}^{kt}+35[\/latex].[latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1165137563332\">We know the initial temperature was 165, so [latex]T\\left(0\\right)=165.[\/latex]<\/p>\r\n&nbsp;\r\n<div id=\"eip-id1165134353692\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}165&amp;=A{e}^{k0}+35&amp;&amp; \\text{Substitute }\\left(0,165\\right).\\hfill \\\\ A&amp;=130&amp;&amp; \\text{Solve for }A.\\hfill \\end{align*}[\/latex]<\/div>\r\n<div><\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165137677897\">We were given another data point, [latex]T\\left(10\\right)=150,[\/latex] which we can use to solve for [latex]k.[\/latex]<\/p>\r\n&nbsp;\r\n<div id=\"eip-id1165134081505\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}150&amp;=130{e}^{k10}+35&amp;&amp;\\text{Substitute (10, 150)}.\\\\115&amp;=130{e}^{k10}&amp;&amp; \\text{Subtract 35}.\\\\\\frac{115}{130}&amp;={e}^{10k}&amp;&amp;\\text{Divide by 130}.\\\\\\mathrm{ln}\\left(\\frac{115}{130}\\right)&amp;=10k&amp;&amp;\\text{Take the natural log of both sides}.\\\\k&amp;=\\frac{\\mathrm{ln}\\left(\\frac{115}{130}\\right)}{10}\\approx -0.0123&amp;&amp;\\text{Divide by the coefficient of }k.\\end{align*}[\/latex]<\/div>\r\n<div><\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165134294911\">This gives us the equation for the cooling of the cheesecake: [latex]T\\left(t\\right)=130{e}^{\u20130.0123t}+35.[\/latex]<\/p>\r\n<p id=\"fs-id1165135194273\">Now we can solve for the time it will take for the temperature to cool to 70 degrees.<\/p>\r\n&nbsp;\r\n<div id=\"eip-id1165135149883\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}70&amp;=130{e}^{-0.0123t}+35&amp;&amp; \\text{Substitute in 70 for }T\\left(t\\right).\\hfill \\\\ 35&amp;=130{e}^{-0.0123t}&amp;&amp; \\text{Subtract 35}.\\hfill \\\\\\frac{35}{130}&amp;={e}^{-0.0123t}&amp;&amp; \\text{Divide by 130}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{35}{130}\\right)&amp;=-0.0123t&amp;&amp; \\text{Take the natural logarithm.}\\hfill \\\\ t&amp;=\\frac{\\mathrm{ln}\\left(\\frac{35}{130}\\right)}{-0.0123}\\approx 106.68&amp;&amp; \\text{Divide by the coefficient of }t.\\hfill \\end{align*}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165135186349\">It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to [latex]\\text{70\u00b0F}\\text{.}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135305911\" class=\"precalculus tryit\">\r\n<h3>Try it #6<\/h3>\r\n<div id=\"ti_04_07_04\">\r\n<div id=\"fs-id1165137686700\">\r\n<p id=\"fs-id1165137686701\">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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137686706\">[reveal-answer q=\"fs-id1165137686706\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137686706\"]\r\n<p id=\"fs-id1165137686707\">6.026 hours<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137737677\" class=\"bc-section section\">\r\n<h3>Using Logistic Growth Models (Optional)<\/h3>\r\n<p id=\"fs-id1165137737682\">Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an <strong><span class=\"no-emphasis\">exponential growth<\/span><\/strong> model, though the exponential growth model is still useful over a short term, before approaching the limiting value.<\/p>\r\n<p id=\"fs-id1165135194441\">The <strong>logistic growth model<\/strong> is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model\u2019s upper bound, called the <strong>carrying capacity<\/strong>. For constants [latex]a, \\text{ }b,[\/latex] and [latex]c,[\/latex] the logistic growth of a population over time [latex]x[\/latex] is represented by the model<\/p>\r\n\r\n<div id=\"eip-391\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}.[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1165135439859\">The graph in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_006\">Figure 8<\/a> shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_07_006\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131426\/CNX_Precalc_Figure_04_07_006.jpg\" alt=\"Graph of f(x)=c\/(1+ae^(-tx)). The carrying capacity is the asymptote at y=c. The initial value of population is (0, c\/(1+a)). The point of maximum growth is (ln(a)\/b, c\/2).\" width=\"487\" height=\"367\" \/> <strong>Figure 8.<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135207602\">\r\n<div class=\"textbox shaded\">\r\n<h3>Logistic Growth<\/h3>\r\n<p id=\"fs-id1165135511485\">The logistic growth model is<\/p>\r\n\r\n<div id=\"eip-id1165134239632\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\r\n<p id=\"eip-id1165135264518\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137526317\">\r\n \t<li>[latex]\\frac{c}{1+a}[\/latex] is the initial value<\/li>\r\n \t<li>[latex]c[\/latex] is the <em>carrying capacity<\/em>, or <em>limiting value,\u00a0<\/em>and<\/li>\r\n \t<li>[latex]b[\/latex] is a constant determined by the rate of growth.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_07_06\" class=\"textbox examples\">\r\n<div id=\"fs-id1165135572080\">\r\n<div id=\"fs-id1165135572082\">\r\n<h3>Example 8:\u00a0 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 a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.\r\n<p id=\"fs-id1165134194949\">For example, at time [latex]t=0[\/latex] 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 [latex]b=0.6030.[\/latex] 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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135471237\">[reveal-answer q=\"fs-id1165135471237\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135471237\"]\r\n<p id=\"fs-id1165135471239\">We substitute the given data into the logistic growth model<\/p>\r\n\r\n<div id=\"eip-id1165133200610\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}.[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1165137698371\">Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is [latex]c=1000.[\/latex] To find [latex]a,[\/latex] we use the formula that the number of cases at time [latex]t=0[\/latex] is [latex]\\frac{c}{1+a}=1,[\/latex] from which it follows that [latex]a=999.[\/latex] This model predicts that, after ten days, the number of people who have had the flu is [latex]f\\left(10\\right)=\\frac{1000}{1+999{e}^{-0.6030\\times 10}}\\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, [latex]c=1000.[\/latex]<\/p>\r\n\r\n<h3>Analysis<\/h3>\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<p id=\"fs-id1165135344064\">The graph in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_007\">Figure 9<\/a>\u00a0gives a good picture of how this model fits the data.<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_07_007\" class=\"medium\">[caption id=\"\" align=\"aligncenter\" width=\"446\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131429\/CNX_Precalc_Figure_04_07_007.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as \u201cCases\u201d and the x-axis labeled as \u201cDays\u201d. There was 1 case on day 0, 20 on day 5, 294 on day 10, and 1000 on day 21.\" width=\"446\" height=\"300\" \/> <strong>Figure 9.\u00a0<\/strong>The graph of [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}[\/latex][\/caption]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135445930\">\r\n<div id=\"CNX_Precalc_Figure_04_07_007\" class=\"medium\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135262651\" class=\"precalculus tryit\">\r\n<h3>Try it #7<\/h3>\r\n<div id=\"ti_04_07_05\">\r\n<div id=\"fs-id1165135262661\">\r\n<p id=\"fs-id1165135689538\">Using the model in <a class=\"autogenerated-content\" href=\"#Example_04_07_06\">Example 7<\/a>, estimate the number of cases of flu on day 15.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135689547\">[reveal-answer q=\"fs-id1165135689547\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135689547\"]\r\n<p id=\"fs-id1165135689548\">895 cases on day 15<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135511570\" class=\"bc-section section\"><\/div>\r\n<div id=\"fs-id1165137749158\" class=\"bc-section section\">\r\n<div id=\"fs-id1165135171587\" class=\"precalculus media\">\r\n<div class=\"textbox shaded\">\r\n<h3>Media:<\/h3>\r\nAccess these online resources for additional instruction and practice with exponential and logarithmic models.\r\n<ul>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/logph\">Logarithm Application \u2013 pH<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/expmodelhalf\">Exponential Model \u2013 Age Using Half-Life<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/newtoncooling\">Newton\u2019s Law of Cooling<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/expgrowthdbl\">Exponential Growth Given Doubling Time<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/initialdouble\">Exponential Growth \u2013 Find Initial Amount Given Doubling Time<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137894245\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\r\n<ul id=\"fs-id1165137894248\">\r\n \t<li>The basic exponential function is [latex]f\\left(x\\right)=a{b}^{x}.[\/latex] If [latex]b&gt;1,[\/latex] we have exponential growth; if [latex]0 \\lt b \\lt 1,[\/latex] we have exponential decay.<\/li>\r\n \t<li>We can also write this formula in terms of continuous growth as [latex]A={A}_{0}{e}^{kx},[\/latex] where [latex]{A}_{0}\\text{ }[\/latex] is the starting value. If [latex]{A}_{0}[\/latex] is positive, then we have exponential growth when [latex]k&gt;0[\/latex] and exponential decay when [latex]k&lt;0.\\text{ }[\/latex]<\/li>\r\n \t<li>In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay.<\/li>\r\n \t<li>Given a substance\u2019s doubling time or half-time, we can find a function that represents its exponential growth or decay.<\/li>\r\n \t<li>We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic (Optional) graphs help us to develop models that best fit our data.<\/li>\r\n \t<li>Any exponential function with the form [latex]y=a{b}^{x}[\/latex] can be rewritten as an equivalent exponential function with the form [latex]y={A}_{0}{e}^{kx}[\/latex] where [latex]k=\\mathrm{ln}\\left(b\\right).[\/latex]<\/li>\r\n \t<li>(Optional) We can use Newton\u2019s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time.<\/li>\r\n \t<li>(Optional) 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<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1165134069388\">\r\n \t<dt>carrying capacity<\/dt>\r\n \t<dd id=\"fs-id1165134069394\">in a logistic model, the limiting value of the output<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134069398\">\r\n \t<dt>doubling time<\/dt>\r\n \t<dd id=\"fs-id1165135528921\">the time it takes for a quantity to double<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135528925\">\r\n \t<dt>half-life<\/dt>\r\n \t<dd id=\"fs-id1165135528931\">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 id=\"fs-id1165135528936\">\r\n \t<dt>logistic growth model<\/dt>\r\n \t<dd id=\"fs-id1165135528941\">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, [latex]c[\/latex] is the carrying capacity, or limiting value, and [latex]b[\/latex] is a constant determined by the rate of growth<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137844266\">\r\n \t<dt>Newton\u2019s Law of Cooling<\/dt>\r\n \t<dd>the scientific formula for temperature as a function of time as an object\u2019s temperature is equalized with the ambient temperature<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137844278\">\r\n \t<dt>order of magnitude<\/dt>\r\n \t<dd id=\"fs-id1165137844283\">the power of ten, when a number is expressed in scientific notation, with one non-zero digit to the left of the decimal<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Model exponential growth and decay.<\/li>\n<li>Doubling time and half-life.<\/li>\n<\/ul>\n<p>Optional:<\/p>\n<ul>\n<li>Use Newton\u2019s Law of Cooling.<\/li>\n<li>Use logistic-growth models.<\/li>\n<\/ul>\n<\/div>\n<div id=\"CNX_Precalc_Figure_04_07_001\" class=\"small\">\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\/3896\/2019\/03\/07131409\/CNX_Precalc_Figure_04_07_001F.jpg\" alt=\"Inside a nuclear research reactor.\" width=\"325\" height=\"409\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 1.\u00a0<\/strong>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<\/div>\n<p id=\"fs-id1165134081045\">We have already explored some basic applications of exponential and logarithmic functions. In this section, we explore some important applications in more depth.<\/p>\n<div id=\"fs-id1165135190498\" class=\"bc-section section\">\n<h3>Modeling Exponential Growth and Decay<\/h3>\n<p id=\"fs-id1165135169375\">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<div class=\"unnumbered\" style=\"text-align: center;\">[latex]y={A}_{0}{e}^{kt}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"eip-292\">where [latex]{A}_{0}[\/latex] is equal to the value at time zero, [latex]e[\/latex] is the natural base (Euler\u2019s constant), and [latex]k[\/latex] is a positive constant that determines the rate (percentage) of continuous growth.\u00a0 We also may use<\/p>\n<p style=\"text-align: center;\">[latex]y={A}_{0}{b}^{t}[\/latex][latex]\\\\[\/latex]<\/p>\n<p>where [latex]{A}_{0}[\/latex] is equal to the value at time zero, and [latex]b[\/latex] is the growth factor which is greater than 1.\u00a0 We may use the <strong><span class=\"no-emphasis\">exponential growth<\/span><\/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 on a doubling time.<\/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><span class=\"no-emphasis\">exponential decay<\/span><\/strong> model. Again, we can use the form [latex]y={A}_{0}{e}^{kt}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, [latex]e[\/latex] is Euler\u2019s constant, and [latex]k[\/latex] is the (negative) continuous decay rate or we can use the form [latex]y={A}_{0}{b}^{t}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, and [latex]b[\/latex] is the decay factor between zero and one.\u00a0 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 id=\"fs-id1165137824748\">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 id=\"CNX_Precalc_Figure_04_07_002\" class=\"small\"><\/div>\n<div id=\"CNX_Precalc_Figure_04_07_003\" class=\"small\"><\/div>\n<div>\n<div class=\"textbox shaded\">\n<h3>Characteristics of the Exponential Function, [latex]y={A}_{0}{e}^{kt}[\/latex] and [latex]y={A}_{0}{b}^{t}[\/latex]<\/h3>\n<p id=\"fs-id1165137551809\">An exponential function with the form [latex]y={A}_{0}{e}^{kt}[\/latex] or [latex]y={A}_{0}{b}^{t}[\/latex] has the following characteristics:<\/p>\n<ul id=\"fs-id1165137530021\">\n<li>one-to-one function<\/li>\n<li>horizontal asymptote: [latex]y=0[\/latex]<\/li>\n<li>domain: [latex]\\left(\u2013\\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 [latex]k>0[\/latex] or [latex]b>1[\/latex] and decreasing if [latex]k \\lt 0[\/latex] or [latex]0 \\lt b \\lt 1[\/latex]<\/li>\n<\/ul>\n<div id=\"CNX_Precalc_Figure_04_07_004\" class=\"medium\">\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131419\/CNX_Precalc_Figure_04_07_004new.jpg\" alt=\"\" width=\"731\" height=\"337\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 2.\u00a0<\/strong>An exponential function models exponential growth when [latex]k&gt;0[\/latex] and exponential decay when [latex]k \\lt 0.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135190498\" class=\"bc-section section\">\n<div id=\"fs-id1165134040514\" class=\"bc-section section\">\n<div id=\"fs-id1165137749158\" class=\"bc-section section\">\n<h3>Expressing an Exponential Model in Base <em><span class=\"e2\">e<\/span><\/em><\/h3>\n<p id=\"fs-id1165137852030\">While powers and logarithms of any base can be used in modeling, the two most common bases are [latex]10[\/latex] and [latex]e.[\/latex] In science and mathematics, the base [latex]e[\/latex] is often preferred. We can use laws of exponents and laws of logarithms to change any base to base [latex]e.[\/latex]<\/p>\n<div id=\"fs-id1165134342616\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137452983\"><strong>Given a model with the form [latex]y=a{b}^{x},[\/latex] change it to the form [latex]y={A}_{0}{e}^{kx}.[\/latex]<\/strong><\/p>\n<ol id=\"fs-id1165134148496\" type=\"1\">\n<li>Since [latex]{b}^{x}={e}^{\\mathrm{ln}\\left({b}^{x}\\right)}[\/latex], rewrite [latex]y=a{b}^{x}[\/latex] as [latex]y=a{e}^{\\mathrm{ln}\\left({b}^{x}\\right)}.[\/latex]<\/li>\n<li>Use the power rule of logarithms to rewrite [latex]y[\/latex] as [latex]y=a{e}^{x\\mathrm{ln}\\left(b\\right)}=a{e}^{\\mathrm{ln}\\left(b\\right)x}.[\/latex]<\/li>\n<li>Note that [latex]a={A}_{0}[\/latex] and [latex]k=\\mathrm{ln}\\left(b\\right)[\/latex] in the equation [latex]y={A}_{0}{e}^{kx}.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_07_08\" class=\"textbox examples\">\n<div id=\"fs-id1165134261257\">\n<div id=\"fs-id1165132957903\">\n<h3>Example 1:\u00a0 Changing to The Natural Base<\/h3>\n<p id=\"fs-id1165132957912\">Change the function [latex]y=2.5{\\left(3.1\\right)}^{x}[\/latex] so that this same function is written in the form [latex]y={A}_{0}{e}^{kx}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137432377\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137432377\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137432377\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137432379\">The formula is derived as follows<\/p>\n<div id=\"eip-id1165134193507\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}y&=2.5{\\left(3.1\\right)}^{x}&& \\hfill \\\\ &=2.5{e}^{\\mathrm{ln}\\left({3.1}^{x}\\right)}&& \\text{Insert exponential and its inverse}\\text{.}\\hfill \\\\ &=2.5{e}^{x\\mathrm{ln}\\left(3.1\\right)}&& \\text{Laws of logs}\\text{.}\\hfill \\\\ &=2.5{e}^{\\left(\\mathrm{ln}\\left(3.1\\right)\\right)}{}^{x}&& \\text{Commutative law of multiplication.}\\hfill \\end{align*}[\/latex]<\/div>\n<div><\/div>\n<div class=\"unnumbered\" style=\"text-align: left;\">In the above equation, we see that [latex]{A}_{0}=2.5,[\/latex] and [latex]k=\\mathrm{ln}\\left(3.1\\right)\\approx 1.1314.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135664951\" class=\"precalculus tryit\">\n<h3>Try it #1<\/h3>\n<div id=\"ti_04_07_07\">\n<div id=\"fs-id1165135664961\">\n<p id=\"fs-id1165135664963\">Change the function [latex]y=3{\\left(0.5\\right)}^{x}[\/latex] to one having [latex]e[\/latex] as the base.<\/p>\n<\/div>\n<div id=\"fs-id1165135333684\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135333684\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135333684\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135333685\">[latex]y=3{e}^{\\left(\\mathrm{ln}\\left(0.5\\right)\\right)x}\\approx3{e}^{-.6931x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h4>Calculating Doubling Time<\/h4>\n<p id=\"fs-id1165137897897\">For growing quantities, we might want to find out how long it takes for a quantity to double. The time it takes for a quantity to double is called the <strong>doubling time<\/strong>.\u00a0 Let [latex]{A}_{0}[\/latex] represent the initial quantity at time zero.\u00a0 We then want to find when the quantity is doubled or equal to [latex]{2A}_{0}.[\/latex]\u00a0 Using the exponential growth equation [latex]f\\left(t\\right)={A}_{0}{b}^{t} \\text{ or } f\\left(t\\right)={A}_{0}{e}^{kt},[\/latex] we solve the equation [latex]{2A}_{0}={A}_{0}{b}^{t}[\/latex] or the equation [latex]{2A}_{0}={A}_{0}{e}^{kt}[\/latex] for [latex]t[\/latex] to find the doubling time [latex]t,[\/latex]\u00a0 if the growth rate or growth factor is known.\u00a0 Similarly, if we know the doubling time but not the growth rate or factor, we can use\u00a0[latex]{2A}_{0}={A}_{0}{e}^{kt}[\/latex] to solve for the continuous growth rate [latex]k[\/latex] or we can use\u00a0[latex]{2A}_{0}={A}_{0}{b}^{t}[\/latex] to solve for the growth factor [latex]b.[\/latex]<\/p>\n<\/div>\n<div>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0 Graphing Exponential Growth<\/h3>\n<div id=\"fs-id1165135190498\" class=\"bc-section section\">\n<div id=\"fs-id1165135190498\" class=\"bc-section section\">\n<div>\n<p id=\"fs-id1165134381668\">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>\n<div id=\"fs-id1165135388513\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q65549\">Show Solution<\/span><\/p>\n<div id=\"q65549\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135261481\">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] and we can use the model [latex]A=10e^{kt}.[\/latex]\u00a0 To find [latex]k,[\/latex] use the fact that after one hour [latex]\\left(t=1\\right)[\/latex] the population doubles from [latex]10[\/latex] to [latex]20,[\/latex] and plug in the point [latex]\\left(1,\\text{ } 20\\right)[\/latex] to the formula. The formula is derived as follows:<\/p>\n<div id=\"fs-id1165137530969\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}20&=10{e}^{k\\cdot 1}&&\\text{ }\\\\2&={e}^{k}&&\\text{Divide by 10.}\\\\ \\mathrm{ln}\\left(2\\right)&=k&&\\text{Take the natural logarithm of both sides.}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id3079167\">Therefore, [latex]k=\\mathrm{ln}\\left(2\\right)\\approx0.6931.[\/latex] Thus the equation we want to graph is [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}[\/latex] or [latex]y\\approx10e^{0.6931t}.[\/latex]\u00a0 We can also write this as, [latex]y=10{\\left({e}^{\\mathrm{ln}2}\\right)}^{t}=10{\\left(2\\right)}^{t}.[\/latex]<\/p>\n<p>The graph is shown in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_005\">Figure 3<\/a>.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"CNX_Precalc_Figure_04_07_005\" class=\"small\">\n<div style=\"width: 303px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131422\/CNX_Precalc_Figure_04_07_005_fixed.jpg\" alt=\"A graph starting at ten on the y-axis and rising rapidly to the right.\" width=\"293\" height=\"263\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 3.\u00a0<\/strong>The graph of [latex]y=10{e}^{\\left(\\mathrm{ln}2\\right)t}[\/latex]<\/p>\n<\/div>\n<h3>Analysis<\/h3>\n<p>The population of bacteria after 10 hours is 10,240.\u00a0 We could describe this amount as being of the order of magnitude [latex]{10}^4.[\/latex]\u00a0 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 order of magnitude in 10 hours.<\/p>\n<\/div>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134040514\" class=\"bc-section section\">\n<div id=\"Example_04_07_04\" class=\"textbox examples\">\n<div id=\"fs-id1165137657275\">\n<div id=\"fs-id1165137464856\">\n<h3>Example 3:\u00a0 Finding a Function That Describes Exponential Growth<\/h3>\n<p id=\"fs-id1165135306923\">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>\n<div id=\"fs-id1165134433337\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134433337\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134433337\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134433339\">Let [latex]{A}_{0}[\/latex] be the quantity of transistors that can currently be put on the computer chip.\u00a0 We will then work with the model [latex]y={A}_{0}{b}^{t}.[\/latex]\u00a0 We know that in two years, the number of transistors on the computer chip will double to be [latex]{2A}_{0}.[\/latex] We have the second point [latex]\\left(2, 2A_{0}\\right)[\/latex] that can be plugged into our model to get:<\/p>\n<div><\/div>\n<div id=\"eip-id1165135333736\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}{2A}_{0}&={A}_{0}{b}^{2} &&\\text{ }\\\\ 2&={b}^{2}&& \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ b&=\\sqrt[\\leftroot{1}\\uproot{2} ]{2}\\approx1.414\\hfill && \\text{Take the square root of both sides }\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165135365796\">Substitute [latex]b[\/latex] into the\u00a0 formula to get the function [latex]y={A}_{0}{1.414}^{t}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135190645\" class=\"precalculus tryit\">\n<h3>Try it #2<\/h3>\n<div id=\"ti_04_07_03\">\n<div id=\"fs-id1165137834355\">\n<p id=\"fs-id1165137758818\">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<p><span style=\"font-size: 1em;\"><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135525924\">Show Solution<\/span><\/span><\/p>\n<\/div>\n<div id=\"fs-id1165135525924\">\n<div id=\"qfs-id1165135525924\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135525926\">[latex]f\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{3}t}\\approx{A}_{0}{e}^{0.2310t}[\/latex] or [latex]f\\left(t\\right)={A}_{0}{1.26}^{t}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137803583\" class=\"bc-section section\">\n<h4>Half-Life<\/h4>\n<div id=\"fs-id1165137803583\" class=\"bc-section section\">\n<p>We now turn to <strong><span class=\"no-emphasis\">exponential decay<\/span><\/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 id=\"eip-346\">To find the half-life of a function describing exponential decay, solve one of the following equations:<\/p>\n<div id=\"eip-942\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}[\/latex]\u00a0 or\u00a0\u00a0[latex]\\frac{1}{2}{A}_{0}={A}_{o}{b}^{t}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1165135177747\">The half-life depends only on the constant [latex]k[\/latex] or [latex]b[\/latex] and not on the starting quantity [latex]{A}_{0}.[\/latex]<\/p>\n<div id=\"fs-id1165137418076\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137528941\"><strong>Given the half-life, find the continuous decay rate.<\/strong><\/p>\n<ol id=\"fs-id1165135186559\" type=\"1\">\n<li>Write [latex]A={A}_{o}{e}^{kt}.[\/latex]<\/li>\n<li>Replace [latex]A[\/latex] by [latex]\\frac{1}{2}{A}_{0}[\/latex] and replace [latex]t[\/latex] by the given half-life.<\/li>\n<li>Solve to find [latex]k.[\/latex] Express [latex]k[\/latex] as an exact value (do not round).<\/li>\n<\/ol>\n<p id=\"fs-id1165135613329\">Recall from Section 2.5, you can change the continuous rate to a noncontinuous rate if that is the desired rate.<\/p>\n<\/div>\n<div id=\"Example_04_07_02\" class=\"textbox examples\">\n<div>\n<div id=\"fs-id1165137732189\">\n<h3>Example 4:\u00a0 Finding the Function that Describes Radioactive Decay<\/h3>\n<p id=\"fs-id1165137501905\">The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, [latex]t.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137812032\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137812032\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137812032\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137805508\">Let [latex]{A}_{0}[\/latex] be the initial amount of carbon-14. We can work with the continuous or noncontinuous growth rate but for this example choose to work with [latex]A={A}_{0}{e}^{kt}.[\/latex] When [latex]t=5730,[\/latex] there will be [latex]\\frac{1}{2}{A}_{0}[\/latex] of the carbon-14 remaining.\u00a0 Substitute [latex]\\left(5730, \\frac{1}{2}{A}_{0}\\right)[\/latex] into the equation\u00a0 and solve as follows.<\/p>\n<div id=\"eip-id1165134317970\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}A&={A}_{0}{e}^{kt}&& \\text{The continuous growth formula}.\\hfill \\\\ 0.5{A}_{0}&={A}_{0}{e}^{k\\cdot 5730}&& \\text{Substitute the half-life for }t\\text{ and }0.5{A}_{0}\\text{ for }A.\\hfill \\\\ 0.5&={e}^{5730k}&& \\text{Divide by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(0.5\\right)&=5730k&& \\text{Take the natural logarithm of both sides}.\\hfill \\\\ k&=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}&& \\text{Divide by the coefficient of }k.\\hfill \\\\ A&={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}&& \\text{Substitute for }k\\text{ in the continuous growth formula}.\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1165137423070\">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 [latex]t,[\/latex] [latex]\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\approx -1.2097\u00d7{10}^{-4}[\/latex] is negative, as expected in the case of exponential decay.<\/p>\n<p>Evaluating [latex]{e}^\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730},[\/latex] we have a growth factor of [latex]b\\approx0.999879[\/latex] and an equivalent equation of\u00a0[latex]f\\left(t\\right)\\approx{A}_{0}{0.999879}^{t}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137639710\" class=\"precalculus tryit\">\n<h3>Try it #3<\/h3>\n<div id=\"ti_04_07_01\">\n<div id=\"fs-id1165135527002\">\n<p id=\"fs-id1165135527004\">The half-life of plutonium-244 is 80,000,000 years. Find function gives the amount of plutonium-244 remaining as a function of time, measured in years.<\/p>\n<\/div>\n<div id=\"fs-id1165134431212\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165134431212\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165134431212\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135484584\">[latex]f\\left(t\\right)={A}_{0}{e}^{-0.0000000087t}[\/latex]\u00a0 or\u00a0\u00a0[latex]f\\left(t\\right)={A}_{0}{0.9999999913}^{t}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137828123\" class=\"bc-section section\">\n<h4><a class=\"autogenerated-content\" style=\"font-size: 1rem; font-weight: normal; text-align: initial;\" href=\"#Table_04_06_001\">Table 1<\/a><span style=\"font-size: 1rem; font-weight: normal; text-align: initial; color: #373d3f;\">\u00a0lists the half-life for several of the more common radioactive substances.<\/span><\/h4>\n<\/div>\n<table id=\"Table_04_06_001\" summary=\"Seven rows and three columns. The first column is labeled, \u201csubstance\u201d, the second column is labeled, \u201cuse\u201d, and the third column is labeled, \u201chalf-life\u201d. Gallium-67 is used for nuclear medicine and has a half-life of 80 hours. Cobalt-60 is used for manufacturing and has a half-life of 5.3 years. Technetium-99m is used for nuclear medicine and has a half-life of 6 hours. Americium-241 is used for construction and has a half-life of 432 years. Carbon-14 is used for archeological dating and has a half-life of 5,715 years. Uranium-235 is used for atomic power and has a half-life of 703,800,000 years.\">\n<caption>Table 1<\/caption>\n<thead>\n<tr>\n<th class=\"border\">Substance<\/th>\n<th class=\"border\">Use<\/th>\n<th class=\"border\">Half-life<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td class=\"border\">gallium-67<\/td>\n<td class=\"border\">nuclear medicine<\/td>\n<td class=\"border\">80 hours<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">cobalt-60<\/td>\n<td class=\"border\">manufacturing<\/td>\n<td class=\"border\">5.3 years<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">technetium-99m<\/td>\n<td class=\"border\">nuclear medicine<\/td>\n<td class=\"border\">6 hours<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">americium-241<\/td>\n<td class=\"border\">construction<\/td>\n<td class=\"border\">432 years<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">carbon-14<\/td>\n<td class=\"border\">archeological dating<\/td>\n<td class=\"border\">5,715 years<\/td>\n<\/tr>\n<tr>\n<td class=\"border\">uranium-235<\/td>\n<td class=\"border\">atomic power<\/td>\n<td class=\"border\">703,800,000 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165135209669\">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.<\/p>\n<div id=\"Example_04_06_13\" class=\"textbox examples\">\n<div id=\"fs-id1165137628651\">\n<div id=\"fs-id1165137628653\">\n<h3>Example 5:\u00a0 Using the Formula for Radioactive Decay to Find the Quantity of a Substance<\/h3>\n<p id=\"fs-id1165137628659\">How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?<\/p>\n<\/div>\n<div id=\"fs-id1165137628663\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137628663\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137628663\" class=\"hidden-answer\" style=\"display: none\">\nWe start with 1000 grams so we let [latex]{A}_{0}=1000.[\/latex]\u00a0 We will use the formula [latex]f\\left(t\\right)={A}_{0} {b}^{t}[\/latex] so [latex]f\\left(t\\right)=1000{b}^{t}.[\/latex]\u00a0<span style=\"font-size: 0.9em;\">The half life of uranium-235 is 703,800,000. Therefore, we will plug in the point (703800000, 500) to the equation to find the growth factor b.<\/span><\/p>\n<div><\/div>\n<div>The initial equation is [latex]500=1000{b}^{703800000}[\/latex] and we divide both sides by 1000 to get [latex]0.5={b}^{703800000}.[\/latex] Take the 703,800,000th root of both sides and we have [latex]b\\approx0.999999999015036[\/latex] and the equation<\/div>\n<div><\/div>\n<div style=\"text-align: center;\">[latex]f\\left(t\\right)=1000\\left(0.999999999015036\\right)^t.[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<div>Now that we have found the model to use, we can turn our attention to the question asked.\u00a0 Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.[latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div id=\"fs-id1165137628651\">\n<div>\n<div><\/div>\n<div>\n<div>Solve [latex]900=\\text{1000}\\left(0.999999999015036\\right)^{t}[\/latex] for [latex]t.[\/latex]\u00a0 We begin by dividing both sides by 1000 to get<\/div>\n<div style=\"text-align: center;\">[latex]0.9=0.999999999015036^t.[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div>Next take the natural logarithm of both sides to get<\/div>\n<div style=\"text-align: center;\">[latex]\\mathrm{ln}\\left(0.9\\right) = \\mathrm{ln}\\left(0.999999999015036^t\\right).[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div>Using the power property of logarithms the t comes out front to give<\/div>\n<div style=\"text-align: center;\">[latex]\\mathrm{ln}\\left(0.9\\right) = t\\mathrm{ln}\\left(0.999999999015036\\right).[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div>Finally, dividing both sides by [latex]\\mathrm{ln}\\left(0.999999999015036\\right)[\/latex] gives<\/div>\n<div style=\"text-align: center;\">[latex]t=\\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.999999999015036\\right)}\\approx\\text{106,968,248}.[\/latex][latex]\\\\[\/latex]<\/div>\n<div class=\"unnumbered\" style=\"text-align: center;\"><\/div>\n<\/div>\n<div>There will be 900 grams of uranium-235 after approximately 107 million years have passed.<\/div>\n<div>\n<div><\/div>\n<\/div>\n<\/div>\n<div class=\"unnumbered\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137453455\">\n<p id=\"fs-id1165137453460\">\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135250678\" class=\"precalculus tryit\">\n<h3>Try it #4<\/h3>\n<div id=\"ti_04_06_13\">\n<div id=\"fs-id1165137426963\">\n<p id=\"fs-id1165137426966\">How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?<\/p>\n<\/div>\n<div id=\"fs-id1165137426971\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137426971\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137426971\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137426973\" style=\"text-align: center;\">[latex]t=703,800,000\u00d7\\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137828123\" class=\"bc-section section\"><\/div>\n<div class=\"bc-section section\">\n<div id=\"fs-id1165135511570\" class=\"bc-section section\">\n<h3>Choosing an Appropriate Model for Data<\/h3>\n<p id=\"fs-id1165135511575\">Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.<\/p>\n<p id=\"fs-id1165135532198\">Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered.<\/p>\n<p id=\"fs-id1165134550686\">In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down.<\/p>\n<p id=\"fs-id1165135511577\">After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.<\/p>\n<div id=\"Example_04_07_07\" class=\"textbox examples\">\n<div id=\"fs-id1165134550694\">\n<div id=\"fs-id1165135196760\">\n<h3>Example 6:\u00a0 Choosing a Mathematical Model<\/h3>\n<p id=\"fs-id1165135196765\">Does a linear, exponential, logarithmic, or quadratic model best fit the values listed in <a class=\"autogenerated-content\" href=\"#Table_04_07_001\">Table 2<\/a>? Find the model, and use a graph to check your choice.<\/p>\n<table id=\"Table_04_07_001\" summary=\"..\">\n<caption>Table 2<\/caption>\n<tbody>\n<tr>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\"><strong>[latex]x[\/latex]<\/strong><\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">1<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">2<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">3<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">4<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">5<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">6<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">7<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">8<\/td>\n<td class=\"border\" style=\"width: 43.6563px; text-align: center;\">9<\/td>\n<\/tr>\n<tr>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\"><strong>[latex]y[\/latex]<\/strong><\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">0<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">1.386<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">2.197<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">2.773<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">3.219<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">3.584<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">3.892<\/td>\n<td class=\"border\" style=\"width: 41.6563px; text-align: center;\">4.159<\/td>\n<td class=\"border\" style=\"width: 43.6563px; text-align: center;\">4.394<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id1165135416447\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135416447\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135416447\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135416449\">First, plot the data on a graph as in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_008\">Figure 4<\/a>. For the purpose of graphing, round the data to two significant digits.<\/p>\n<div id=\"CNX_Precalc_Figure_04_07_008\" class=\"small\">\n<div style=\"width: 319px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131432\/CNX_Precalc_Figure_04_07_008.jpg\" alt=\"Graph of the previous table\u2019s values.\" width=\"309\" height=\"302\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 4.<\/strong><\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try [latex]y=a\\mathrm{ln}\\left(bx\\right).[\/latex] Plugging in the first point, [latex]\\left(\\text{1,0}\\right)\\text{,}[\/latex] gives [latex]0=a\\mathrm{ln}b.[\/latex] We reject the case that [latex]a=0[\/latex] (if it were, all outputs would be 0), so we know [latex]\\mathrm{ln}\\left(b\\right)=0.[\/latex] Thus, [latex]b=1[\/latex] and [latex]y=a\\mathrm{ln}\\left(\\text{x}\\right).[\/latex] Next, we can use the point [latex]\\left(\\text{9,4}\\text{.394}\\right)[\/latex] to solve for [latex]a:[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div id=\"eip-id1165134187109\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}y&=a\\mathrm{ln}\\left(x\\right)\\hfill \\\\ 4.394&=a\\mathrm{ln}\\left(9\\right)\\hfill \\\\a&=\\frac{4.394}{\\mathrm{ln}\\left(9\\right)}\\hfill \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<div id=\"fs-id1165134550694\">\n<div id=\"fs-id1165135416447\">\n<p id=\"fs-id1165135536301\">Because [latex]a=\\frac{4.394}{\\mathrm{ln}\\left(9\\right)}\\approx 2,[\/latex] an appropriate model for the data is [latex]y=2\\mathrm{ln}\\left(x\\right).[\/latex]<\/p>\n<p>To check the accuracy of the model, we graph the function together with the given points as in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_009\">Figure 5<\/a>.<\/p>\n<div id=\"CNX_Precalc_Figure_04_07_009\" class=\"small\">\n<div style=\"width: 315px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131436\/CNX_Precalc_Figure_04_07_009a.jpg\" alt=\"Graph of previous table\u2019s values showing that it fits the function y=2ln(x) with an asymptote at x=0.\" width=\"305\" height=\"298\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 5.\u00a0<\/strong>The graph of [latex]y=2\\mathrm{ln}\\left(x\\right).[\/latex]<\/p>\n<\/div>\n<\/div>\n<div><\/div>\n<div>\n<p id=\"fs-id1165135369595\">We can conclude that the model is a good fit to the data.<\/p>\n<p id=\"fs-id1165135369598\">Compare <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_009\">Figure 5<\/a>\u00a0to the graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] shown in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_009a\">Figure 6<\/a>.<\/p>\n<div id=\"CNX_Precalc_Figure_04_07_009a\" class=\"small\">\n<div style=\"width: 320px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131439\/CNX_Precalc_Figure_04_07_009b.jpg\" alt=\"Graph of previous table\u2019s values showing that it fits the function y=2ln(x) with an asymptote at x=0.\" width=\"310\" height=\"303\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 6.\u00a0<\/strong>The graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right).[\/latex]<\/p>\n<\/div>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<p id=\"fs-id1165137836798\">The graphs appear to be identical when [latex]x>0.[\/latex] A quick check confirms this conclusion:\u00a0 [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)=2\\mathrm{ln}\\left(x\\right)[\/latex] for [latex]x>0.[\/latex]<\/p>\n<p id=\"fs-id1165137771398\">However, if [latex]x<0,[\/latex] the graph of [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] includes a \u201cextra\u201d branch, as shown in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_010\">Figure 9<\/a>. This occurs because, while [latex]y=2\\mathrm{ln}\\left(x\\right)[\/latex] cannot have negative values in the domain (as such values would force the argument to be negative), the function [latex]y=\\mathrm{ln}\\left({x}^{2}\\right)[\/latex] can have negative domain values.<\/p>\n<div id=\"CNX_Precalc_Figure_04_07_010\" class=\"small\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131442\/CNX_Precalc_Figure_04_07_010.jpg\" alt=\"Graph of y=ln(x^2).\" width=\"487\" height=\"216\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 7.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"small\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165134054938\" class=\"precalculus tryit\">\n<h3>Try it #5<\/h3>\n<div id=\"ti_04_07_06\">\n<div id=\"fs-id1165137611513\">\n<p id=\"fs-id1165137611515\">Does a linear, exponential, or logarithmic model best fit the data in <a class=\"autogenerated-content\" href=\"#Table_04_07_02\">Table 3<\/a>? Find the model.<\/p>\n<table id=\"Table_04_07_02\" style=\"height: 24px;\" summary=\"..\">\n<caption>Table 3<\/caption>\n<tbody>\n<tr style=\"height: 12px;\">\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\"><strong>[latex]x[\/latex]<\/strong><\/td>\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">1<\/td>\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">2<\/td>\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">3<\/td>\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">4<\/td>\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">5<\/td>\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">6<\/td>\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">7<\/td>\n<td class=\"border\" style=\"height: 12px; width: 46.5px; text-align: center;\">8<\/td>\n<td class=\"border\" style=\"height: 12px; width: 46.5px; text-align: center;\">9<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td class=\"border\" style=\"height: 12px; width: 86.5px; text-align: center;\"><strong>[latex]y[\/latex]<\/strong><\/td>\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">3.297<\/td>\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">5.437<\/td>\n<td class=\"border\" style=\"height: 12px; width: 32.5px; text-align: center;\">8.963<\/td>\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">14.778<\/td>\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">24.365<\/td>\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">40.172<\/td>\n<td class=\"border\" style=\"height: 12px; width: 39.5px; text-align: center;\">66.231<\/td>\n<td class=\"border\" style=\"height: 12px; width: 46.5px; text-align: center;\">109.196<\/td>\n<td class=\"border\" style=\"height: 12px; width: 46.5px; text-align: center;\">180.034<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div>\n<p id=\"fs-id1165135173377\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q165360\">Show Solution<\/span><\/p>\n<div id=\"q165360\" class=\"hidden-answer\" style=\"display: none\">\n<p>Exponential.\u00a0 [latex]y=2{e}^{0.5x}.[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137749158\" class=\"bc-section section\"><\/div>\n<h3>Using Newton\u2019s Law of Cooling (Optional)<\/h3>\n<p id=\"fs-id1165137854986\">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><span class=\"no-emphasis\">vertical shift<\/span><\/strong> of the generic <strong><span class=\"no-emphasis\">exponential decay<\/span><\/strong> function. 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<div id=\"eip-809\" class=\"unnumbered\" style=\"text-align: center;\">[latex]T\\left(t\\right)=a{e}^{kt}+{T}_{s}[\/latex].<\/div>\n<div id=\"eip-741\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div id=\"fs-id1165135629984\">\n<div class=\"textbox shaded\">\n<h3>Newton\u2019s Law of Cooling<\/h3>\n<p id=\"fs-id1165137644683\">The temperature of an object, [latex]T,[\/latex] in surrounding air with temperature [latex]{T}_{s}[\/latex] will behave according to the formula<\/p>\n<div id=\"fs-id1165137660934\" style=\"text-align: center;\">[latex]T\\left(t\\right)=A{e}^{kt}+{T}_{s}[\/latex]<\/div>\n<p>where<\/p>\n<ul id=\"fs-id1165137415924\">\n<li>[latex]t[\/latex] is time,<\/li>\n<li>[latex]A[\/latex] is the difference between the initial temperature of the object and the surroundings, and<\/li>\n<li>[latex]k[\/latex] is a constant, the continuous rate of cooling of the object.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137667467\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"eip-id2434562\"><strong>Given a set of conditions, apply Newton\u2019s Law of Cooling.<\/strong><\/p>\n<ol id=\"fs-id1165137563116\" type=\"1\">\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 [latex]A[\/latex] and [latex]k.[\/latex]<\/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 id=\"Example_04_07_05\" class=\"textbox examples\">\n<div id=\"fs-id1165135422904\">\n<div id=\"fs-id1165135422906\">\n<h3>\u00a0Example 7:\u00a0 Using Newton\u2019s Law of Cooling<\/h3>\n<p id=\"fs-id1165135439846\">A cheesecake is taken out of the oven with an ideal internal temperature of [latex]\\text{165\u00b0F,}[\/latex] and is placed into a [latex]35\u00b0F[\/latex] refrigerator. After 10 minutes, the cheesecake has cooled to [latex]\\text{150\u00b0F}.[\/latex] If we must wait until the cheesecake has cooled to [latex]\\text{70\u00b0F}[\/latex] before we eat it, how long will we have to wait?<\/p>\n<\/div>\n<div id=\"fs-id1165137705784\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137705784\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137705784\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137705786\">Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake\u2019s temperature will decay exponentially toward 35, we have the following equation<\/p>\n<div id=\"eip-id1165134081701\" class=\"unnumbered\" style=\"text-align: center;\">[latex]T\\left(t\\right)=A{e}^{kt}+35[\/latex].[latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1165137563332\">We know the initial temperature was 165, so [latex]T\\left(0\\right)=165.[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div id=\"eip-id1165134353692\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}165&=A{e}^{k0}+35&& \\text{Substitute }\\left(0,165\\right).\\hfill \\\\ A&=130&& \\text{Solve for }A.\\hfill \\end{align*}[\/latex]<\/div>\n<div><\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165137677897\">We were given another data point, [latex]T\\left(10\\right)=150,[\/latex] which we can use to solve for [latex]k.[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div id=\"eip-id1165134081505\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}150&=130{e}^{k10}+35&&\\text{Substitute (10, 150)}.\\\\115&=130{e}^{k10}&& \\text{Subtract 35}.\\\\\\frac{115}{130}&={e}^{10k}&&\\text{Divide by 130}.\\\\\\mathrm{ln}\\left(\\frac{115}{130}\\right)&=10k&&\\text{Take the natural log of both sides}.\\\\k&=\\frac{\\mathrm{ln}\\left(\\frac{115}{130}\\right)}{10}\\approx -0.0123&&\\text{Divide by the coefficient of }k.\\end{align*}[\/latex]<\/div>\n<div><\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165134294911\">This gives us the equation for the cooling of the cheesecake: [latex]T\\left(t\\right)=130{e}^{\u20130.0123t}+35.[\/latex]<\/p>\n<p id=\"fs-id1165135194273\">Now we can solve for the time it will take for the temperature to cool to 70 degrees.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"eip-id1165135149883\" class=\"unnumbered\" style=\"text-align: center;\">[latex]\\begin{align*}70&=130{e}^{-0.0123t}+35&& \\text{Substitute in 70 for }T\\left(t\\right).\\hfill \\\\ 35&=130{e}^{-0.0123t}&& \\text{Subtract 35}.\\hfill \\\\\\frac{35}{130}&={e}^{-0.0123t}&& \\text{Divide by 130}.\\hfill \\\\ \\mathrm{ln}\\left(\\frac{35}{130}\\right)&=-0.0123t&& \\text{Take the natural logarithm.}\\hfill \\\\ t&=\\frac{\\mathrm{ln}\\left(\\frac{35}{130}\\right)}{-0.0123}\\approx 106.68&& \\text{Divide by the coefficient of }t.\\hfill \\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165135186349\">It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to [latex]\\text{70\u00b0F}\\text{.}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135305911\" class=\"precalculus tryit\">\n<h3>Try it #6<\/h3>\n<div id=\"ti_04_07_04\">\n<div id=\"fs-id1165137686700\">\n<p id=\"fs-id1165137686701\">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>\n<div id=\"fs-id1165137686706\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137686706\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137686706\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137686707\">6.026 hours<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137737677\" class=\"bc-section section\">\n<h3>Using Logistic Growth Models (Optional)<\/h3>\n<p id=\"fs-id1165137737682\">Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an <strong><span class=\"no-emphasis\">exponential growth<\/span><\/strong> model, though the exponential growth model is still useful over a short term, before approaching the limiting value.<\/p>\n<p id=\"fs-id1165135194441\">The <strong>logistic growth model<\/strong> is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model\u2019s upper bound, called the <strong>carrying capacity<\/strong>. For constants [latex]a, \\text{ }b,[\/latex] and [latex]c,[\/latex] the logistic growth of a population over time [latex]x[\/latex] is represented by the model<\/p>\n<div id=\"eip-391\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}.[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1165135439859\">The graph in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_006\">Figure 8<\/a> shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.<\/p>\n<div id=\"CNX_Precalc_Figure_04_07_006\" class=\"small\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131426\/CNX_Precalc_Figure_04_07_006.jpg\" alt=\"Graph of f(x)=c\/(1+ae^(-tx)). The carrying capacity is the asymptote at y=c. The initial value of population is (0, c\/(1+a)). The point of maximum growth is (ln(a)\/b, c\/2).\" width=\"487\" height=\"367\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 8.<\/strong><\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135207602\">\n<div class=\"textbox shaded\">\n<h3>Logistic Growth<\/h3>\n<p id=\"fs-id1165135511485\">The logistic growth model is<\/p>\n<div id=\"eip-id1165134239632\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\n<p id=\"eip-id1165135264518\">where<\/p>\n<ul id=\"fs-id1165137526317\">\n<li>[latex]\\frac{c}{1+a}[\/latex] is the initial value<\/li>\n<li>[latex]c[\/latex] is the <em>carrying capacity<\/em>, or <em>limiting value,\u00a0<\/em>and<\/li>\n<li>[latex]b[\/latex] is a constant determined by the rate of growth.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"Example_04_07_06\" class=\"textbox examples\">\n<div id=\"fs-id1165135572080\">\n<div id=\"fs-id1165135572082\">\n<h3>Example 8:\u00a0 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 a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.<\/p>\n<p id=\"fs-id1165134194949\">For example, at time [latex]t=0[\/latex] 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 [latex]b=0.6030.[\/latex] 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>\n<div id=\"fs-id1165135471237\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135471237\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135471237\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135471239\">We substitute the given data into the logistic growth model<\/p>\n<div id=\"eip-id1165133200610\" class=\"unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}.[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1165137698371\">Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is [latex]c=1000.[\/latex] To find [latex]a,[\/latex] we use the formula that the number of cases at time [latex]t=0[\/latex] is [latex]\\frac{c}{1+a}=1,[\/latex] from which it follows that [latex]a=999.[\/latex] This model predicts that, after ten days, the number of people who have had the flu is [latex]f\\left(10\\right)=\\frac{1000}{1+999{e}^{-0.6030\\times 10}}\\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, [latex]c=1000.[\/latex]<\/p>\n<h3>Analysis<\/h3>\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 id=\"fs-id1165135344064\">The graph in <a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_07_007\">Figure 9<\/a>\u00a0gives a good picture of how this model fits the data.<\/p>\n<div id=\"CNX_Precalc_Figure_04_07_007\" class=\"medium\">\n<div style=\"width: 456px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07131429\/CNX_Precalc_Figure_04_07_007.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as \u201cCases\u201d and the x-axis labeled as \u201cDays\u201d. There was 1 case on day 0, 20 on day 5, 294 on day 10, and 1000 on day 21.\" width=\"446\" height=\"300\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 9.\u00a0<\/strong>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>\n<div id=\"fs-id1165135445930\">\n<div id=\"CNX_Precalc_Figure_04_07_007\" class=\"medium\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135262651\" class=\"precalculus tryit\">\n<h3>Try it #7<\/h3>\n<div id=\"ti_04_07_05\">\n<div id=\"fs-id1165135262661\">\n<p id=\"fs-id1165135689538\">Using the model in <a class=\"autogenerated-content\" href=\"#Example_04_07_06\">Example 7<\/a>, estimate the number of cases of flu on day 15.<\/p>\n<\/div>\n<div id=\"fs-id1165135689547\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135689547\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135689547\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135689548\">895 cases on day 15<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135511570\" class=\"bc-section section\"><\/div>\n<div id=\"fs-id1165137749158\" class=\"bc-section section\">\n<div id=\"fs-id1165135171587\" class=\"precalculus media\">\n<div class=\"textbox shaded\">\n<h3>Media:<\/h3>\n<p>Access these online resources for additional instruction and practice with exponential and logarithmic models.<\/p>\n<ul>\n<li><a href=\"http:\/\/openstax.org\/l\/logph\">Logarithm Application \u2013 pH<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/expmodelhalf\">Exponential Model \u2013 Age Using Half-Life<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/newtoncooling\">Newton\u2019s Law of Cooling<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/expgrowthdbl\">Exponential Growth Given Doubling Time<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/initialdouble\">Exponential Growth \u2013 Find Initial Amount Given Doubling Time<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137894245\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1165137894248\">\n<li>The basic exponential function is [latex]f\\left(x\\right)=a{b}^{x}.[\/latex] If [latex]b>1,[\/latex] we have exponential growth; if [latex]0 \\lt b \\lt 1,[\/latex] we have exponential decay.<\/li>\n<li>We can also write this formula in terms of continuous growth as [latex]A={A}_{0}{e}^{kx},[\/latex] where [latex]{A}_{0}\\text{ }[\/latex] is the starting value. If [latex]{A}_{0}[\/latex] is positive, then we have exponential growth when [latex]k>0[\/latex] and exponential decay when [latex]k<0.\\text{ }[\/latex]<\/li>\n<li>In general, we solve problems involving exponential growth or decay in two steps. First, we set up a model and use the model to find the parameters. Then we use the formula with these parameters to predict growth and decay.<\/li>\n<li>Given a substance\u2019s doubling time or half-time, we can find a function that represents its exponential growth or decay.<\/li>\n<li>We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic (Optional) graphs help us to develop models that best fit our data.<\/li>\n<li>Any exponential function with the form [latex]y=a{b}^{x}[\/latex] can be rewritten as an equivalent exponential function with the form [latex]y={A}_{0}{e}^{kx}[\/latex] where [latex]k=\\mathrm{ln}\\left(b\\right).[\/latex]<\/li>\n<li>(Optional) We can use Newton\u2019s Law of Cooling to find how long it will take for a cooling object to reach a desired temperature, or to find what temperature an object will be after a given time.<\/li>\n<li>(Optional) 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<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165134069388\">\n<dt>carrying capacity<\/dt>\n<dd id=\"fs-id1165134069394\">in a logistic model, the limiting value of the output<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134069398\">\n<dt>doubling time<\/dt>\n<dd id=\"fs-id1165135528921\">the time it takes for a quantity to double<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135528925\">\n<dt>half-life<\/dt>\n<dd id=\"fs-id1165135528931\">the length of time it takes for a substance to exponentially decay to half of its original quantity<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135528936\">\n<dt>logistic growth model<\/dt>\n<dd id=\"fs-id1165135528941\">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, [latex]c[\/latex] is the carrying capacity, or limiting value, and [latex]b[\/latex] is a constant determined by the rate of growth<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137844266\">\n<dt>Newton\u2019s Law of Cooling<\/dt>\n<dd>the scientific formula for temperature as a function of time as an object\u2019s temperature is equalized with the ambient temperature<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137844278\">\n<dt>order of magnitude<\/dt>\n<dd id=\"fs-id1165137844283\">the power of ten, when a number is expressed in scientific notation, with one non-zero digit to the left of the decimal<\/dd>\n<\/dl>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-476\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Exponential and Logarithmic Models. <strong>Authored by<\/strong>: Douglas Hoffman. <strong>Provided by<\/strong>: Openstax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:_tqWoaDz@13\/Exponential-and-Logarithmic-Models\">https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:_tqWoaDz@13\/Exponential-and-Logarithmic-Models<\/a>. <strong>Project<\/strong>: Essential Precalcus, Part 1. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":311,"menu_order":13,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Exponential and Logarithmic Models\",\"author\":\"Douglas Hoffman\",\"organization\":\"Openstax\",\"url\":\"https:\/\/cnx.org\/contents\/l3_8ZlRi@1.94:_tqWoaDz@13\/Exponential-and-Logarithmic-Models\",\"project\":\"Essential Precalcus, Part 1\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-476","chapter","type-chapter","status-publish","hentry"],"part":223,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/476","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":29,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/476\/revisions"}],"predecessor-version":[{"id":3284,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/476\/revisions\/3284"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/parts\/223"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/476\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/media?parent=476"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=476"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/contributor?post=476"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/license?post=476"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}