{"id":13724,"date":"2018-08-24T17:39:37","date_gmt":"2018-08-24T17:39:37","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalcone\/?post_type=chapter&#038;p=13724"},"modified":"2020-05-21T05:01:04","modified_gmt":"2020-05-21T05:01:04","slug":"exponential-and-logarithmic-models","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/chapter\/exponential-and-logarithmic-models\/","title":{"raw":"Section 3.5: Exponential and Logarithmic Models","rendered":"Section 3.5: Exponential and Logarithmic Models"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Model exponential growth and decay.<\/li>\r\n \t<li>Use logistic-growth models.<\/li>\r\n \t<li>Express an exponential model in base <em>e<\/em>.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<figure 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-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010832\/CNX_Precalc_Figure_04_07_001F2.jpg\" alt=\"Inside a nuclear research reactor.\" width=\"325\" height=\"409\" \/> <b>Figure 1.<\/b> A nuclear research reactor inside the Neely Nuclear Research Center on the Georgia Institute of Technology campus (credit: Georgia Tech Research Institute)[\/caption]<\/figure>\r\n<p id=\"fs-id1165137748523\">Because the output of exponential functions increases very rapidly, the term \"exponential growth\" is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.<\/p>\r\n\r\n<div id=\"fs-id1165137564690\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Exponential Growth<\/h3>\r\n<p id=\"fs-id1165137834019\">A function that models <strong>exponential growth<\/strong> grows by a rate proportional to the amount present. For any real number <em>x<\/em>\u00a0and any positive real numbers <em>a\u00a0<\/em>and <em>b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form<\/p>\r\n\r\n<div id=\"fs-id1165137851784\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\text{ }f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\r\n<p id=\"eip-626\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137863819\">\r\n \t<li><em>a<\/em>\u00a0is the initial or starting value of the function.<\/li>\r\n \t<li><em>b<\/em>\u00a0is the growth factor or growth multiplier per unit <em>x<\/em>.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165137644244\">In more general terms, we have an <em>exponential function<\/em>, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let\u2019s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function [latex]A\\left(x\\right)=100+50x[\/latex]. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex].<\/p>\r\n<p id=\"fs-id1165135512493\">A few years of growth for these companies are illustrated below.<\/p>\r\n\r\n<table id=\"Table_04_01_05\" summary=\"Six rows and three columns. The first column is labeled, \">\r\n<thead>\r\n<tr>\r\n<th>Year,\u00a0<em>x<\/em><\/th>\r\n<th>Stores, Company A<\/th>\r\n<th>Stores, Company B<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>100 + 50(0) = 100<\/td>\r\n<td>100(1 + 0.5)<sup>0<\/sup> = 100<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1<\/td>\r\n<td>100 + 50(1) = 150<\/td>\r\n<td>100(1 + 0.5)<sup>1<\/sup> = 150<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>100 + 50(2) = 200<\/td>\r\n<td>100(1 + 0.5)<sup>2<\/sup> = 225<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>100 + 50(3) = 250<\/td>\r\n<td>100(1 + 0.5)<sup>3<\/sup> =\u00a0337.5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>x<\/em><\/td>\r\n<td><em>A<\/em>(<em>x<\/em>) = 100 + 50x<\/td>\r\n<td><em>B<\/em>(<em>x<\/em>) = 100(1 + 0.5)<sup><em>x<\/em><\/sup><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe graphs comparing the number of stores for each company over a five-year period are shown in below<strong>.<\/strong> We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010809\/CNX_Precalc_Figure_04_01_0012.jpg\" alt=\"Graph of Companies A and B\u2019s functions, which values are found in the previous table.\" width=\"487\" height=\"845\" \/> <b>Figure 2.<\/b> The graph shows the numbers of stores Companies A and B opened over a five-year period.[\/caption]\r\n<p id=\"fs-id1165135209682\">Notice that the domain for both functions is [latex]\\left[0,\\infty \\right)[\/latex], and the range for both functions is [latex]\\left[100,\\infty \\right)[\/latex]. After year 1, Company B always has more stores than Company A.<\/p>\r\n<p id=\"fs-id1165137836429\">Now we will turn our attention to the function representing the number of stores for Company B, [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]. In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex], where 100 is the initial value, 1.5 is called the <em>base<\/em>, and <em>x<\/em>\u00a0is called the <em>exponent<\/em>.<\/p>\r\n\r\n<div id=\"Example_04_01_03\" class=\"example\">\r\n<div id=\"fs-id1165137535640\" class=\"exercise\">\r\n<div id=\"fs-id1165137535642\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Evaluating a Real-World Exponential Model<\/h3>\r\n<p id=\"fs-id1165135541867\">At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with an annual growth rate of about 1.2%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t}[\/latex], where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?<\/p>\r\n[reveal-answer q=\"807378\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"807378\"]\r\n<p id=\"fs-id1165137786635\">To estimate the population in 2031, we evaluate the models for <em>t\u00a0<\/em>= 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth,<\/p>\r\n<p style=\"text-align: center\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/p>\r\n<p id=\"fs-id1165135394343\">There will be about 1.549 billion people in India in the year 2031.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137635314\">The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about 0.6%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.39{\\left(1.006\\right)}^{t}[\/latex], where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example 2?<\/p>\r\n[reveal-answer q=\"740298\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"740298\"]\r\n\r\nAbout 1.548 billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Find the Equation of an Exponential Function<\/h2>\r\n<section id=\"fs-id1165135526980\">\r\n<p id=\"fs-id1165135526985\">In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants <em>a<\/em>\u00a0and <em>b<\/em>, and evaluate the function.<\/p>\r\n\r\n<div id=\"fs-id1165135369632\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135180102\">How To: Given two data points, write an exponential model.<\/h3>\r\n<ol id=\"fs-id1165135180107\">\r\n \t<li>If one of the data points has the form [latex]\\left(0,a\\right)[\/latex], then <em>a<\/em>\u00a0is the initial value. Using <em>a<\/em>, substitute the second point into the equation [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex], and solve for <em>b<\/em>.<\/li>\r\n \t<li>If neither of the data points have the form [latex]\\left(0,a\\right)[\/latex], substitute both points into two equations with the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex]. Solve the resulting system of two equations in two unknowns to find <em>a<\/em>\u00a0and <em>b<\/em>.<\/li>\r\n \t<li>Using the <em>a<\/em>\u00a0and <em>b<\/em>\u00a0found in the steps above, write the exponential function in the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_01_04\" class=\"example\">\r\n<div id=\"fs-id1165137580876\" class=\"exercise\">\r\n<div id=\"fs-id1165137580878\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Writing an Exponential Model When the Initial Value Is Known<\/h3>\r\n<p id=\"fs-id1165137667588\">In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function <em>N<\/em>(<em>t<\/em>) representing the population <em>N<\/em>\u00a0of deer over time <em>t<\/em>.<\/p>\r\n[reveal-answer q=\"518916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"518916\"]\r\n<p id=\"fs-id1165135188418\">We let our independent variable <em>t<\/em>\u00a0be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, <em>a\u00a0<\/em>= 80. We can now substitute the second point into the equation [latex]N\\left(t\\right)=80{b}^{t}[\/latex] to find <em>b<\/em>:<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}N\\left(t\\right)&amp; =80{b}^{t} \\\\ 180&amp; =80{b}^{6} &amp;&amp; \\text{Substitute using point }\\left(6, 180\\right). \\\\ \\frac{9}{4} &amp; ={b}^{6} &amp;&amp; \\text{Divide and write in lowest terms}. \\\\ b &amp; ={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}} &amp;&amp; \\text{Isolate }b\\text{ using properties of exponents}. \\\\ b &amp; \\approx 1.1447 &amp;&amp; \\text{Round to 4 decimal places}. \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135193230\"><strong>NOTE:<\/strong> <em>Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.<\/em><\/p>\r\n<p id=\"fs-id1165137705073\">The exponential model for the population of deer is [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex]. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)<\/p>\r\nWe can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(0,\\text{ 8}0\\right)[\/latex] and [latex]\\left(\\text{6},\\text{ 18}0\\right)[\/latex]. We can also see that the domain for the function is [latex]\\left[0,\\infty \\right)[\/latex], and the range for the function is [latex]\\left[80,\\infty \\right)[\/latex].\r\n\r\n<span id=\"fs-id1165137598921\"> <img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010810\/CNX_Precalc_Figure_04_01_0022.jpg\" alt=\"Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).\" \/><\/span>\r\n<p style=\"text-align: center\"><strong>Figure 3.\u00a0<\/strong>Graph showing the population of deer over time, [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex], <em>t<\/em>\u00a0years after 2006<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135496547\">A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013 the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population <em>N<\/em>\u00a0of wolves over time <em>t<\/em>.<\/p>\r\n[reveal-answer q=\"901847\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"901847\"]\r\n\r\n[latex]\\left(0,129\\right)[\/latex] and [latex]\\left(2,236\\right);N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Model exponential growth and decay<\/h2>\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 id=\"eip-10\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]A(t)={A}_{0}{e}^{kt}[\/latex]<\/div>\r\n<p id=\"eip-292\">where [latex]{A}_{0}[\/latex] is equal to the value at time zero, <em>e<\/em>\u00a0is Euler\u2019s constant, and <em>k<\/em>\u00a0is a positive constant that determines the rate (percentage) of growth. We may use the <strong>exponential growth<\/strong> function in applications involving <strong>doubling time<\/strong>, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.<\/p>\r\n<p id=\"fs-id1165137416167\">On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the <strong>exponential decay<\/strong> model. Again, we have the form [latex]A(t)={A}_{0}{e}^{kt}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, and <em>e<\/em>\u00a0is Euler\u2019s constant. Now <em>k<\/em>\u00a0is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating <strong>half-life<\/strong>, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.<\/p>\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 Figure 4 and Figure 5. 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<figure class=\"small\"><img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010832\/CNX_Precalc_Figure_04_07_0022.jpg\" alt=\"Graph of y=2e^(3x) with the labeled points (-1\/3, 2\/e), (0, 2), and (1\/3, 2e) and with the asymptote at y=0.\" width=\"1059\" height=\"709\" \/><\/figure>\r\n<figure id=\"CNX_Precalc_Figure_04_07_002\" class=\"small\"><figcaption><strong>Figure 4.<\/strong> A graph showing exponential growth. The equation is [latex]y=2{e}^{3x}[\/latex].<\/figcaption><span id=\"fs-id1165137565761\"><img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010833\/CNX_Precalc_Figure_04_07_0032.jpg\" alt=\"Graph of y=3e^(-2x) with the labeled points (-1\/2, 3e), (0, 3), and (1\/2, 3\/e) and with the asymptote at y=0.\" \/><\/span><\/figure>\r\n<figure id=\"CNX_Precalc_Figure_04_07_003\" class=\"small\"><figcaption><strong>Figure 5.\u00a0<\/strong>A graph showing exponential decay. The equation is [latex]y=3{e}^{-2x}[\/latex].<\/figcaption><\/figure>\r\n<p id=\"fs-id1165137643044\">Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The <strong>order of magnitude<\/strong> is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star, <strong>Proxima Centauri<\/strong>, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is [latex]4.01134972\\times {10}^{13}[\/latex]. So, we could describe this number as having order of magnitude [latex]{10}^{13}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165137414501\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Characteristics of the Exponential Function, [latex]y=A_{0}e^{kt}[\/latex]<\/h3>\r\n<p id=\"fs-id1165137551809\">An exponential function with the form [latex]A(t)={A}_{0}{e}^{kt}[\/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: <em>y\u00a0<\/em>= 0<\/li>\r\n \t<li>domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\r\n \t<li>range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\r\n \t<li>x intercept: none<\/li>\r\n \t<li>y-intercept: [latex]\\left(0,{A}_{0}\\right)[\/latex]<\/li>\r\n \t<li>increasing if <em>k\u00a0<\/em>&gt; 0<\/li>\r\n \t<li>decreasing if <em>k\u00a0<\/em>&lt; 0<\/li>\r\n<\/ul>\r\n<figure id=\"CNX_Precalc_Figure_04_07_004\" class=\"medium\"><span id=\"fs-id1165135511596\"> <img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010833\/CNX_Precalc_Figure_04_07_004new2.jpg\" alt=\"Two graphs of y=(A_0)(e^(kt)) with the asymptote at y=0. The first graph is of when k&gt;0 and with the labeled points (1\/k, (A_0)e), (0, A_0), and (-1\/k, (A_0)\/e). The second graph is of when k&lt;0 and with the labeled points (-1\/k, (A_0)e), (0, A_0), and (1\/k, (A_0)\/e).\" \/><\/span><\/figure>\r\n<figure class=\"medium\"><span id=\"fs-id1165135511596\"><strong>Figure 6.\u00a0<\/strong>An exponential function models exponential growth when <em>k\u00a0<\/em>&gt; 0 and exponential decay when <em>k\u00a0<\/em>&lt; 0.<\/span><\/figure>\r\n<\/div>\r\n<div id=\"Example_04_07_01\" class=\"example\">\r\n<div id=\"fs-id1165137584968\" class=\"exercise\">\r\n<div id=\"fs-id1165137399318\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Exponential Growth<\/h3>\r\n<p id=\"fs-id1165134381668\">A population of bacteria doubles every hour. If the culture started with 10 bacteria, answer the following:<\/p>\r\na) What is the exponential growth formula?\r\nb) How many bacteria are present after 4 hours?\r\nc) When will the population reach 640 bacteria?\r\n\r\n[reveal-answer q=\"644853\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"644853\"]\r\n<p id=\"fs-id1165135261481\"><strong>a)<\/strong> When an amount grows at a fixed percent per unit time, the growth is exponential. To find [latex]{A}_{0}[\/latex] we use the fact that [latex]{A}_{0}[\/latex] is the amount at time zero, so [latex]{A}_{0}=10[\/latex]. To find <em>k<\/em>, use the fact that after one hour [latex]\\left(t=1\\right)[\/latex] the population doubles from 10 to 20. The formula is derived as follows<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;20=10{e}^{k\\cdot 1} \\\\ &amp;2={e}^{k}&amp;&amp; \\text{Divide by 10} \\\\ &amp;\\mathrm{ln}2=k&amp;&amp; \\text{Take the natural logarithm} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id3079167\">so [latex]k=\\mathrm{ln}\\left(2\\right)[\/latex]. Thus the equation we want to graph is [latex]A(t)=10{e}^{\\left(\\mathrm{ln}2\\right)t}=10{\\left({e}^{\\mathrm{ln}2}\\right)}^{t}=10\\cdot {2}^{t}[\/latex].<\/p>\r\n<strong>b)<\/strong> To find the number of bacteria after 4 hours, we will substitute 4 in for t in our formula:\u00a0\u00a0[latex]A(t)=10\\cdot {2}^{t} \\\\A(4)=10\\cdot {2}^{4} \\\\ A(4) = 160\\text{ bacteria.}[\/latex]\r\n<strong>c)\u00a0<\/strong>To find when the bacteria population will reach 640, we will substitute 640 for [latex]A(t)[\/latex]: [latex]\\text{ }A(t)=10\\cdot {2}^{t} \\\\640=10\\cdot {2}^{x} \\\\ 64={2}^{x} \\\\ x=6\\text{ hours.}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_07_01\" class=\"example\">\r\n<div id=\"fs-id1165137584968\" class=\"exercise\">\r\n<div id=\"fs-id1165137399318\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Exponential Growth<\/h3>\r\n<p id=\"fs-id1165134381668\">At the start of an experiment, there are 100 cells. Three hours later, there were 250 cells. Answer the following:<\/p>\r\na) What is the exponential growth formula?\r\nb) How many bacteria are present after 4 hours?\r\nc) When will the population reach 848 cells?\r\n\r\n[reveal-answer q=\"644860\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"644860\"]\r\n<p id=\"fs-id1165135261481\"><strong>a)<\/strong> 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}=100[\/latex]. To find <em>k<\/em>, use the fact that after three hours [latex]\\left(t=3\\right)[\/latex] the population grows from 100 to 250. The formula is derived as follows<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;250=100{e}^{k\\cdot 3} \\\\ &amp;2.5={e}^{3k}&amp;&amp; \\text{Divide by 100} \\\\ &amp;\\mathrm{ln}(2.5)=3k&amp;&amp; \\text{Take the natural logarithm} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id3079167\">so [latex]k=\\dfrac{\\mathrm{ln}\\left(2.5\\right)}{3}\\approx 0.3054[\/latex]. Thus the equation we want to graph is [latex]A(t)=100{e}^{0.3054t}[\/latex].<\/p>\r\n<strong>b)<\/strong> To find the number of cells after 4 hours, we will substitute 4 in for t in our formula:\u00a0\u00a0[latex]A(t)=100{e}^{0.3054t} \\\\ A(4)=100{e}^{0.3054(4)} \\\\ A(4)\\approx 339\\text{ cells.}[\/latex]\r\n<strong>c)\u00a0<\/strong>To find when the bacteria population will reach 848, we will substitute 848 for [latex]A(t)[\/latex]: [latex]A(t)=100{e}^{0.3054t} \\\\848=100{e}^{0.3054t} \\\\ 8.48={e}^{0.3054t} \\\\ t = \\dfrac{\\mathrm{ln}(8.48)}{0.3054}\\approx 7\\text{ hours.}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<section id=\"fs-id1165137803583\">\r\n<h2>Half-Life<\/h2>\r\n<section id=\"fs-id1165137828382\">\r\n<p id=\"fs-id1165137828387\">In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/p>\r\n<p id=\"fs-id1165134192326\">One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its <strong>half-life<\/strong>. The table below\u00a0lists the half-life for several of the more common radioactive substances.<\/p>\r\n\r\n<table id=\"Table_04_06_001\" summary=\"Seven rows and three columns. The first column is labeled, \">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center\">Substance<\/th>\r\n<th style=\"text-align: center\">Use<\/th>\r\n<th style=\"text-align: center\">Half-life<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>gallium-67<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>80 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>cobalt-60<\/td>\r\n<td>manufacturing<\/td>\r\n<td>5.3 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>technetium-99m<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>6 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>americium-241<\/td>\r\n<td>construction<\/td>\r\n<td>432 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>carbon-14<\/td>\r\n<td>archeological dating<\/td>\r\n<td>5,715 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>uranium-235<\/td>\r\n<td>atomic power<\/td>\r\n<td>703,800,000 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\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<p id=\"eip-346\">To find the half-life of a function describing exponential decay, solve the following equation:<\/p>\r\n\r\n<div id=\"eip-942\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}[\/latex]<\/div>\r\n<p id=\"fs-id1165135177747\">We find that the half-life depends only on the constant <em>k<\/em>\u00a0and not on the starting quantity [latex]{A}_{0}[\/latex].<\/p>\r\n<p id=\"fs-id1165137454141\">The formula is derived as follows<\/p>\r\n\r\n<div id=\"eip-167\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}\\frac{1}{2}{A}_{0}&amp;={A}_{o}{e}^{kt} \\\\ \\frac{1}{2}&amp;={e}^{kt}&amp;&amp; \\text{Divide by }{A}_{0}. \\\\ \\mathrm{ln}\\left(\\frac{1}{2}\\right)&amp;=kt&amp;&amp; \\text{Take the natural log}. \\\\ -\\mathrm{ln}\\left(2\\right)&amp;=kt&amp;&amp; \\text{Apply laws of logarithms}. \\\\ -\\frac{\\mathrm{ln}\\left(2\\right)}{t}&amp;=k&amp;&amp; \\text{Divide by }t. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137423324\">Since <em>t<\/em>, the time, is positive, <em>k<\/em>\u00a0must, as expected, be negative. This gives us the half-life formula<\/p>\r\n\r\n<div id=\"fs-id1165137697986\" class=\"equation\" style=\"text-align: center\">[latex]k=-\\frac{\\mathrm{ln}\\left(2\\right)}{t}[\/latex]<\/div>\r\n<div id=\"fs-id1165137418076\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137528941\">How To: Given the half-life, find the decay rate.<\/h3>\r\n<ol id=\"fs-id1165135186559\">\r\n \t<li>Write [latex]A={A}_{o}{e}^{kt}[\/latex].<\/li>\r\n \t<li>Replace <em>A<\/em>\u00a0by [latex]\\frac{1}{2}{A}_{0}[\/latex] and replace <em>t<\/em>\u00a0by the given half-life.<\/li>\r\n \t<li>Solve to find <em>k<\/em>. Express <em>k<\/em>\u00a0as an exact value (do not round).<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1165135613329\">Note: <em>It is also possible to find the decay rate using<\/em> [latex]k=-\\frac{\\mathrm{ln}\\left(2\\right)}{t}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_04_07_02\" class=\"example\">\r\n<div id=\"fs-id1165137827442\" class=\"exercise\">\r\n<div id=\"fs-id1165137732189\" class=\"problem textbox shaded\">\r\n<h3>Example 5: Finding the Function that Describes Radioactive Decay<\/h3>\r\n<p id=\"fs-id1165137501905\">The half-life of Sodium-24, an isotope, is 15 years. Write the decay function that estimates the amount of Sodium-24 after t years given that there is initially 10 grams.\u00a0 How much is left after 20 years, rounded to the nearest gram?<\/p>\r\n[reveal-answer q=\"365040\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"365040\"]\r\n<p id=\"fs-id1165137805508\">This formula is derived as follows.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}A(t)&amp;={A}_{0}{e}^{kt}&amp;&amp; \\text{The continuous growth\/decay formula}.\\\\ k&amp;=\\frac{-\\mathrm{ln}(2)}{15}&amp;&amp; \\text{Substitute the half-life for t and initial amount for }A_{0} \\\\ A(t)&amp;=10{e}^{\\frac{-\\mathrm{ln}(2)t}{15}}&amp;&amp;\\text{This is the decay function} \\\\ A(20)&amp;=10{e}^{\\frac{-\\mathrm{ln}(2)\\cdot 20}{15}}&amp;&amp;\\text{Now plug in 20 for t} \\\\ A(20)&amp;\\approx 4\\text{ grams.} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135527004\">The half-life of Bismuth-212, an isotope, is 60.5 seconds. Write the decay function that estimates the amount of Bismuth-212 after t years given that there is initially 5 grams.\u00a0 How much is left after 80 seconds, rounded to the nearest gram?<\/p>\r\n[reveal-answer q=\"258944\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"258944\"]\r\n\r\nThe decay function is: [latex]\\text{ }A(t)=5{e}^{\\frac{-\\mathrm{ln}(2)t}{60.5}}[\/latex]\r\n\r\nAfter 80 seconds, approximately 2 grams remain.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137828123\">\r\n<h2>Radiocarbon Dating<\/h2>\r\n<p id=\"fs-id1165135154026\">The formula for radioactive decay is important in <strong>radiocarbon dating<\/strong>, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.<\/p>\r\n<p id=\"fs-id1165137731568\">Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates\u2014although the ratio has changed slightly over the centuries.<\/p>\r\n<p id=\"fs-id1165137409468\">As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.<\/p>\r\n<p id=\"fs-id1165135193799\">Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after <em>t<\/em>\u00a0years is<\/p>\r\n\r\n<div id=\"eip-866\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]A\\approx {A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]<\/div>\r\n<p id=\"eip-692\">where<\/p>\r\n\r\n<div id=\"eip-id1165137849256\">\r\n<ul>\r\n \t<li><em>A<\/em>\u00a0is the amount of carbon-14 remaining<\/li>\r\n \t<li>[latex]{A}_{0}[\/latex] is the amount of carbon-14 when the plant or animal began decaying.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165137634094\">This formula is derived as follows:<\/p>\r\n\r\n<div id=\"eip-777\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}A&amp;={A}_{0}{e}^{kt}&amp;&amp; \\text{The continuous growth formula}.\\\\ 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 }f\\left(t\\right). \\\\ 0.5&amp;={e}^{5730k}&amp;&amp; \\text{Divide by }{A}_{0}. \\\\ \\mathrm{ln}\\left(0.5\\right)&amp;=5730k&amp;&amp; \\text{Take the natural log of both sides}. \\\\ k&amp;=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}&amp;&amp; \\text{Divide by the coefficient of }k. \\\\ A&amp;={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}&amp;&amp; \\text{Substitute for }r\\text{ in the continuous growth formula}. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137600416\">To find the age of an object, we solve this equation for <em>t<\/em>:<\/p>\r\n\r\n<div id=\"eip-934\" class=\"equation\" style=\"text-align: center\">[latex]t=\\frac{\\mathrm{ln}\\left(\\frac{A}{{A}_{0}}\\right)}{-0.000121}[\/latex]<\/div>\r\n<p id=\"fs-id1165137841700\">Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let <em>r<\/em>\u00a0be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation [latex]A\\approx {A}_{0}{e}^{-0.000121t}[\/latex] we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is [latex]r=\\frac{A}{{A}_{0}}\\approx {e}^{-0.000121t}[\/latex]. We solve this equation for <em>t<\/em>, to get<\/p>\r\n\r\n<div id=\"eip-453\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex]<\/div>\r\n<div id=\"fs-id1165137807119\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165134086082\">How To: Given the percentage of carbon-14 in an object, determine its age.<\/h3>\r\n<ol id=\"fs-id1165135209232\">\r\n \t<li>Express the given percentage of carbon-14 as an equivalent decimal, <em>k<\/em>.<\/li>\r\n \t<li>Substitute for <em>k<\/em> in the equation [latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex] and solve for the age, <em>t<\/em>.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_07_03\" class=\"example\">\r\n<div id=\"fs-id1165137885960\" class=\"exercise\">\r\n<div id=\"fs-id1165137750064\" class=\"problem textbox shaded\">\r\n<h3>Example 6: Finding the Age of a Bone<\/h3>\r\n<p id=\"fs-id1165137771807\">A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?<\/p>\r\n[reveal-answer q=\"433778\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"433778\"]\r\n<p id=\"fs-id1165135445864\">We substitute 20% = 0.20 for <em>k<\/em>\u00a0in the equation and solve for <em>t<\/em>:<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}t&amp;=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}&amp;&amp; \\text{Use the general form of the equation}. \\\\ &amp;=\\frac{\\mathrm{ln}\\left(0.20\\right)}{-0.000121}&amp;&amp; \\text{Substitute for }r. \\\\ &amp;\\approx 13301&amp;&amp; \\text{Round to the nearest year}. \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135512179\">The bone fragment is about 13,301 years old.<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135205625\">The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as [latex]\\text{13,301 years}\\pm \\text{1% or 13,301 years}\\pm \\text{133 years}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137892459\">Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?<\/p>\r\n[reveal-answer q=\"131296\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"131296\"]less than 230 years, 229.3157 to be exact[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]15893[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Calculating Doubling Time<\/span>\r\n\r\n<\/section><section id=\"fs-id1165134040514\">\r\n<p id=\"fs-id1165137897897\">For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the <strong>doubling time<\/strong>.<\/p>\r\n<p id=\"fs-id1165137447183\">Given the basic <strong>exponential growth<\/strong> equation [latex]A={A}_{0}{e}^{kt}[\/latex], doubling time can be found by solving for when the original quantity has doubled, that is, by solving [latex]2{A}_{0}={A}_{0}{e}^{kt}[\/latex].<\/p>\r\n<p id=\"eip-853\">The formula is derived as follows:<\/p>\r\n\r\n<div id=\"eip-244\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}2{A}_{0}&amp;={A}_{0}{e}^{kt} \\\\ 2&amp;={e}^{kt}&amp;&amp; \\text{Divide by }{A}_{0}. \\\\ \\mathrm{ln}2&amp;=kt&amp;&amp; \\text{Take the natural logarithm}. \\\\ t&amp;=\\frac{\\mathrm{ln}2}{k}&amp;&amp; \\text{Divide by the coefficient of }t. \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137722416\">Thus the doubling time is<\/p>\r\n\r\n<div id=\"eip-495\" class=\"equation\" style=\"text-align: center\">[latex]t=\\frac{\\mathrm{ln}2}{k}[\/latex]<\/div>\r\n<div id=\"Example_04_07_04\" class=\"example\">\r\n<div id=\"fs-id1165137657275\" class=\"exercise\">\r\n<div id=\"fs-id1165137464856\" class=\"problem textbox shaded\">\r\n<h3>Example 7: 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[reveal-answer q=\"854608\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"854608\"]\r\n<p id=\"fs-id1165134433339\">The formula is derived as follows:<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}t&amp;=\\frac{\\mathrm{ln}2}{k}&amp;&amp; \\text{The doubling time formula}.\\\\ 2&amp;=\\frac{\\mathrm{ln}2}{k}&amp;&amp; \\text{Use a doubling time of two years}. \\\\ k&amp;=\\frac{\\mathrm{ln}2}{2}&amp;&amp; \\text{Multiply by }k\\text{ and divide by 2}. \\\\ A&amp;={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}&amp;&amp; \\text{Substitute }k\\text{ into the continuous growth formula}. \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135365796\">The function is [latex]A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\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[reveal-answer q=\"447089\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"447089\"]\r\n\r\n[latex]f\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{3}t}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><\/section><\/div>\r\n<\/div>\r\n<h2>Use logistic-growth models<\/h2>\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>exponential growth<\/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 <em>a<\/em>, <em>b<\/em>, and <em>c<\/em>, the logistic growth of a population over time <em>x<\/em>\u00a0is represented by the model<\/p>\r\n\r\n<div id=\"eip-391\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]f\\left(x\\right)=\\dfrac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\r\n<p id=\"fs-id1165135439859\">Figure 7\u00a0shows 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\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010833\/CNX_Precalc_Figure_04_07_0062.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\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\n<div id=\"fs-id1165135207602\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Logistic Growth<\/h3>\r\n<p id=\"fs-id1165135511485\">The logistic growth model is<\/p>\r\n\r\n<div id=\"eip-id1165134239632\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]f\\left(x\\right)=\\dfrac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\r\n<p id=\"eip-id1165135264518\">where<\/p>\r\n\r\n<div id=\"fs-id1165137526317\">\r\n<ul>\r\n \t<li>[latex]\\frac{c}{1+a}[\/latex] is the initial value<\/li>\r\n \t<li><em>c<\/em>\u00a0is the <em>carrying capacity<\/em>, or <em>limiting value<\/em><\/li>\r\n \t<li><em>b<\/em>\u00a0is 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=\"example\">\r\n<div id=\"fs-id1165135572080\" class=\"exercise\">\r\n<div id=\"fs-id1165135572082\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Using the Logistic-Growth Model<\/h3>\r\n<p id=\"fs-id1165135572087\">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>\r\n<p id=\"fs-id1165134194949\">For example, at time <em>t\u00a0<\/em>= 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is <em>b\u00a0<\/em>= 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.<\/p>\r\n[reveal-answer q=\"788840\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"788840\"]\r\n\r\nWe substitute the given data into the logistic growth model\r\n<p style=\"text-align: center\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\r\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 <em>c\u00a0<\/em>= 1000. To find <em>a<\/em>, we use the formula that the number of cases at time <em>t\u00a0<\/em>= 0 is [latex]\\frac{c}{1+a}=1[\/latex], from which it follows that <em>a\u00a0<\/em>= 999. This model predicts that, after ten days, the number of people who have had the flu is [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}\\approx 293.8[\/latex]. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, <em>c\u00a0<\/em>= 1000.<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135445936\" style=\"text-align: left\">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>\r\n<p id=\"fs-id1165135344064\" style=\"text-align: left\">Figure 8\u00a0gives a good picture of how this model fits the data.<\/p>\r\n\r\n<figure class=\"medium\"><img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010834\/CNX_Precalc_Figure_04_07_0072.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as \" \/><\/figure>\r\n<figure id=\"CNX_Precalc_Figure_04_07_007\" class=\"medium\"><figcaption><strong>Figure 8.\u00a0<\/strong>The graph of [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}[\/latex]<\/figcaption><\/figure>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135689538\">Using the model in Example 5, estimate the number of cases of flu on day 15.<\/p>\r\n[reveal-answer q=\"325943\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"325943\"]\r\n\r\n895 cases on day 15\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]54629[\/ohm_question]\r\n\r\n<\/div>\r\n<section id=\"fs-id1165135511570\"><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Expressing an Exponential Model in Base <\/span><em style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">e<\/em><\/section><section id=\"fs-id1165137749158\">\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 <em>e<\/em>\u00a0is often preferred. We can use laws of exponents and laws of logarithms to change any base to base <em>e<\/em>.<\/p>\r\n\r\n<div id=\"fs-id1165134342616\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137452983\">How To: Given a model with the form [latex]y=a{b}^{x}[\/latex], change it to the form [latex]y={A}_{0}{e}^{kx}[\/latex].<\/h3>\r\n<ol id=\"fs-id1165134148496\">\r\n \t<li>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 y 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=\"example\">\r\n<div id=\"fs-id1165134261257\" class=\"exercise\">\r\n<div id=\"fs-id1165132957903\" class=\"problem textbox shaded\">\r\n<h3>Example 9: Changing to base <em>e<\/em><\/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[reveal-answer q=\"865754\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"865754\"]\r\n<p id=\"fs-id1165137432379\">The formula is derived as follows<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}y&amp;=2.5{\\left(3.1\\right)}^{x} \\\\ &amp;=2.5{e}^{\\mathrm{ln}\\left({3.1}^{x}\\right)}&amp;&amp; \\text{Insert exponential and its inverse.} \\\\ &amp;=2.5{e}^{x\\mathrm{ln}3.1}&amp;&amp; \\text{Laws of logs.} \\\\ &amp;=2.5{e}^{\\left(\\mathrm{ln}3.1\\right)}{}^{x}&amp;&amp; \\text{Commutative law of multiplication} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135664963\">Change the function [latex]y=3{\\left(0.5\\right)}^{x}[\/latex] to one having <i>e<\/i>\u00a0as the base.<\/p>\r\n[reveal-answer q=\"843558\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"843558\"]\r\n\r\n[latex]y=3{e}^{\\left(\\mathrm{ln}0.5\\right)x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]15878[\/ohm_question]\r\n\r\n<\/div>\r\n<span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Key Equations<\/span>\r\n\r\n<section id=\"fs-id1165135208780\" class=\"key-equations\">\r\n<table id=\"fs-id2347772\" summary=\"..\">\r\n<tbody>\r\n<tr>\r\n<td>Half-life formula<\/td>\r\n<td>If [latex]\\text{ }A={A}_{0}{e}^{kt}[\/latex], <em>k\u00a0<\/em>&lt; 0, the half-life is [latex]t=-\\frac{\\mathrm{ln}\\left(2\\right)}{k}[\/latex].<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Carbon-14 dating<\/td>\r\n<td>[latex]t=\\frac{\\mathrm{ln}\\left(\\frac{A}{{A}_{0}}\\right)}{-0.000121}[\/latex].[latex]{A}_{0}[\/latex] <em>A<\/em>\u00a0is the amount of carbon-14 when the plant or animal died\r\n\r\n<em>t<\/em>\u00a0is the amount of carbon-14 remaining today\r\n\r\nis the age of the fossil in years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Doubling time formula<\/td>\r\n<td>If [latex]A={A}_{0}{e}^{kt}[\/latex], <em>k\u00a0<\/em>&gt; 0, the doubling time is [latex]t=\\frac{\\mathrm{ln}2}{k}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Newton\u2019s Law of Cooling<\/td>\r\n<td>[latex]T\\left(t\\right)=A{e}^{kt}+{T}_{s}[\/latex], where [latex]{T}_{s}[\/latex] is the ambient temperature, [latex]A=T\\left(0\\right)-{T}_{s}[\/latex], and <em>k<\/em> is the continuous rate of cooling.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section><section id=\"fs-id1165137894245\" class=\"key-concepts\">\r\n<h2>Key Concepts<\/h2>\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 <em>b\u00a0<\/em>&gt; 1, we have exponential growth; if 0 &lt; <em>b\u00a0<\/em>&lt; 1, we have exponential decay.<\/li>\r\n \t<li>We can also write this formula in terms of continuous growth as [latex]A={A}_{0}{e}^{kx}[\/latex], where [latex]{A}_{0}[\/latex] is the starting value. If [latex]{A}_{0}[\/latex] is positive, then we have exponential growth when <em>k\u00a0<\/em>&gt; 0 and exponential decay when <em>k\u00a0<\/em>&lt; 0.<\/li>\r\n \t<li>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>We can find the age, <em>t<\/em>, of an organic artifact by measuring the amount, <em>k<\/em>, of carbon-14 remaining in the artifact and using the formula [latex]t=\\frac{\\mathrm{ln}\\left(k\\right)}{-0.000121}[\/latex] to solve for <em>t<\/em>.<\/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 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>We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors.<\/li>\r\n \t<li>We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data.<\/li>\r\n \t<li>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}b[\/latex].<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165134069388\" class=\"definition\">\r\n \t<dt><strong>carrying capacity<\/strong><\/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\" class=\"definition\">\r\n \t<dt><strong>doubling time<\/strong><\/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\" class=\"definition\">\r\n \t<dt><strong>half-life<\/strong><\/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\" class=\"definition\">\r\n \t<dt><strong>logistic growth model<\/strong><\/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, <em>c<\/em>\u00a0is the carrying capacity, or limiting value, and <em>b<\/em>\u00a0is a constant determined by the rate of growth<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137844266\" class=\"definition\">\r\n \t<dt><strong>Newton\u2019s Law of Cooling<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137844271\">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\" class=\"definition\">\r\n \t<dt><strong>order of magnitude<\/strong><\/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<\/section><\/div>\r\n<\/div>\r\n<\/section>&nbsp;\r\n<h2 style=\"text-align: center\">Section 3.5 Homework Exercises<\/h2>\r\n1. With what kind of exponential model would half-life be associated? What role does half-life play in these models?\r\n\r\n2. What is carbon dating? Why does it work? Give an example in which carbon dating would be useful.\r\n\r\n3. With what kind of exponential model would doubling time be associated? What role does doubling time play in these models?\r\n\r\n4. Why use a logistic model instead of an exponential model?\r\n\r\n5. The population of a colony of flies fits an exponential growth model. If there are 1000 flies initially and there are 1800 after 1 day, answer the following:\r\na) What is the exponential growth model where t represents the number of days?\r\nb) What is the size of the colony after 4 days? Round to the nearest whole number.\r\nc) How long is it until there are 30,000 flies? Round to the nearest tenth.\r\n\r\n6. The population of a colony of gnats fits an exponential growth model. If there are 500 gnats initially and there are 800 after 1 day, answer the following:\r\na) What is the exponential growth model where t represents the number of days?\r\nb) What is the size of the colony after 4 days? Round to the nearest whole number.\r\nc) How long is it until there are 90,000 gnats? Round to the nearest tenth.\r\n\r\nFor the following exercises, use the logistic growth model [latex]f\\left(x\\right)=\\frac{150}{1+8{e}^{-2x}}[\/latex].\r\n\r\n7. Find and interpret [latex]f\\left(0\\right)[\/latex]. Round to the nearest tenth.\r\n\r\n8.\u00a0Find and interpret [latex]f\\left(4\\right)[\/latex]. Round to the nearest tenth.\r\n\r\n9. Find the carrying capacity.\r\n\r\n10.\u00a0Graph the model.\r\n\r\n11. Determine whether the data from the table could best be represented as a function that is linear, exponential, or logarithmic. Then write a formula for a model that represents the data.\r\n\r\nx f\u00a0(x)\r\n\u20132 0.694\r\n\u20131 0.833\r\n0 1\r\n1 1.2\r\n2 1.44\r\n3 1.728\r\n4 2.074\r\n5 2.488\r\n\r\n12. Rewrite [latex]f\\left(x\\right)=1.68{\\left(0.65\\right)}^{x}[\/latex] as an exponential equation with base e\u00a0to five significant digits.\r\n\r\nFor the following exercises, use a graphing calculator and this scenario: the population of a fish farm in t\u00a0years is modeled by the equation [latex]P\\left(t\\right)=\\frac{1000}{1+9{e}^{-0.6t}}[\/latex].\r\n\r\n13. Graph the function.\r\n\r\n14.\u00a0What is the initial population of fish?\r\n\r\n15. To the nearest tenth, what is the doubling time for the fish population?\r\n\r\n16.\u00a0To the nearest whole number, what will the fish population be after 2 years?\r\n\r\n17. To the nearest tenth, how long will it take for the population to reach 900?\r\n\r\n18.\u00a0What is the carrying capacity for the fish population? Justify your answer using the graph of P.\r\n\r\n19. A substance has a half-life of 2.045 minutes. If the initial amount of the substance was 132.8 grams, how many half-lives will have passed before the substance decays to 8.3 grams? What is the total time of decay?\r\n\r\n20.\u00a0The formula for an increasing population is given by [latex]P\\left(t\\right)={P}_{0}{e}^{rt}[\/latex] where [latex]{P}_{0}[\/latex] is the initial population and r\u00a0&gt; 0. Derive a general formula for the time t it takes for the population to increase by a factor of M.\r\n\r\n21. Recall the formula for calculating the magnitude of an earthquake, [latex]M=\\frac{2}{3}\\mathrm{log}\\left(\\frac{S}{{S}_{0}}\\right)[\/latex]. Show each step for solving this equation algebraically for the seismic moment S.\r\n\r\n22. What is the y-intercept of the logistic growth model [latex]y=\\frac{c}{1+a{e}^{-rx}}[\/latex]? Show the steps for calculation. What does this point tell us about the population?\r\n\r\n23. Prove that [latex]{b}^{x}={e}^{x\\mathrm{ln}\\left(b\\right)}[\/latex] for positive [latex]b\\ne 1[\/latex].\r\n\r\nFor the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour.\r\n\r\n24. To the nearest hour, what is the half-life of the drug?\r\n\r\n25. Write an exponential model representing the amount of the drug remaining in the patient\u2019s system after t\u00a0hours. Then use the formula to find the amount of the drug that would remain in the patient\u2019s system after 3 hours. Round to the nearest milligram.\r\n\r\n26.\u00a0Using the model found in the previous exercise, find [latex]f\\left(10\\right)[\/latex] and interpret the result. Round to the nearest hundredth.\r\n\r\nFor the following exercises, use this scenario: A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day.\r\n\r\n27. To the nearest day, how long will it take for half of the Iodine-125 to decay?\r\n\r\n28.\u00a0Write an exponential model representing the amount of Iodine-125 remaining in the tumor after t\u00a0days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 60 days. Round to the nearest tenth of a gram.\r\n\r\n29. A scientist begins with 250 grams of a radioactive substance. After 250 minutes, the sample has decayed to 32 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance?\r\n\r\n30.\u00a0The half-life of Radium-226 is 1590 years. What is the annual decay rate? Express the decimal result to four significant digits and the percentage to two significant digits.\r\n\r\n31. The half-life of Erbium-165 is 10.4 hours. What is the hourly decay rate? Express the decimal result to four significant digits and the percentage to two significant digits.\r\n\r\n32.\u00a0A wooden artifact from an archeological dig contains 60 percent of the carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of carbon-14 is 5730 years.)\r\n\r\n33. A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?\r\n\r\nFor the following exercises, use this scenario: A biologist recorded a count of 360 bacteria present in a culture after 5 minutes and 1000 bacteria present after 20 minutes.\r\n\r\n34. To the nearest whole number, what was the initial population in the culture?\r\n\r\n35. Rounding to six significant digits, write an exponential equation representing this situation. To the nearest minute, how long did it take the population to double?\r\n\r\n36. Plot each set of approximate values of intensity of sounds on a logarithmic scale: Whisper: [latex]{10}^{-10} \\frac{W}{{m}^{2}}[\/latex], Vacuum: [latex]{10}^{-4}\\frac{W}{{m}^{2}}[\/latex], Jet: [latex]{10}^{2} \\frac{W}{{m}^{2}}[\/latex]\r\n\r\n37. Recall the formula for calculating the magnitude of an earthquake, [latex]M=\\frac{2}{3}\\mathrm{log}\\left(\\frac{S}{{S}_{0}}\\right)[\/latex]. One earthquake has magnitude 3.9 on the MMS scale. If a second earthquake has 750 times as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth.\r\n\r\nFor the following exercises, use this scenario: The equation [latex]N\\left(t\\right)=\\frac{500}{1+49{e}^{-0.7t}}[\/latex] models the number of people in a town who have heard a rumor after t days.\r\n\r\n38. How many people started the rumor?\r\n\r\n39. To the nearest whole number, how many people will have heard the rumor after 3 days?\r\n\r\n40.\u00a0As t\u00a0increases without bound, what value does N(t) approach? Interpret your answer.\r\n\r\nFor the following exercise, choose the correct answer choice.\r\n\r\n41. A doctor and injects a patient with 13 milligrams of radioactive dye that decays exponentially. After 12 minutes, there are 4.75 milligrams of dye remaining in the patient\u2019s system. Which is an appropriate model for this situation?\r\n\r\nA. [latex]f\\left(t\\right)=13{\\left(0.0805\\right)}^{t}[\/latex]\r\nB. [latex]f\\left(t\\right)=13{e}^{0.9195t}[\/latex]\r\nC. [latex]f\\left(t\\right)=13{e}^{\\left(-0.0839t\\right)}[\/latex]\r\nD. [latex]f\\left(t\\right)=\\frac{4.75}{1+13{e}^{-0.83925t}}[\/latex]","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Model exponential growth and decay.<\/li>\n<li>Use logistic-growth models.<\/li>\n<li>Express an exponential model in base <em>e<\/em>.<\/li>\n<\/ul>\n<\/div>\n<figure 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-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010832\/CNX_Precalc_Figure_04_07_001F2.jpg\" alt=\"Inside a nuclear research reactor.\" width=\"325\" height=\"409\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1.<\/b> 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<\/figure>\n<p id=\"fs-id1165137748523\">Because the output of exponential functions increases very rapidly, the term &#8220;exponential growth&#8221; is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.<\/p>\n<div id=\"fs-id1165137564690\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Exponential Growth<\/h3>\n<p id=\"fs-id1165137834019\">A function that models <strong>exponential growth<\/strong> grows by a rate proportional to the amount present. For any real number <em>x<\/em>\u00a0and any positive real numbers <em>a\u00a0<\/em>and <em>b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form<\/p>\n<div id=\"fs-id1165137851784\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\text{ }f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\n<p id=\"eip-626\">where<\/p>\n<ul id=\"fs-id1165137863819\">\n<li><em>a<\/em>\u00a0is the initial or starting value of the function.<\/li>\n<li><em>b<\/em>\u00a0is the growth factor or growth multiplier per unit <em>x<\/em>.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137644244\">In more general terms, we have an <em>exponential function<\/em>, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let\u2019s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function [latex]A\\left(x\\right)=100+50x[\/latex]. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex].<\/p>\n<p id=\"fs-id1165135512493\">A few years of growth for these companies are illustrated below.<\/p>\n<table id=\"Table_04_01_05\" summary=\"Six rows and three columns. The first column is labeled,\">\n<thead>\n<tr>\n<th>Year,\u00a0<em>x<\/em><\/th>\n<th>Stores, Company A<\/th>\n<th>Stores, Company B<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>100 + 50(0) = 100<\/td>\n<td>100(1 + 0.5)<sup>0<\/sup> = 100<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>100 + 50(1) = 150<\/td>\n<td>100(1 + 0.5)<sup>1<\/sup> = 150<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>100 + 50(2) = 200<\/td>\n<td>100(1 + 0.5)<sup>2<\/sup> = 225<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>100 + 50(3) = 250<\/td>\n<td>100(1 + 0.5)<sup>3<\/sup> =\u00a0337.5<\/td>\n<\/tr>\n<tr>\n<td><em>x<\/em><\/td>\n<td><em>A<\/em>(<em>x<\/em>) = 100 + 50x<\/td>\n<td><em>B<\/em>(<em>x<\/em>) = 100(1 + 0.5)<sup><em>x<\/em><\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The graphs comparing the number of stores for each company over a five-year period are shown in below<strong>.<\/strong> We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010809\/CNX_Precalc_Figure_04_01_0012.jpg\" alt=\"Graph of Companies A and B\u2019s functions, which values are found in the previous table.\" width=\"487\" height=\"845\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2.<\/b> The graph shows the numbers of stores Companies A and B opened over a five-year period.<\/p>\n<\/div>\n<p id=\"fs-id1165135209682\">Notice that the domain for both functions is [latex]\\left[0,\\infty \\right)[\/latex], and the range for both functions is [latex]\\left[100,\\infty \\right)[\/latex]. After year 1, Company B always has more stores than Company A.<\/p>\n<p id=\"fs-id1165137836429\">Now we will turn our attention to the function representing the number of stores for Company B, [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]. In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex], where 100 is the initial value, 1.5 is called the <em>base<\/em>, and <em>x<\/em>\u00a0is called the <em>exponent<\/em>.<\/p>\n<div id=\"Example_04_01_03\" class=\"example\">\n<div id=\"fs-id1165137535640\" class=\"exercise\">\n<div id=\"fs-id1165137535642\" class=\"problem textbox shaded\">\n<h3>Example 1: Evaluating a Real-World Exponential Model<\/h3>\n<p id=\"fs-id1165135541867\">At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with an annual growth rate of about 1.2%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t}[\/latex], where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q807378\">Show Solution<\/span><\/p>\n<div id=\"q807378\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137786635\">To estimate the population in 2031, we evaluate the models for <em>t\u00a0<\/em>= 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth,<\/p>\n<p style=\"text-align: center\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/p>\n<p id=\"fs-id1165135394343\">There will be about 1.549 billion people in India in the year 2031.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137635314\">The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about 0.6%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.39{\\left(1.006\\right)}^{t}[\/latex], where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example 2?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q740298\">Show Solution<\/span><\/p>\n<div id=\"q740298\" class=\"hidden-answer\" style=\"display: none\">\n<p>About 1.548 billion people; by the year 2031, India\u2019s population will exceed China\u2019s by about 0.001 billion, or 1 million people.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Find the Equation of an Exponential Function<\/h2>\n<section id=\"fs-id1165135526980\">\n<p id=\"fs-id1165135526985\">In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants <em>a<\/em>\u00a0and <em>b<\/em>, and evaluate the function.<\/p>\n<div id=\"fs-id1165135369632\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135180102\">How To: Given two data points, write an exponential model.<\/h3>\n<ol id=\"fs-id1165135180107\">\n<li>If one of the data points has the form [latex]\\left(0,a\\right)[\/latex], then <em>a<\/em>\u00a0is the initial value. Using <em>a<\/em>, substitute the second point into the equation [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex], and solve for <em>b<\/em>.<\/li>\n<li>If neither of the data points have the form [latex]\\left(0,a\\right)[\/latex], substitute both points into two equations with the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex]. Solve the resulting system of two equations in two unknowns to find <em>a<\/em>\u00a0and <em>b<\/em>.<\/li>\n<li>Using the <em>a<\/em>\u00a0and <em>b<\/em>\u00a0found in the steps above, write the exponential function in the form [latex]f\\left(x\\right)=a{\\left(b\\right)}^{x}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_01_04\" class=\"example\">\n<div id=\"fs-id1165137580876\" class=\"exercise\">\n<div id=\"fs-id1165137580878\" class=\"problem textbox shaded\">\n<h3>Example 2: Writing an Exponential Model When the Initial Value Is Known<\/h3>\n<p id=\"fs-id1165137667588\">In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function <em>N<\/em>(<em>t<\/em>) representing the population <em>N<\/em>\u00a0of deer over time <em>t<\/em>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q518916\">Show Solution<\/span><\/p>\n<div id=\"q518916\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135188418\">We let our independent variable <em>t<\/em>\u00a0be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, <em>a\u00a0<\/em>= 80. We can now substitute the second point into the equation [latex]N\\left(t\\right)=80{b}^{t}[\/latex] to find <em>b<\/em>:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}N\\left(t\\right)& =80{b}^{t} \\\\ 180& =80{b}^{6} && \\text{Substitute using point }\\left(6, 180\\right). \\\\ \\frac{9}{4} & ={b}^{6} && \\text{Divide and write in lowest terms}. \\\\ b & ={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}} && \\text{Isolate }b\\text{ using properties of exponents}. \\\\ b & \\approx 1.1447 && \\text{Round to 4 decimal places}. \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135193230\"><strong>NOTE:<\/strong> <em>Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.<\/em><\/p>\n<p id=\"fs-id1165137705073\">The exponential model for the population of deer is [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex]. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)<\/p>\n<p>We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(0,\\text{ 8}0\\right)[\/latex] and [latex]\\left(\\text{6},\\text{ 18}0\\right)[\/latex]. We can also see that the domain for the function is [latex]\\left[0,\\infty \\right)[\/latex], and the range for the function is [latex]\\left[80,\\infty \\right)[\/latex].<\/p>\n<p><span id=\"fs-id1165137598921\"> <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010810\/CNX_Precalc_Figure_04_01_0022.jpg\" alt=\"Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).\" \/><\/span><\/p>\n<p style=\"text-align: center\"><strong>Figure 3.\u00a0<\/strong>Graph showing the population of deer over time, [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex], <em>t<\/em>\u00a0years after 2006<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135496547\">A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013 the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population <em>N<\/em>\u00a0of wolves over time <em>t<\/em>.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q901847\">Show Solution<\/span><\/p>\n<div id=\"q901847\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(0,129\\right)[\/latex] and [latex]\\left(2,236\\right);N\\left(t\\right)=129{\\left(\\text{1}\\text{.3526}\\right)}^{t}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Model exponential growth and decay<\/h2>\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 id=\"eip-10\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]A(t)={A}_{0}{e}^{kt}[\/latex]<\/div>\n<p id=\"eip-292\">where [latex]{A}_{0}[\/latex] is equal to the value at time zero, <em>e<\/em>\u00a0is Euler\u2019s constant, and <em>k<\/em>\u00a0is a positive constant that determines the rate (percentage) of growth. We may use the <strong>exponential growth<\/strong> function in applications involving <strong>doubling time<\/strong>, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time. In some applications, however, as we will see when we discuss the logistic equation, the logistic model sometimes fits the data better than the exponential model.<\/p>\n<p id=\"fs-id1165137416167\">On the other hand, if a quantity is falling rapidly toward zero, without ever reaching zero, then we should probably choose the <strong>exponential decay<\/strong> model. Again, we have the form [latex]A(t)={A}_{0}{e}^{kt}[\/latex] where [latex]{A}_{0}[\/latex] is the starting value, and <em>e<\/em>\u00a0is Euler\u2019s constant. Now <em>k<\/em>\u00a0is a negative constant that determines the rate of decay. We may use the exponential decay model when we are calculating <strong>half-life<\/strong>, or the time it takes for a substance to exponentially decay to half of its original quantity. We use half-life in applications involving radioactive isotopes.<\/p>\n<p 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 Figure 4 and Figure 5. 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<figure class=\"small\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010832\/CNX_Precalc_Figure_04_07_0022.jpg\" alt=\"Graph of y=2e^(3x) with the labeled points (-1\/3, 2\/e), (0, 2), and (1\/3, 2e) and with the asymptote at y=0.\" width=\"1059\" height=\"709\" \/><\/figure>\n<figure id=\"CNX_Precalc_Figure_04_07_002\" class=\"small\"><figcaption><strong>Figure 4.<\/strong> A graph showing exponential growth. The equation is [latex]y=2{e}^{3x}[\/latex].<\/figcaption><span id=\"fs-id1165137565761\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010833\/CNX_Precalc_Figure_04_07_0032.jpg\" alt=\"Graph of y=3e^(-2x) with the labeled points (-1\/2, 3e), (0, 3), and (1\/2, 3\/e) and with the asymptote at y=0.\" \/><\/span><\/figure>\n<figure id=\"CNX_Precalc_Figure_04_07_003\" class=\"small\"><figcaption><strong>Figure 5.\u00a0<\/strong>A graph showing exponential decay. The equation is [latex]y=3{e}^{-2x}[\/latex].<\/figcaption><\/figure>\n<p id=\"fs-id1165137643044\">Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often use orders of magnitude. The <strong>order of magnitude<\/strong> is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star, <strong>Proxima Centauri<\/strong>, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is [latex]4.01134972\\times {10}^{13}[\/latex]. So, we could describe this number as having order of magnitude [latex]{10}^{13}[\/latex].<\/p>\n<div id=\"fs-id1165137414501\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Characteristics of the Exponential Function, [latex]y=A_{0}e^{kt}[\/latex]<\/h3>\n<p id=\"fs-id1165137551809\">An exponential function with the form [latex]A(t)={A}_{0}{e}^{kt}[\/latex] has the following characteristics:<\/p>\n<ul id=\"fs-id1165137530021\">\n<li>one-to-one function<\/li>\n<li>horizontal asymptote: <em>y\u00a0<\/em>= 0<\/li>\n<li>domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\n<li>range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\n<li>x intercept: none<\/li>\n<li>y-intercept: [latex]\\left(0,{A}_{0}\\right)[\/latex]<\/li>\n<li>increasing if <em>k\u00a0<\/em>&gt; 0<\/li>\n<li>decreasing if <em>k\u00a0<\/em>&lt; 0<\/li>\n<\/ul>\n<figure id=\"CNX_Precalc_Figure_04_07_004\" class=\"medium\"><span id=\"fs-id1165135511596\"> <img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010833\/CNX_Precalc_Figure_04_07_004new2.jpg\" alt=\"Two graphs of y=(A_0)(e^(kt)) with the asymptote at y=0. The first graph is of when k&gt;0 and with the labeled points (1\/k, (A_0)e), (0, A_0), and (-1\/k, (A_0)\/e). The second graph is of when k&lt;0 and with the labeled points (-1\/k, (A_0)e), (0, A_0), and (1\/k, (A_0)\/e).\" \/><\/span><\/figure>\n<figure class=\"medium\"><span id=\"fs-id1165135511596\"><strong>Figure 6.\u00a0<\/strong>An exponential function models exponential growth when <em>k\u00a0<\/em>&gt; 0 and exponential decay when <em>k\u00a0<\/em>&lt; 0.<\/span><\/figure>\n<\/div>\n<div id=\"Example_04_07_01\" class=\"example\">\n<div id=\"fs-id1165137584968\" class=\"exercise\">\n<div id=\"fs-id1165137399318\" class=\"problem textbox shaded\">\n<h3>Example 3: Exponential Growth<\/h3>\n<p id=\"fs-id1165134381668\">A population of bacteria doubles every hour. If the culture started with 10 bacteria, answer the following:<\/p>\n<p>a) What is the exponential growth formula?<br \/>\nb) How many bacteria are present after 4 hours?<br \/>\nc) When will the population reach 640 bacteria?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q644853\">Show Solution<\/span><\/p>\n<div id=\"q644853\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135261481\"><strong>a)<\/strong> When an amount grows at a fixed percent per unit time, the growth is exponential. To find [latex]{A}_{0}[\/latex] we use the fact that [latex]{A}_{0}[\/latex] is the amount at time zero, so [latex]{A}_{0}=10[\/latex]. To find <em>k<\/em>, use the fact that after one hour [latex]\\left(t=1\\right)[\/latex] the population doubles from 10 to 20. The formula is derived as follows<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&20=10{e}^{k\\cdot 1} \\\\ &2={e}^{k}&& \\text{Divide by 10} \\\\ &\\mathrm{ln}2=k&& \\text{Take the natural logarithm} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id3079167\">so [latex]k=\\mathrm{ln}\\left(2\\right)[\/latex]. Thus the equation we want to graph is [latex]A(t)=10{e}^{\\left(\\mathrm{ln}2\\right)t}=10{\\left({e}^{\\mathrm{ln}2}\\right)}^{t}=10\\cdot {2}^{t}[\/latex].<\/p>\n<p><strong>b)<\/strong> To find the number of bacteria after 4 hours, we will substitute 4 in for t in our formula:\u00a0\u00a0[latex]A(t)=10\\cdot {2}^{t} \\\\A(4)=10\\cdot {2}^{4} \\\\ A(4) = 160\\text{ bacteria.}[\/latex]<br \/>\n<strong>c)\u00a0<\/strong>To find when the bacteria population will reach 640, we will substitute 640 for [latex]A(t)[\/latex]: [latex]\\text{ }A(t)=10\\cdot {2}^{t} \\\\640=10\\cdot {2}^{x} \\\\ 64={2}^{x} \\\\ x=6\\text{ hours.}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_07_01\" class=\"example\">\n<div id=\"fs-id1165137584968\" class=\"exercise\">\n<div id=\"fs-id1165137399318\" class=\"problem textbox shaded\">\n<h3>Example 4: Exponential Growth<\/h3>\n<p id=\"fs-id1165134381668\">At the start of an experiment, there are 100 cells. Three hours later, there were 250 cells. Answer the following:<\/p>\n<p>a) What is the exponential growth formula?<br \/>\nb) How many bacteria are present after 4 hours?<br \/>\nc) When will the population reach 848 cells?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q644860\">Show Solution<\/span><\/p>\n<div id=\"q644860\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135261481\"><strong>a)<\/strong> 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}=100[\/latex]. To find <em>k<\/em>, use the fact that after three hours [latex]\\left(t=3\\right)[\/latex] the population grows from 100 to 250. The formula is derived as follows<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&250=100{e}^{k\\cdot 3} \\\\ &2.5={e}^{3k}&& \\text{Divide by 100} \\\\ &\\mathrm{ln}(2.5)=3k&& \\text{Take the natural logarithm} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id3079167\">so [latex]k=\\dfrac{\\mathrm{ln}\\left(2.5\\right)}{3}\\approx 0.3054[\/latex]. Thus the equation we want to graph is [latex]A(t)=100{e}^{0.3054t}[\/latex].<\/p>\n<p><strong>b)<\/strong> To find the number of cells after 4 hours, we will substitute 4 in for t in our formula:\u00a0\u00a0[latex]A(t)=100{e}^{0.3054t} \\\\ A(4)=100{e}^{0.3054(4)} \\\\ A(4)\\approx 339\\text{ cells.}[\/latex]<br \/>\n<strong>c)\u00a0<\/strong>To find when the bacteria population will reach 848, we will substitute 848 for [latex]A(t)[\/latex]: [latex]A(t)=100{e}^{0.3054t} \\\\848=100{e}^{0.3054t} \\\\ 8.48={e}^{0.3054t} \\\\ t = \\dfrac{\\mathrm{ln}(8.48)}{0.3054}\\approx 7\\text{ hours.}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1165137803583\">\n<h2>Half-Life<\/h2>\n<section id=\"fs-id1165137828382\">\n<p id=\"fs-id1165137828387\">In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/p>\n<p id=\"fs-id1165134192326\">One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its <strong>half-life<\/strong>. The table below\u00a0lists the half-life for several of the more common radioactive substances.<\/p>\n<table id=\"Table_04_06_001\" summary=\"Seven rows and three columns. The first column is labeled,\">\n<thead>\n<tr>\n<th style=\"text-align: center\">Substance<\/th>\n<th style=\"text-align: center\">Use<\/th>\n<th style=\"text-align: center\">Half-life<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>gallium-67<\/td>\n<td>nuclear medicine<\/td>\n<td>80 hours<\/td>\n<\/tr>\n<tr>\n<td>cobalt-60<\/td>\n<td>manufacturing<\/td>\n<td>5.3 years<\/td>\n<\/tr>\n<tr>\n<td>technetium-99m<\/td>\n<td>nuclear medicine<\/td>\n<td>6 hours<\/td>\n<\/tr>\n<tr>\n<td>americium-241<\/td>\n<td>construction<\/td>\n<td>432 years<\/td>\n<\/tr>\n<tr>\n<td>carbon-14<\/td>\n<td>archeological dating<\/td>\n<td>5,715 years<\/td>\n<\/tr>\n<tr>\n<td>uranium-235<\/td>\n<td>atomic power<\/td>\n<td>703,800,000 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p 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<p id=\"eip-346\">To find the half-life of a function describing exponential decay, solve the following equation:<\/p>\n<div id=\"eip-942\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}[\/latex]<\/div>\n<p id=\"fs-id1165135177747\">We find that the half-life depends only on the constant <em>k<\/em>\u00a0and not on the starting quantity [latex]{A}_{0}[\/latex].<\/p>\n<p id=\"fs-id1165137454141\">The formula is derived as follows<\/p>\n<div id=\"eip-167\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}\\frac{1}{2}{A}_{0}&={A}_{o}{e}^{kt} \\\\ \\frac{1}{2}&={e}^{kt}&& \\text{Divide by }{A}_{0}. \\\\ \\mathrm{ln}\\left(\\frac{1}{2}\\right)&=kt&& \\text{Take the natural log}. \\\\ -\\mathrm{ln}\\left(2\\right)&=kt&& \\text{Apply laws of logarithms}. \\\\ -\\frac{\\mathrm{ln}\\left(2\\right)}{t}&=k&& \\text{Divide by }t. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137423324\">Since <em>t<\/em>, the time, is positive, <em>k<\/em>\u00a0must, as expected, be negative. This gives us the half-life formula<\/p>\n<div id=\"fs-id1165137697986\" class=\"equation\" style=\"text-align: center\">[latex]k=-\\frac{\\mathrm{ln}\\left(2\\right)}{t}[\/latex]<\/div>\n<div id=\"fs-id1165137418076\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137528941\">How To: Given the half-life, find the decay rate.<\/h3>\n<ol id=\"fs-id1165135186559\">\n<li>Write [latex]A={A}_{o}{e}^{kt}[\/latex].<\/li>\n<li>Replace <em>A<\/em>\u00a0by [latex]\\frac{1}{2}{A}_{0}[\/latex] and replace <em>t<\/em>\u00a0by the given half-life.<\/li>\n<li>Solve to find <em>k<\/em>. Express <em>k<\/em>\u00a0as an exact value (do not round).<\/li>\n<\/ol>\n<p id=\"fs-id1165135613329\">Note: <em>It is also possible to find the decay rate using<\/em> [latex]k=-\\frac{\\mathrm{ln}\\left(2\\right)}{t}[\/latex].<\/p>\n<\/div>\n<div id=\"Example_04_07_02\" class=\"example\">\n<div id=\"fs-id1165137827442\" class=\"exercise\">\n<div id=\"fs-id1165137732189\" class=\"problem textbox shaded\">\n<h3>Example 5: Finding the Function that Describes Radioactive Decay<\/h3>\n<p id=\"fs-id1165137501905\">The half-life of Sodium-24, an isotope, is 15 years. Write the decay function that estimates the amount of Sodium-24 after t years given that there is initially 10 grams.\u00a0 How much is left after 20 years, rounded to the nearest gram?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q365040\">Show Solution<\/span><\/p>\n<div id=\"q365040\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137805508\">This formula is derived as follows.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}A(t)&={A}_{0}{e}^{kt}&& \\text{The continuous growth\/decay formula}.\\\\ k&=\\frac{-\\mathrm{ln}(2)}{15}&& \\text{Substitute the half-life for t and initial amount for }A_{0} \\\\ A(t)&=10{e}^{\\frac{-\\mathrm{ln}(2)t}{15}}&&\\text{This is the decay function} \\\\ A(20)&=10{e}^{\\frac{-\\mathrm{ln}(2)\\cdot 20}{15}}&&\\text{Now plug in 20 for t} \\\\ A(20)&\\approx 4\\text{ grams.} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135527004\">The half-life of Bismuth-212, an isotope, is 60.5 seconds. Write the decay function that estimates the amount of Bismuth-212 after t years given that there is initially 5 grams.\u00a0 How much is left after 80 seconds, rounded to the nearest gram?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q258944\">Show Solution<\/span><\/p>\n<div id=\"q258944\" class=\"hidden-answer\" style=\"display: none\">\n<p>The decay function is: [latex]\\text{ }A(t)=5{e}^{\\frac{-\\mathrm{ln}(2)t}{60.5}}[\/latex]<\/p>\n<p>After 80 seconds, approximately 2 grams remain.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137828123\">\n<h2>Radiocarbon Dating<\/h2>\n<p id=\"fs-id1165135154026\">The formula for radioactive decay is important in <strong>radiocarbon dating<\/strong>, which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.<\/p>\n<p id=\"fs-id1165137731568\">Carbon-14 is a radioactive isotope of carbon that has a half-life of 5,730 years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12, which has an atomic weight of 12 and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last 60,000 years, using tree rings and other organic samples of known dates\u2014although the ratio has changed slightly over the centuries.<\/p>\n<p id=\"fs-id1165137409468\">As long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.<\/p>\n<p id=\"fs-id1165135193799\">Since the half-life of carbon-14 is 5,730 years, the formula for the amount of carbon-14 remaining after <em>t<\/em>\u00a0years is<\/p>\n<div id=\"eip-866\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]A\\approx {A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]<\/div>\n<p id=\"eip-692\">where<\/p>\n<div id=\"eip-id1165137849256\">\n<ul>\n<li><em>A<\/em>\u00a0is the amount of carbon-14 remaining<\/li>\n<li>[latex]{A}_{0}[\/latex] is the amount of carbon-14 when the plant or animal began decaying.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137634094\">This formula is derived as follows:<\/p>\n<div id=\"eip-777\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}A&={A}_{0}{e}^{kt}&& \\text{The continuous growth formula}.\\\\ 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 }f\\left(t\\right). \\\\ 0.5&={e}^{5730k}&& \\text{Divide by }{A}_{0}. \\\\ \\mathrm{ln}\\left(0.5\\right)&=5730k&& \\text{Take the natural log of both sides}. \\\\ k&=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}&& \\text{Divide by the coefficient of }k. \\\\ A&={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}&& \\text{Substitute for }r\\text{ in the continuous growth formula}. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137600416\">To find the age of an object, we solve this equation for <em>t<\/em>:<\/p>\n<div id=\"eip-934\" class=\"equation\" style=\"text-align: center\">[latex]t=\\frac{\\mathrm{ln}\\left(\\frac{A}{{A}_{0}}\\right)}{-0.000121}[\/latex]<\/div>\n<p id=\"fs-id1165137841700\">Out of necessity, we neglect here the many details that a scientist takes into consideration when doing carbon-14 dating, and we only look at the basic formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let <em>r<\/em>\u00a0be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated, determined by a method called liquid scintillation. From the equation [latex]A\\approx {A}_{0}{e}^{-0.000121t}[\/latex] we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is [latex]r=\\frac{A}{{A}_{0}}\\approx {e}^{-0.000121t}[\/latex]. We solve this equation for <em>t<\/em>, to get<\/p>\n<div id=\"eip-453\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex]<\/div>\n<div id=\"fs-id1165137807119\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165134086082\">How To: Given the percentage of carbon-14 in an object, determine its age.<\/h3>\n<ol id=\"fs-id1165135209232\">\n<li>Express the given percentage of carbon-14 as an equivalent decimal, <em>k<\/em>.<\/li>\n<li>Substitute for <em>k<\/em> in the equation [latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex] and solve for the age, <em>t<\/em>.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_07_03\" class=\"example\">\n<div id=\"fs-id1165137885960\" class=\"exercise\">\n<div id=\"fs-id1165137750064\" class=\"problem textbox shaded\">\n<h3>Example 6: Finding the Age of a Bone<\/h3>\n<p id=\"fs-id1165137771807\">A bone fragment is found that contains 20% of its original carbon-14. To the nearest year, how old is the bone?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q433778\">Show Solution<\/span><\/p>\n<div id=\"q433778\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135445864\">We substitute 20% = 0.20 for <em>k<\/em>\u00a0in the equation and solve for <em>t<\/em>:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}t&=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}&& \\text{Use the general form of the equation}. \\\\ &=\\frac{\\mathrm{ln}\\left(0.20\\right)}{-0.000121}&& \\text{Substitute for }r. \\\\ &\\approx 13301&& \\text{Round to the nearest year}. \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135512179\">The bone fragment is about 13,301 years old.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135205625\">The instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about 1%, so this age should be given as [latex]\\text{13,301 years}\\pm \\text{1% or 13,301 years}\\pm \\text{133 years}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137892459\">Cesium-137 has a half-life of about 30 years. If we begin with 200 mg of cesium-137, will it take more or less than 230 years until only 1 milligram remains?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q131296\">Show Solution<\/span><\/p>\n<div id=\"q131296\" class=\"hidden-answer\" style=\"display: none\">less than 230 years, 229.3157 to be exact<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm15893\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15893&theme=oea&iframe_resize_id=ohm15893\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Calculating Doubling Time<\/span><\/p>\n<\/section>\n<section id=\"fs-id1165134040514\">\n<p id=\"fs-id1165137897897\">For decaying quantities, we determined how long it took for half of a substance to decay. For growing quantities, we might want to find out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the <strong>doubling time<\/strong>.<\/p>\n<p id=\"fs-id1165137447183\">Given the basic <strong>exponential growth<\/strong> equation [latex]A={A}_{0}{e}^{kt}[\/latex], doubling time can be found by solving for when the original quantity has doubled, that is, by solving [latex]2{A}_{0}={A}_{0}{e}^{kt}[\/latex].<\/p>\n<p id=\"eip-853\">The formula is derived as follows:<\/p>\n<div id=\"eip-244\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}2{A}_{0}&={A}_{0}{e}^{kt} \\\\ 2&={e}^{kt}&& \\text{Divide by }{A}_{0}. \\\\ \\mathrm{ln}2&=kt&& \\text{Take the natural logarithm}. \\\\ t&=\\frac{\\mathrm{ln}2}{k}&& \\text{Divide by the coefficient of }t. \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137722416\">Thus the doubling time is<\/p>\n<div id=\"eip-495\" class=\"equation\" style=\"text-align: center\">[latex]t=\\frac{\\mathrm{ln}2}{k}[\/latex]<\/div>\n<div id=\"Example_04_07_04\" class=\"example\">\n<div id=\"fs-id1165137657275\" class=\"exercise\">\n<div id=\"fs-id1165137464856\" class=\"problem textbox shaded\">\n<h3>Example 7: 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q854608\">Show Solution<\/span><\/p>\n<div id=\"q854608\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165134433339\">The formula is derived as follows:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}t&=\\frac{\\mathrm{ln}2}{k}&& \\text{The doubling time formula}.\\\\ 2&=\\frac{\\mathrm{ln}2}{k}&& \\text{Use a doubling time of two years}. \\\\ k&=\\frac{\\mathrm{ln}2}{2}&& \\text{Multiply by }k\\text{ and divide by 2}. \\\\ A&={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}&& \\text{Substitute }k\\text{ into the continuous growth formula}. \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135365796\">The function is [latex]A={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{2}t}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q447089\">Show Solution<\/span><\/p>\n<div id=\"q447089\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{3}t}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/section>\n<\/div>\n<\/div>\n<h2>Use logistic-growth models<\/h2>\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>exponential growth<\/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 <em>a<\/em>, <em>b<\/em>, and <em>c<\/em>, the logistic growth of a population over time <em>x<\/em>\u00a0is represented by the model<\/p>\n<div id=\"eip-391\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]f\\left(x\\right)=\\dfrac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\n<p id=\"fs-id1165135439859\">Figure 7\u00a0shows 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 style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010833\/CNX_Precalc_Figure_04_07_0062.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\"><b>Figure 7<\/b><\/p>\n<\/div>\n<div id=\"fs-id1165135207602\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Logistic Growth<\/h3>\n<p id=\"fs-id1165135511485\">The logistic growth model is<\/p>\n<div id=\"eip-id1165134239632\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]f\\left(x\\right)=\\dfrac{c}{1+a{e}^{-bx}}[\/latex]<\/div>\n<p id=\"eip-id1165135264518\">where<\/p>\n<div id=\"fs-id1165137526317\">\n<ul>\n<li>[latex]\\frac{c}{1+a}[\/latex] is the initial value<\/li>\n<li><em>c<\/em>\u00a0is the <em>carrying capacity<\/em>, or <em>limiting value<\/em><\/li>\n<li><em>b<\/em>\u00a0is a constant determined by the rate of growth.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"Example_04_07_06\" class=\"example\">\n<div id=\"fs-id1165135572080\" class=\"exercise\">\n<div id=\"fs-id1165135572082\" class=\"problem textbox shaded\">\n<h3>Example 8: Using the Logistic-Growth Model<\/h3>\n<p id=\"fs-id1165135572087\">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 <em>t\u00a0<\/em>= 0 there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is <em>b\u00a0<\/em>= 0.6030. Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q788840\">Show Solution<\/span><\/p>\n<div id=\"q788840\" class=\"hidden-answer\" style=\"display: none\">\n<p>We substitute the given data into the logistic growth model<\/p>\n<p style=\"text-align: center\">[latex]f\\left(x\\right)=\\frac{c}{1+a{e}^{-bx}}[\/latex]<\/p>\n<p 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 <em>c\u00a0<\/em>= 1000. To find <em>a<\/em>, we use the formula that the number of cases at time <em>t\u00a0<\/em>= 0 is [latex]\\frac{c}{1+a}=1[\/latex], from which it follows that <em>a\u00a0<\/em>= 999. This model predicts that, after ten days, the number of people who have had the flu is [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}\\approx 293.8[\/latex]. Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, <em>c\u00a0<\/em>= 1000.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135445936\" style=\"text-align: left\">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\" style=\"text-align: left\">Figure 8\u00a0gives a good picture of how this model fits the data.<\/p>\n<figure class=\"medium\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010834\/CNX_Precalc_Figure_04_07_0072.jpg\" alt=\"Graph of f(x)=1000\/(1+999e^(-0.5030x)) with the y-axis labeled as\" \/><\/figure>\n<figure id=\"CNX_Precalc_Figure_04_07_007\" class=\"medium\"><figcaption><strong>Figure 8.\u00a0<\/strong>The graph of [latex]f\\left(x\\right)=\\frac{1000}{1+999{e}^{-0.6030x}}[\/latex]<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135689538\">Using the model in Example 5, estimate the number of cases of flu on day 15.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q325943\">Show Solution<\/span><\/p>\n<div id=\"q325943\" class=\"hidden-answer\" style=\"display: none\">\n<p>895 cases on day 15<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm54629\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=54629&theme=oea&iframe_resize_id=ohm54629\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<section id=\"fs-id1165135511570\"><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Expressing an Exponential Model in Base <\/span><em style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">e<\/em><\/section>\n<section id=\"fs-id1165137749158\">\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 <em>e<\/em>\u00a0is often preferred. We can use laws of exponents and laws of logarithms to change any base to base <em>e<\/em>.<\/p>\n<div id=\"fs-id1165134342616\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137452983\">How To: Given a model with the form [latex]y=a{b}^{x}[\/latex], change it to the form [latex]y={A}_{0}{e}^{kx}[\/latex].<\/h3>\n<ol id=\"fs-id1165134148496\">\n<li>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 y 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=\"example\">\n<div id=\"fs-id1165134261257\" class=\"exercise\">\n<div id=\"fs-id1165132957903\" class=\"problem textbox shaded\">\n<h3>Example 9: Changing to base <em>e<\/em><\/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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q865754\">Show Solution<\/span><\/p>\n<div id=\"q865754\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137432379\">The formula is derived as follows<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}y&=2.5{\\left(3.1\\right)}^{x} \\\\ &=2.5{e}^{\\mathrm{ln}\\left({3.1}^{x}\\right)}&& \\text{Insert exponential and its inverse.} \\\\ &=2.5{e}^{x\\mathrm{ln}3.1}&& \\text{Laws of logs.} \\\\ &=2.5{e}^{\\left(\\mathrm{ln}3.1\\right)}{}^{x}&& \\text{Commutative law of multiplication} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135664963\">Change the function [latex]y=3{\\left(0.5\\right)}^{x}[\/latex] to one having <i>e<\/i>\u00a0as the base.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q843558\">Show Solution<\/span><\/p>\n<div id=\"q843558\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=3{e}^{\\left(\\mathrm{ln}0.5\\right)x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm15878\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15878&theme=oea&iframe_resize_id=ohm15878\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p><span style=\"color: #077fab;font-size: 1.15em;font-weight: 600\">Key Equations<\/span><\/p>\n<section id=\"fs-id1165135208780\" class=\"key-equations\">\n<table id=\"fs-id2347772\" summary=\"..\">\n<tbody>\n<tr>\n<td>Half-life formula<\/td>\n<td>If [latex]\\text{ }A={A}_{0}{e}^{kt}[\/latex], <em>k\u00a0<\/em>&lt; 0, the half-life is [latex]t=-\\frac{\\mathrm{ln}\\left(2\\right)}{k}[\/latex].<\/td>\n<\/tr>\n<tr>\n<td>Carbon-14 dating<\/td>\n<td>[latex]t=\\frac{\\mathrm{ln}\\left(\\frac{A}{{A}_{0}}\\right)}{-0.000121}[\/latex].[latex]{A}_{0}[\/latex] <em>A<\/em>\u00a0is the amount of carbon-14 when the plant or animal died<\/p>\n<p><em>t<\/em>\u00a0is the amount of carbon-14 remaining today<\/p>\n<p>is the age of the fossil in years<\/td>\n<\/tr>\n<tr>\n<td>Doubling time formula<\/td>\n<td>If [latex]A={A}_{0}{e}^{kt}[\/latex], <em>k\u00a0<\/em>&gt; 0, the doubling time is [latex]t=\\frac{\\mathrm{ln}2}{k}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Newton\u2019s Law of Cooling<\/td>\n<td>[latex]T\\left(t\\right)=A{e}^{kt}+{T}_{s}[\/latex], where [latex]{T}_{s}[\/latex] is the ambient temperature, [latex]A=T\\left(0\\right)-{T}_{s}[\/latex], and <em>k<\/em> is the continuous rate of cooling.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section id=\"fs-id1165137894245\" class=\"key-concepts\">\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165137894248\">\n<li>The basic exponential function is [latex]f\\left(x\\right)=a{b}^{x}[\/latex]. If <em>b\u00a0<\/em>&gt; 1, we have exponential growth; if 0 &lt; <em>b\u00a0<\/em>&lt; 1, we have exponential decay.<\/li>\n<li>We can also write this formula in terms of continuous growth as [latex]A={A}_{0}{e}^{kx}[\/latex], where [latex]{A}_{0}[\/latex] is the starting value. If [latex]{A}_{0}[\/latex] is positive, then we have exponential growth when <em>k\u00a0<\/em>&gt; 0 and exponential decay when <em>k\u00a0<\/em>&lt; 0.<\/li>\n<li>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>We can find the age, <em>t<\/em>, of an organic artifact by measuring the amount, <em>k<\/em>, of carbon-14 remaining in the artifact and using the formula [latex]t=\\frac{\\mathrm{ln}\\left(k\\right)}{-0.000121}[\/latex] to solve for <em>t<\/em>.<\/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 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>We can use logistic growth functions to model real-world situations where the rate of growth changes over time, such as population growth, spread of disease, and spread of rumors.<\/li>\n<li>We can use real-world data gathered over time to observe trends. Knowledge of linear, exponential, logarithmic, and logistic graphs help us to develop models that best fit our data.<\/li>\n<li>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}b[\/latex].<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165134069388\" class=\"definition\">\n<dt><strong>carrying capacity<\/strong><\/dt>\n<dd id=\"fs-id1165134069394\">in a logistic model, the limiting value of the output<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134069398\" class=\"definition\">\n<dt><strong>doubling time<\/strong><\/dt>\n<dd id=\"fs-id1165135528921\">the time it takes for a quantity to double<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135528925\" class=\"definition\">\n<dt><strong>half-life<\/strong><\/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\" class=\"definition\">\n<dt><strong>logistic growth model<\/strong><\/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, <em>c<\/em>\u00a0is the carrying capacity, or limiting value, and <em>b<\/em>\u00a0is a constant determined by the rate of growth<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137844266\" class=\"definition\">\n<dt><strong>Newton\u2019s Law of Cooling<\/strong><\/dt>\n<dd id=\"fs-id1165137844271\">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\" class=\"definition\">\n<dt><strong>order of magnitude<\/strong><\/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<\/section>\n<\/div>\n<\/div>\n<\/section>\n<p>&nbsp;<\/p>\n<h2 style=\"text-align: center\">Section 3.5 Homework Exercises<\/h2>\n<p>1. With what kind of exponential model would half-life be associated? What role does half-life play in these models?<\/p>\n<p>2. What is carbon dating? Why does it work? Give an example in which carbon dating would be useful.<\/p>\n<p>3. With what kind of exponential model would doubling time be associated? What role does doubling time play in these models?<\/p>\n<p>4. Why use a logistic model instead of an exponential model?<\/p>\n<p>5. The population of a colony of flies fits an exponential growth model. If there are 1000 flies initially and there are 1800 after 1 day, answer the following:<br \/>\na) What is the exponential growth model where t represents the number of days?<br \/>\nb) What is the size of the colony after 4 days? Round to the nearest whole number.<br \/>\nc) How long is it until there are 30,000 flies? Round to the nearest tenth.<\/p>\n<p>6. The population of a colony of gnats fits an exponential growth model. If there are 500 gnats initially and there are 800 after 1 day, answer the following:<br \/>\na) What is the exponential growth model where t represents the number of days?<br \/>\nb) What is the size of the colony after 4 days? Round to the nearest whole number.<br \/>\nc) How long is it until there are 90,000 gnats? Round to the nearest tenth.<\/p>\n<p>For the following exercises, use the logistic growth model [latex]f\\left(x\\right)=\\frac{150}{1+8{e}^{-2x}}[\/latex].<\/p>\n<p>7. Find and interpret [latex]f\\left(0\\right)[\/latex]. Round to the nearest tenth.<\/p>\n<p>8.\u00a0Find and interpret [latex]f\\left(4\\right)[\/latex]. Round to the nearest tenth.<\/p>\n<p>9. Find the carrying capacity.<\/p>\n<p>10.\u00a0Graph the model.<\/p>\n<p>11. Determine whether the data from the table could best be represented as a function that is linear, exponential, or logarithmic. Then write a formula for a model that represents the data.<\/p>\n<p>x f\u00a0(x)<br \/>\n\u20132 0.694<br \/>\n\u20131 0.833<br \/>\n0 1<br \/>\n1 1.2<br \/>\n2 1.44<br \/>\n3 1.728<br \/>\n4 2.074<br \/>\n5 2.488<\/p>\n<p>12. Rewrite [latex]f\\left(x\\right)=1.68{\\left(0.65\\right)}^{x}[\/latex] as an exponential equation with base e\u00a0to five significant digits.<\/p>\n<p>For the following exercises, use a graphing calculator and this scenario: the population of a fish farm in t\u00a0years is modeled by the equation [latex]P\\left(t\\right)=\\frac{1000}{1+9{e}^{-0.6t}}[\/latex].<\/p>\n<p>13. Graph the function.<\/p>\n<p>14.\u00a0What is the initial population of fish?<\/p>\n<p>15. To the nearest tenth, what is the doubling time for the fish population?<\/p>\n<p>16.\u00a0To the nearest whole number, what will the fish population be after 2 years?<\/p>\n<p>17. To the nearest tenth, how long will it take for the population to reach 900?<\/p>\n<p>18.\u00a0What is the carrying capacity for the fish population? Justify your answer using the graph of P.<\/p>\n<p>19. A substance has a half-life of 2.045 minutes. If the initial amount of the substance was 132.8 grams, how many half-lives will have passed before the substance decays to 8.3 grams? What is the total time of decay?<\/p>\n<p>20.\u00a0The formula for an increasing population is given by [latex]P\\left(t\\right)={P}_{0}{e}^{rt}[\/latex] where [latex]{P}_{0}[\/latex] is the initial population and r\u00a0&gt; 0. Derive a general formula for the time t it takes for the population to increase by a factor of M.<\/p>\n<p>21. Recall the formula for calculating the magnitude of an earthquake, [latex]M=\\frac{2}{3}\\mathrm{log}\\left(\\frac{S}{{S}_{0}}\\right)[\/latex]. Show each step for solving this equation algebraically for the seismic moment S.<\/p>\n<p>22. What is the y-intercept of the logistic growth model [latex]y=\\frac{c}{1+a{e}^{-rx}}[\/latex]? Show the steps for calculation. What does this point tell us about the population?<\/p>\n<p>23. Prove that [latex]{b}^{x}={e}^{x\\mathrm{ln}\\left(b\\right)}[\/latex] for positive [latex]b\\ne 1[\/latex].<\/p>\n<p>For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour.<\/p>\n<p>24. To the nearest hour, what is the half-life of the drug?<\/p>\n<p>25. Write an exponential model representing the amount of the drug remaining in the patient\u2019s system after t\u00a0hours. Then use the formula to find the amount of the drug that would remain in the patient\u2019s system after 3 hours. Round to the nearest milligram.<\/p>\n<p>26.\u00a0Using the model found in the previous exercise, find [latex]f\\left(10\\right)[\/latex] and interpret the result. Round to the nearest hundredth.<\/p>\n<p>For the following exercises, use this scenario: A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day.<\/p>\n<p>27. To the nearest day, how long will it take for half of the Iodine-125 to decay?<\/p>\n<p>28.\u00a0Write an exponential model representing the amount of Iodine-125 remaining in the tumor after t\u00a0days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 60 days. Round to the nearest tenth of a gram.<\/p>\n<p>29. A scientist begins with 250 grams of a radioactive substance. After 250 minutes, the sample has decayed to 32 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance?<\/p>\n<p>30.\u00a0The half-life of Radium-226 is 1590 years. What is the annual decay rate? Express the decimal result to four significant digits and the percentage to two significant digits.<\/p>\n<p>31. The half-life of Erbium-165 is 10.4 hours. What is the hourly decay rate? Express the decimal result to four significant digits and the percentage to two significant digits.<\/p>\n<p>32.\u00a0A wooden artifact from an archeological dig contains 60 percent of the carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of carbon-14 is 5730 years.)<\/p>\n<p>33. A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?<\/p>\n<p>For the following exercises, use this scenario: A biologist recorded a count of 360 bacteria present in a culture after 5 minutes and 1000 bacteria present after 20 minutes.<\/p>\n<p>34. To the nearest whole number, what was the initial population in the culture?<\/p>\n<p>35. Rounding to six significant digits, write an exponential equation representing this situation. To the nearest minute, how long did it take the population to double?<\/p>\n<p>36. Plot each set of approximate values of intensity of sounds on a logarithmic scale: Whisper: [latex]{10}^{-10} \\frac{W}{{m}^{2}}[\/latex], Vacuum: [latex]{10}^{-4}\\frac{W}{{m}^{2}}[\/latex], Jet: [latex]{10}^{2} \\frac{W}{{m}^{2}}[\/latex]<\/p>\n<p>37. Recall the formula for calculating the magnitude of an earthquake, [latex]M=\\frac{2}{3}\\mathrm{log}\\left(\\frac{S}{{S}_{0}}\\right)[\/latex]. One earthquake has magnitude 3.9 on the MMS scale. If a second earthquake has 750 times as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth.<\/p>\n<p>For the following exercises, use this scenario: The equation [latex]N\\left(t\\right)=\\frac{500}{1+49{e}^{-0.7t}}[\/latex] models the number of people in a town who have heard a rumor after t days.<\/p>\n<p>38. How many people started the rumor?<\/p>\n<p>39. To the nearest whole number, how many people will have heard the rumor after 3 days?<\/p>\n<p>40.\u00a0As t\u00a0increases without bound, what value does N(t) approach? Interpret your answer.<\/p>\n<p>For the following exercise, choose the correct answer choice.<\/p>\n<p>41. A doctor and injects a patient with 13 milligrams of radioactive dye that decays exponentially. After 12 minutes, there are 4.75 milligrams of dye remaining in the patient\u2019s system. Which is an appropriate model for this situation?<\/p>\n<p>A. [latex]f\\left(t\\right)=13{\\left(0.0805\\right)}^{t}[\/latex]<br \/>\nB. [latex]f\\left(t\\right)=13{e}^{0.9195t}[\/latex]<br \/>\nC. [latex]f\\left(t\\right)=13{e}^{\\left(-0.0839t\\right)}[\/latex]<br \/>\nD. [latex]f\\left(t\\right)=\\frac{4.75}{1+13{e}^{-0.83925t}}[\/latex]<\/p>\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-13724\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":97803,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"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-13724","chapter","type-chapter","status-publish","hentry"],"part":13696,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13724","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/users\/97803"}],"version-history":[{"count":50,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13724\/revisions"}],"predecessor-version":[{"id":17586,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13724\/revisions\/17586"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/parts\/13696"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13724\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/media?parent=13724"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=13724"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/contributor?post=13724"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/license?post=13724"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}