{"id":1446,"date":"2025-07-25T01:29:25","date_gmt":"2025-07-25T01:29:25","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1446"},"modified":"2026-03-24T07:32:57","modified_gmt":"2026-03-24T07:32:57","slug":"exponential-and-logarithmic-models-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/exponential-and-logarithmic-models-fresh-take\/","title":{"raw":"Exponential and Logarithmic Models: Fresh Take","rendered":"Exponential and Logarithmic Models: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Model exponential growth and decay.<\/li>\r\n \t<li>Use Newton\u2019s Law of Cooling.<\/li>\r\n \t<li>Use logistic-growth models.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Exponential Growth and Decay<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\r\n<\/strong>\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\">Exponential Growth: [latex]A = A_0e^{rt}[\/latex]\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]A_0[\/latex]: Initial amount<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]r[\/latex]: Growth rate (positive)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]t[\/latex]: Time<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Exponential Decay: [latex]A = A_0e^{-kt}[\/latex]\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]k[\/latex]: Decay rate (positive)<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Doubling Time: [latex]t = \\frac{\\ln(2)}{r}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Half-life: [latex]t = \\frac{\\ln(2)}{k} = \\frac{\\ln(1\/2)}{-k}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"font-600 text-xl font-bold\"><strong>Characteristics of Exponential Functions<\/strong><\/p>\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">Domain: All real numbers<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Range: [latex](0, \u221e)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">y-intercept: [latex](0, A_0)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Horizontal asymptote: [latex]y = 0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Increasing if [latex]r &gt; 0[\/latex], decreasing if [latex]r &lt; 0[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"font-600 text-xl font-bold\"><strong>Applications<\/strong><\/p>\r\n\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">Population Growth<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Compound Interest<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Moore's Law (computing power)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Radioactive Decay<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Radiocarbon Dating<\/li>\r\n<\/ol>\r\n<p class=\"font-600 text-xl font-bold\"><strong>Problem-Solving Approach<\/strong><\/p>\r\n\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">Identify whether it's growth or decay<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Determine the initial amount ([latex]A_0[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Calculate or identify the rate ([latex]r[\/latex] or [latex]k[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Apply the appropriate formula<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Solve for the unknown variable (often time [latex]t[\/latex] or final amount [latex]A[\/latex])<\/li>\r\n<\/ol>\r\n<p class=\"font-600 text-xl font-bold\"><strong>Key Formulas for Radioactive Decay<\/strong><\/p>\r\n\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">General form: [latex]A(t) = A_0(\\frac{1}{2})^{t\/T}[\/latex] where [latex]T[\/latex] is the half-life<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Radiocarbon dating: [latex]t = \\frac{\\ln(A\/A_0)}{-0.000121}[\/latex] where [latex]A\/A_0[\/latex] is the ratio of current to initial carbon-14<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">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.[reveal-answer q=\"271465\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"271465\"][latex]f\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{3}t}[\/latex][\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">Uranium-235 has a half-life of [latex]703,800,000[\/latex] years. How long will it take before twenty percent of our [latex]1000[\/latex]-gram sample of uranium-235 has decayed?[reveal-answer q=\"923702\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"923702\"][latex]t=703,800,000\\times \\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}[\/latex].[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">The half-life of plutonium-244 is [latex]80,000,000[\/latex] years. Find a function that gives the amount of plutonium-244 remaining as a function of time measured in years.[reveal-answer q=\"267261\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"267261\"][latex]f\\left(t\\right)={A}_{0}{e}^{-0.0000000087t}[\/latex][\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">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?[reveal-answer q=\"758746\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"758746\"]less than 230 years; 229.3157 to be exact[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\"><script src=\"https:\/\/www.youtube.com\/iframe_api \" type=\"text\/javascript\"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-daedchab-YK7rERyFlOM\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/YK7rERyFlOM?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-daedchab-YK7rERyFlOM\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&amp;cc_minimizable=1&amp;cc_minimize_on_load=0&amp;cc_multi_text_track=0&amp;cc_overlay=1&amp;cc_searchable=0&amp;embed=ajax&amp;mf=12850422&amp;p3sdk_version=1.11.7&amp;p=20361&amp;player_type=youtube&amp;plugin_skin=dark&amp;target=3p-plugin-target-daedchab-YK7rERyFlOM&amp;vembed=0&amp;video_id=YK7rERyFlOM&amp;video_target=tpm-plugin-daedchab-YK7rERyFlOM\" type=\"text\/javascript\"><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/College+Algebra+Corequisite\/Transcripts\/Exponential+Growth+App+(y%3Dab%5Et)+-+Find+Initial+Amount+Given+Doubling+Time_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExponential Growth App (y=ab^t) - Find Initial Amount Given Doubling Time\u201d here (opens in new window).<\/a>\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\"><script src=\"https:\/\/www.youtube.com\/iframe_api \" type=\"text\/javascript\"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-hdhcgggc-eSOhLhSz9pk\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/eSOhLhSz9pk?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-hdhcgggc-eSOhLhSz9pk\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&amp;cc_minimizable=1&amp;cc_minimize_on_load=0&amp;cc_multi_text_track=0&amp;cc_overlay=1&amp;cc_searchable=0&amp;embed=ajax&amp;mf=12850423&amp;p3sdk_version=1.11.7&amp;p=20361&amp;player_type=youtube&amp;plugin_skin=dark&amp;target=3p-plugin-target-hdhcgggc-eSOhLhSz9pk&amp;vembed=0&amp;video_id=eSOhLhSz9pk&amp;video_target=tpm-plugin-hdhcgggc-eSOhLhSz9pk\" type=\"text\/javascript\"><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/College+Algebra+Corequisite\/Transcripts\/Exponential+Growth+App+(y%3Dab%5Et)+-+Given+Doubling+Time_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExponential Growth App (y=ab^t) - Given Doubling Time\u201d here (opens in new window).<\/a>\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\"><script src=\"https:\/\/www.youtube.com\/iframe_api \" type=\"text\/javascript\"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-cegdbbdc-HYHzK6kF0ts\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/HYHzK6kF0ts?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-cegdbbdc-HYHzK6kF0ts\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&amp;cc_minimizable=1&amp;cc_minimize_on_load=0&amp;cc_multi_text_track=0&amp;cc_overlay=1&amp;cc_searchable=0&amp;embed=ajax&amp;mf=12850424&amp;p3sdk_version=1.11.7&amp;p=20361&amp;player_type=youtube&amp;plugin_skin=dark&amp;target=3p-plugin-target-cegdbbdc-HYHzK6kF0ts&amp;vembed=0&amp;video_id=HYHzK6kF0ts&amp;video_target=tpm-plugin-cegdbbdc-HYHzK6kF0ts\" type=\"text\/javascript\"><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/College+Algebra+Corequisite\/Transcripts\/Ex+-+Exponential+Model+-+Determine+Age+Using+Carbon-14+Given+Half+Life_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Exponential Model - Determine Age Using Carbon-14 Given Half Life\u201d here (opens in new window).<\/a>\r\n\r\n<\/section>\r\n<h2>Newton\u2019s Law of Cooling<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\r\n<\/strong>\r\n<p class=\"whitespace-pre-wrap break-words\">Newton's Law of Cooling describes how an object's temperature changes exponentially as it approaches the ambient temperature:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">[latex]T(t) = A e^{kt} + T_s[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Where:<\/p>\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]T(t)[\/latex] is the temperature at time [latex]t[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]A[\/latex] is the initial temperature difference (object - surroundings)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]k[\/latex] is the cooling rate (negative for cooling)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]T_s[\/latex] is the surrounding temperature<\/li>\r\n<\/ul>\r\n<p class=\"font-600 text-xl font-bold\"><strong>Characteristics<\/strong><\/p>\r\n\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">Exponential decay towards ambient temperature<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Vertical shift of the exponential decay function<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Horizontal asymptote at [latex]T_s[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"font-600 text-xl font-bold\"><strong>Problem-Solving Approach<\/strong><\/p>\r\n\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">Identify [latex]T_s[\/latex] (ambient temperature)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Find [latex]A[\/latex] by subtracting [latex]T_s[\/latex] from initial temperature<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Use a second data point to solve for [latex]k[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Apply the formula to find unknown time or temperature<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?[reveal-answer q=\"846066\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"846066\"]6.026 hours[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\"><script src=\"https:\/\/www.youtube.com\/iframe_api \" type=\"text\/javascript\"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-bdgbgece-F4sr2jUIHZI\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/F4sr2jUIHZI?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-bdgbgece-F4sr2jUIHZI\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&amp;cc_minimizable=1&amp;cc_minimize_on_load=0&amp;cc_multi_text_track=0&amp;cc_overlay=1&amp;cc_searchable=0&amp;embed=ajax&amp;mf=12850425&amp;p3sdk_version=1.11.7&amp;p=20361&amp;player_type=youtube&amp;plugin_skin=dark&amp;target=3p-plugin-target-bdgbgece-F4sr2jUIHZI&amp;vembed=0&amp;video_id=F4sr2jUIHZI&amp;video_target=tpm-plugin-bdgbgece-F4sr2jUIHZI\" type=\"text\/javascript\"><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/College+Algebra+Corequisite\/Transcripts\/Ex+-+Newton's+Law+of+Cooling+-+Exponential+Function+App_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Newton's Law of Cooling - Exponential Function App\u201d here (opens in new window).<\/a>\r\n\r\n<\/section>\r\n<h2>Logistic Growth<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\r\n<\/strong>\r\n<p class=\"whitespace-pre-wrap break-words\">The logistic growth model describes growth with a limiting factor:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">[latex]f(x) = \\frac{c}{1 + ae^{-bx}}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Where:<\/p>\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]c[\/latex] is the carrying capacity (upper limit)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]a[\/latex] affects the initial value (f(0) = c\/(1+a))<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]b[\/latex] is the growth rate<\/li>\r\n<\/ul>\r\n<p class=\"font-600 text-xl font-bold\"><strong>Characteristics<\/strong><\/p>\r\n\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">S-shaped curve<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Initially similar to exponential growth<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Growth rate decreases as it approaches carrying capacity<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Horizontal asymptote at [latex]y = c[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"font-600 text-xl font-bold\"><strong>Comparison to Exponential Growth<\/strong><\/p>\r\n\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">Exponential: Unlimited growth<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Logistic: Limited by carrying capacity<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Exponential: Constant relative growth rate<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Logistic: Decreasing relative growth rate<\/li>\r\n<\/ol>\r\n<p class=\"font-600 text-xl font-bold\"><strong>Applications<\/strong><\/p>\r\n\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">Population growth with limited resources<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Spread of infectious diseases<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Technology adoption<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Product sales over time<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">An influenza epidemic spreads through a population rapidly at a rate that depends on two factors. The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model good for studying the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.For example, at time [latex]t = 0[\/latex] there is one person in a community of [latex]1,000[\/latex] people who has the flu. So, in that community, at most [latex]1,000[\/latex] people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is [latex]b = 0.6030[\/latex]. Estimate the number of cases of flu on day [latex]15[\/latex].[reveal-answer q=\"421768\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"421768\"]895 cases on day 15[\/hidden-answer]<\/section>\r\n<h2>Exponential Regression<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\r\n<\/strong>\r\n<p class=\"whitespace-pre-wrap break-words\">Exponential regression is used to model data that shows exponential growth or decay. The general form of the model is:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">[latex]y = ab^x[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Where:<\/p>\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]a[\/latex] is the [latex]y[\/latex]-intercept (initial value)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]b[\/latex] is the base (growth\/decay factor)<\/li>\r\n<\/ul>\r\n<p class=\"font-600 text-xl font-bold\"><strong>Characteristics<\/strong><\/p>\r\n\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">For growth: [latex]b &gt; 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For decay: [latex]0 &lt; b &lt; 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]a[\/latex] must be positive<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Initial growth\/decay is slow, then accelerates<\/li>\r\n<\/ol>\r\n<p class=\"font-600 text-xl font-bold\"><strong>When to Use Exponential Regression<\/strong><\/p>\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">Data increases\/decreases by a constant percentage<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Growth starts slow, then accelerates rapidly<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Decay is rapid at first, then slows down approaching zero<\/li>\r\n<\/ul>\r\n<p class=\"font-600 text-xl font-bold\"><strong>Regression Process<\/strong><\/p>\r\n\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n \t<li class=\"whitespace-normal break-words\">Enter data into a graphing utility<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Create a scatter plot<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Perform exponential regression<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Analyze the fit (r\u00b2 value)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Graph the model with the data points<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">The table below shows a recent graduate\u2019s credit card balance each month after graduation.\r\n<table summary=\"Two rows and ten columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><strong>Month<\/strong><\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<td>4<\/td>\r\n<td>5<\/td>\r\n<td>6<\/td>\r\n<td>7<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>Debt ($)<\/strong><\/td>\r\n<td>620.00<\/td>\r\n<td>761.88<\/td>\r\n<td>899.80<\/td>\r\n<td>1039.93<\/td>\r\n<td>1270.63<\/td>\r\n<td>1589.04<\/td>\r\n<td>1851.31<\/td>\r\n<td>2154.92<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Use exponential regression to fit a model to these data.<\/li>\r\n \t<li>If spending continues at this rate, what will the graduate\u2019s credit card debt be one year after graduating?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"99197\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"99197\"]\r\n\r\na.\r\n\r\nUsing an online graphing tool, the exponential regression model that fits these data is [latex]y=528.25{\\left(1.1943\\right)}^{x}[\/latex].\r\n\r\nUsing a graphing calculator, the exponential regression model that fits these data is [latex]y=522.88585984{\\left(1.19645256\\right)}^{x}[\/latex].\r\n\r\nb.\r\n\r\nUsing an online graphing tool, if spending continues at this rate, the graduate\u2019s credit card debt will be $4,448.37 after one year.\r\n\r\nUsing a graphing calculator, if spending continues at this rate, the graduate\u2019s credit card debt will be $4,499.38 after one year.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\"><script src=\"https:\/\/www.youtube.com\/iframe_api \" type=\"text\/javascript\"><\/script>\r\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-cgcaahhg-QfBvEjBz_1s\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/QfBvEjBz_1s?enablejsapi=1 \" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\r\n\r\n<div id=\"3p-plugin-target-cgcaahhg-QfBvEjBz_1s\" class=\"p3sdk-target\"><\/div>\r\n<p class=\"cc-media-iframe-container\"><script src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&amp;cc_minimizable=1&amp;cc_minimize_on_load=0&amp;cc_multi_text_track=0&amp;cc_overlay=1&amp;cc_searchable=0&amp;embed=ajax&amp;mf=12850478&amp;p3sdk_version=1.11.7&amp;p=20361&amp;player_type=youtube&amp;plugin_skin=dark&amp;target=3p-plugin-target-cgcaahhg-QfBvEjBz_1s&amp;vembed=0&amp;video_id=QfBvEjBz_1s&amp;video_target=tpm-plugin-cgcaahhg-QfBvEjBz_1s\" type=\"text\/javascript\"><\/script><\/p>\r\nYou can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/College+Algebra+Corequisite\/Transcripts\/Ex+-+Perform+Exponential+Regression+on+a+Graphing+Calculator_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Perform Exponential Regression on a Graphing Calculator\u201d here (opens in new window).<\/a>\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Model exponential growth and decay.<\/li>\n<li>Use Newton\u2019s Law of Cooling.<\/li>\n<li>Use logistic-growth models.<\/li>\n<\/ul>\n<\/section>\n<h2>Exponential Growth and Decay<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<br \/>\n<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Exponential Growth: [latex]A = A_0e^{rt}[\/latex]\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]A_0[\/latex]: Initial amount<\/li>\n<li class=\"whitespace-normal break-words\">[latex]r[\/latex]: Growth rate (positive)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]t[\/latex]: Time<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Exponential Decay: [latex]A = A_0e^{-kt}[\/latex]\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]k[\/latex]: Decay rate (positive)<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Doubling Time: [latex]t = \\frac{\\ln(2)}{r}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Half-life: [latex]t = \\frac{\\ln(2)}{k} = \\frac{\\ln(1\/2)}{-k}[\/latex]<\/li>\n<\/ul>\n<p class=\"font-600 text-xl font-bold\"><strong>Characteristics of Exponential Functions<\/strong><\/p>\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Domain: All real numbers<\/li>\n<li class=\"whitespace-normal break-words\">Range: [latex](0, \u221e)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">y-intercept: [latex](0, A_0)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Horizontal asymptote: [latex]y = 0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Increasing if [latex]r > 0[\/latex], decreasing if [latex]r < 0[\/latex]<\/li>\n<\/ul>\n<p class=\"font-600 text-xl font-bold\"><strong>Applications<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Population Growth<\/li>\n<li class=\"whitespace-normal break-words\">Compound Interest<\/li>\n<li class=\"whitespace-normal break-words\">Moore&#8217;s Law (computing power)<\/li>\n<li class=\"whitespace-normal break-words\">Radioactive Decay<\/li>\n<li class=\"whitespace-normal break-words\">Radiocarbon Dating<\/li>\n<\/ol>\n<p class=\"font-600 text-xl font-bold\"><strong>Problem-Solving Approach<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Identify whether it&#8217;s growth or decay<\/li>\n<li class=\"whitespace-normal break-words\">Determine the initial amount ([latex]A_0[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">Calculate or identify the rate ([latex]r[\/latex] or [latex]k[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">Apply the appropriate formula<\/li>\n<li class=\"whitespace-normal break-words\">Solve for the unknown variable (often time [latex]t[\/latex] or final amount [latex]A[\/latex])<\/li>\n<\/ol>\n<p class=\"font-600 text-xl font-bold\"><strong>Key Formulas for Radioactive Decay<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">General form: [latex]A(t) = A_0(\\frac{1}{2})^{t\/T}[\/latex] where [latex]T[\/latex] is the half-life<\/li>\n<li class=\"whitespace-normal break-words\">Radiocarbon dating: [latex]t = \\frac{\\ln(A\/A_0)}{-0.000121}[\/latex] where [latex]A\/A_0[\/latex] is the ratio of current to initial carbon-14<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q271465\">Show Solution<\/button><\/p>\n<div id=\"q271465\" class=\"hidden-answer\" style=\"display: none\">[latex]f\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}2}{3}t}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Uranium-235 has a half-life of [latex]703,800,000[\/latex] years. How long will it take before twenty percent of our [latex]1000[\/latex]-gram sample of uranium-235 has decayed?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q923702\">Show Solution<\/button><\/p>\n<div id=\"q923702\" class=\"hidden-answer\" style=\"display: none\">[latex]t=703,800,000\\times \\frac{\\mathrm{ln}\\left(0.8\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{ years }\\approx \\text{ }226,572,993\\text{ years}[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">The half-life of plutonium-244 is [latex]80,000,000[\/latex] years. Find a function that gives the amount of plutonium-244 remaining as a function of time measured in years.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q267261\">Show Solution<\/button><\/p>\n<div id=\"q267261\" class=\"hidden-answer\" style=\"display: none\">[latex]f\\left(t\\right)={A}_{0}{e}^{-0.0000000087t}[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q758746\">Show Solution<\/button><\/p>\n<div id=\"q758746\" class=\"hidden-answer\" style=\"display: none\">less than 230 years; 229.3157 to be exact<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\"><script src=\"https:\/\/www.youtube.com\/iframe_api\" type=\"text\/javascript\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-daedchab-YK7rERyFlOM\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/YK7rERyFlOM?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-daedchab-YK7rERyFlOM\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&amp;cc_minimizable=1&amp;cc_minimize_on_load=0&amp;cc_multi_text_track=0&amp;cc_overlay=1&amp;cc_searchable=0&amp;embed=ajax&amp;mf=12850422&amp;p3sdk_version=1.11.7&amp;p=20361&amp;player_type=youtube&amp;plugin_skin=dark&amp;target=3p-plugin-target-daedchab-YK7rERyFlOM&amp;vembed=0&amp;video_id=YK7rERyFlOM&amp;video_target=tpm-plugin-daedchab-YK7rERyFlOM\" type=\"text\/javascript\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/College+Algebra+Corequisite\/Transcripts\/Exponential+Growth+App+(y%3Dab%5Et)+-+Find+Initial+Amount+Given+Doubling+Time_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExponential Growth App (y=ab^t) &#8211; Find Initial Amount Given Doubling Time\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\"><script src=\"https:\/\/www.youtube.com\/iframe_api\" type=\"text\/javascript\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-hdhcgggc-eSOhLhSz9pk\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/eSOhLhSz9pk?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-hdhcgggc-eSOhLhSz9pk\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&amp;cc_minimizable=1&amp;cc_minimize_on_load=0&amp;cc_multi_text_track=0&amp;cc_overlay=1&amp;cc_searchable=0&amp;embed=ajax&amp;mf=12850423&amp;p3sdk_version=1.11.7&amp;p=20361&amp;player_type=youtube&amp;plugin_skin=dark&amp;target=3p-plugin-target-hdhcgggc-eSOhLhSz9pk&amp;vembed=0&amp;video_id=eSOhLhSz9pk&amp;video_target=tpm-plugin-hdhcgggc-eSOhLhSz9pk\" type=\"text\/javascript\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/College+Algebra+Corequisite\/Transcripts\/Exponential+Growth+App+(y%3Dab%5Et)+-+Given+Doubling+Time_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExponential Growth App (y=ab^t) &#8211; Given Doubling Time\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\"><script src=\"https:\/\/www.youtube.com\/iframe_api\" type=\"text\/javascript\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-cegdbbdc-HYHzK6kF0ts\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/HYHzK6kF0ts?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-cegdbbdc-HYHzK6kF0ts\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&amp;cc_minimizable=1&amp;cc_minimize_on_load=0&amp;cc_multi_text_track=0&amp;cc_overlay=1&amp;cc_searchable=0&amp;embed=ajax&amp;mf=12850424&amp;p3sdk_version=1.11.7&amp;p=20361&amp;player_type=youtube&amp;plugin_skin=dark&amp;target=3p-plugin-target-cegdbbdc-HYHzK6kF0ts&amp;vembed=0&amp;video_id=HYHzK6kF0ts&amp;video_target=tpm-plugin-cegdbbdc-HYHzK6kF0ts\" type=\"text\/javascript\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/College+Algebra+Corequisite\/Transcripts\/Ex+-+Exponential+Model+-+Determine+Age+Using+Carbon-14+Given+Half+Life_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Exponential Model &#8211; Determine Age Using Carbon-14 Given Half Life\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<h2>Newton\u2019s Law of Cooling<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<br \/>\n<\/strong><\/p>\n<p class=\"whitespace-pre-wrap break-words\">Newton&#8217;s Law of Cooling describes how an object&#8217;s temperature changes exponentially as it approaches the ambient temperature:<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[latex]T(t) = A e^{kt} + T_s[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Where:<\/p>\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]T(t)[\/latex] is the temperature at time [latex]t[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]A[\/latex] is the initial temperature difference (object &#8211; surroundings)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]k[\/latex] is the cooling rate (negative for cooling)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]T_s[\/latex] is the surrounding temperature<\/li>\n<\/ul>\n<p class=\"font-600 text-xl font-bold\"><strong>Characteristics<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Exponential decay towards ambient temperature<\/li>\n<li class=\"whitespace-normal break-words\">Vertical shift of the exponential decay function<\/li>\n<li class=\"whitespace-normal break-words\">Horizontal asymptote at [latex]T_s[\/latex]<\/li>\n<\/ol>\n<p class=\"font-600 text-xl font-bold\"><strong>Problem-Solving Approach<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Identify [latex]T_s[\/latex] (ambient temperature)<\/li>\n<li class=\"whitespace-normal break-words\">Find [latex]A[\/latex] by subtracting [latex]T_s[\/latex] from initial temperature<\/li>\n<li class=\"whitespace-normal break-words\">Use a second data point to solve for [latex]k[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Apply the formula to find unknown time or temperature<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q846066\">Show Solution<\/button><\/p>\n<div id=\"q846066\" class=\"hidden-answer\" style=\"display: none\">6.026 hours<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\"><script src=\"https:\/\/www.youtube.com\/iframe_api\" type=\"text\/javascript\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-bdgbgece-F4sr2jUIHZI\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/F4sr2jUIHZI?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-bdgbgece-F4sr2jUIHZI\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&amp;cc_minimizable=1&amp;cc_minimize_on_load=0&amp;cc_multi_text_track=0&amp;cc_overlay=1&amp;cc_searchable=0&amp;embed=ajax&amp;mf=12850425&amp;p3sdk_version=1.11.7&amp;p=20361&amp;player_type=youtube&amp;plugin_skin=dark&amp;target=3p-plugin-target-bdgbgece-F4sr2jUIHZI&amp;vembed=0&amp;video_id=F4sr2jUIHZI&amp;video_target=tpm-plugin-bdgbgece-F4sr2jUIHZI\" type=\"text\/javascript\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/College+Algebra+Corequisite\/Transcripts\/Ex+-+Newton's+Law+of+Cooling+-+Exponential+Function+App_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Newton&#8217;s Law of Cooling &#8211; Exponential Function App\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<h2>Logistic Growth<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<br \/>\n<\/strong><\/p>\n<p class=\"whitespace-pre-wrap break-words\">The logistic growth model describes growth with a limiting factor:<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[latex]f(x) = \\frac{c}{1 + ae^{-bx}}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Where:<\/p>\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]c[\/latex] is the carrying capacity (upper limit)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]a[\/latex] affects the initial value (f(0) = c\/(1+a))<\/li>\n<li class=\"whitespace-normal break-words\">[latex]b[\/latex] is the growth rate<\/li>\n<\/ul>\n<p class=\"font-600 text-xl font-bold\"><strong>Characteristics<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">S-shaped curve<\/li>\n<li class=\"whitespace-normal break-words\">Initially similar to exponential growth<\/li>\n<li class=\"whitespace-normal break-words\">Growth rate decreases as it approaches carrying capacity<\/li>\n<li class=\"whitespace-normal break-words\">Horizontal asymptote at [latex]y = c[\/latex]<\/li>\n<\/ol>\n<p class=\"font-600 text-xl font-bold\"><strong>Comparison to Exponential Growth<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Exponential: Unlimited growth<\/li>\n<li class=\"whitespace-normal break-words\">Logistic: Limited by carrying capacity<\/li>\n<li class=\"whitespace-normal break-words\">Exponential: Constant relative growth rate<\/li>\n<li class=\"whitespace-normal break-words\">Logistic: Decreasing relative growth rate<\/li>\n<\/ol>\n<p class=\"font-600 text-xl font-bold\"><strong>Applications<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Population growth with limited resources<\/li>\n<li class=\"whitespace-normal break-words\">Spread of infectious diseases<\/li>\n<li class=\"whitespace-normal break-words\">Technology adoption<\/li>\n<li class=\"whitespace-normal break-words\">Product sales over time<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">An influenza epidemic spreads through a population rapidly at a rate that depends on two factors. The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model good for studying the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.For example, at time [latex]t = 0[\/latex] there is one person in a community of [latex]1,000[\/latex] people who has the flu. So, in that community, at most [latex]1,000[\/latex] people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is [latex]b = 0.6030[\/latex]. Estimate the number of cases of flu on day [latex]15[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q421768\">Show Solution<\/button><\/p>\n<div id=\"q421768\" class=\"hidden-answer\" style=\"display: none\">895 cases on day 15<\/div>\n<\/div>\n<\/section>\n<h2>Exponential Regression<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea<br \/>\n<\/strong><\/p>\n<p class=\"whitespace-pre-wrap break-words\">Exponential regression is used to model data that shows exponential growth or decay. The general form of the model is:<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[latex]y = ab^x[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Where:<\/p>\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]a[\/latex] is the [latex]y[\/latex]-intercept (initial value)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]b[\/latex] is the base (growth\/decay factor)<\/li>\n<\/ul>\n<p class=\"font-600 text-xl font-bold\"><strong>Characteristics<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">For growth: [latex]b > 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">For decay: [latex]0 < b < 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]a[\/latex] must be positive<\/li>\n<li class=\"whitespace-normal break-words\">Initial growth\/decay is slow, then accelerates<\/li>\n<\/ol>\n<p class=\"font-600 text-xl font-bold\"><strong>When to Use Exponential Regression<\/strong><\/p>\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Data increases\/decreases by a constant percentage<\/li>\n<li class=\"whitespace-normal break-words\">Growth starts slow, then accelerates rapidly<\/li>\n<li class=\"whitespace-normal break-words\">Decay is rapid at first, then slows down approaching zero<\/li>\n<\/ul>\n<p class=\"font-600 text-xl font-bold\"><strong>Regression Process<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Enter data into a graphing utility<\/li>\n<li class=\"whitespace-normal break-words\">Create a scatter plot<\/li>\n<li class=\"whitespace-normal break-words\">Perform exponential regression<\/li>\n<li class=\"whitespace-normal break-words\">Analyze the fit (r\u00b2 value)<\/li>\n<li class=\"whitespace-normal break-words\">Graph the model with the data points<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">The table below shows a recent graduate\u2019s credit card balance each month after graduation.<\/p>\n<table summary=\"Two rows and ten columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><strong>Month<\/strong><\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<td>4<\/td>\n<td>5<\/td>\n<td>6<\/td>\n<td>7<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td><strong>Debt ($)<\/strong><\/td>\n<td>620.00<\/td>\n<td>761.88<\/td>\n<td>899.80<\/td>\n<td>1039.93<\/td>\n<td>1270.63<\/td>\n<td>1589.04<\/td>\n<td>1851.31<\/td>\n<td>2154.92<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Use exponential regression to fit a model to these data.<\/li>\n<li>If spending continues at this rate, what will the graduate\u2019s credit card debt be one year after graduating?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q99197\">Show Solution<\/button><\/p>\n<div id=\"q99197\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.<\/p>\n<p>Using an online graphing tool, the exponential regression model that fits these data is [latex]y=528.25{\\left(1.1943\\right)}^{x}[\/latex].<\/p>\n<p>Using a graphing calculator, the exponential regression model that fits these data is [latex]y=522.88585984{\\left(1.19645256\\right)}^{x}[\/latex].<\/p>\n<p>b.<\/p>\n<p>Using an online graphing tool, if spending continues at this rate, the graduate\u2019s credit card debt will be $4,448.37 after one year.<\/p>\n<p>Using a graphing calculator, if spending continues at this rate, the graduate\u2019s credit card debt will be $4,499.38 after one year.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\"><script src=\"https:\/\/www.youtube.com\/iframe_api\" type=\"text\/javascript\"><\/script><\/p>\n<p class=\"cc-media-iframe-container\"><iframe id=\"tpm-plugin-cgcaahhg-QfBvEjBz_1s\" class=\"cc-media-iframe\" src=\"https:\/\/www.youtube.com\/embed\/QfBvEjBz_1s?enablejsapi=1\" frameborder=\"0\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<div id=\"3p-plugin-target-cgcaahhg-QfBvEjBz_1s\" class=\"p3sdk-target\"><\/div>\n<p class=\"cc-media-iframe-container\"><script src=\"\/\/plugin.3playmedia.com\/ajax.js?cc=1&amp;cc_minimizable=1&amp;cc_minimize_on_load=0&amp;cc_multi_text_track=0&amp;cc_overlay=1&amp;cc_searchable=0&amp;embed=ajax&amp;mf=12850478&amp;p3sdk_version=1.11.7&amp;p=20361&amp;player_type=youtube&amp;plugin_skin=dark&amp;target=3p-plugin-target-cgcaahhg-QfBvEjBz_1s&amp;vembed=0&amp;video_id=QfBvEjBz_1s&amp;video_target=tpm-plugin-cgcaahhg-QfBvEjBz_1s\" type=\"text\/javascript\"><\/script><\/p>\n<p>You can view the\u00a0<a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/College+Algebra+Corequisite\/Transcripts\/Ex+-+Perform+Exponential+Regression+on+a+Graphing+Calculator_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cEx: Perform Exponential Regression on a Graphing Calculator\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n","protected":false},"author":67,"menu_order":24,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"Exponential Growth App (y=ab^t) - Find Initial Amount Given Doubling Time\",\"author\":\"\",\"organization\":\"Mathispower4u\",\"url\":\"https:\/\/youtu.be\/YK7rERyFlOM\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"copyrighted_video\",\"description\":\"Exponential Growth App (y=ab^t) - Given Doubling Time\",\"author\":\"\",\"organization\":\"Mathispower4u\",\"url\":\"https:\/\/youtu.be\/eSOhLhSz9pk\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"copyrighted_video\",\"description\":\"Ex: Exponential Model - Determine Age Using Carbon-14 Given Half Life\",\"author\":\"\",\"organization\":\"Mathispower4u\",\"url\":\"https:\/\/youtu.be\/HYHzK6kF0ts\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard 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