{"id":1238,"date":"2017-02-01T00:08:50","date_gmt":"2017-02-01T00:08:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/?post_type=chapter&#038;p=1238"},"modified":"2019-10-03T21:03:33","modified_gmt":"2019-10-03T21:03:33","slug":"introduction-exponential-and-logistic-growth","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wmopen-mathforliberalarts\/chapter\/introduction-exponential-and-logistic-growth\/","title":{"raw":"Logarithms and Logistic Growth","rendered":"Logarithms and Logistic Growth"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Evaluate and rewrite logarithms using the properties of logarithms<\/li>\r\n \t<li>Use the properties of logarithms to solve exponential models\u00a0for time<\/li>\r\n \t<li>Identify the carrying capacity in a logistic growth model<\/li>\r\n \t<li>Use a logistic growth model to predict growth<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn a confined environment the growth rate of a population may not remain constant. In a lake, for example, there is some <em>maximum sustainable population<\/em> of fish, also called a <strong>carrying capacity<\/strong>. In this section, we will develop a model that contains a carrying capacity term, and use it to predict growth under constraints. \u00a0Because resources are typically limited in systems, these types of models are much more common than linear or geometric growth.\r\n\r\n[caption id=\"attachment_1279\" align=\"aligncenter\" width=\"534\"]<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/02000229\/Mandel_zoom_08_satellite_antenna.jpg\"><img class=\"wp-image-1279\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/02000229\/Mandel_zoom_08_satellite_antenna-300x225.jpg\" alt=\"The famous Mandelbrot set, a fractal whose growth is constrained.\" width=\"534\" height=\"402\" \/><\/a> The famous Mandelbrot set, a fractal whose growth is constrained.[\/caption]\r\n\r\n&nbsp;\r\n<h2>Solve Exponentials for Time: Logarithms<\/h2>\r\n<h3>Reversing an Exponent<\/h3>\r\nEarlier, we found that since Olympia, WA had a population of 245 thousand in 2008 and had been growing at 3% per year, the population could be modeled by the equation\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = (1+0.03)<sup><em>n <\/em><\/sup>(245,000), or equivalently, <em>P<sub>\u00adn<\/sub><\/em> = 245,000(1.03)<sup><em>n<\/em><\/sup>.<\/p>\r\nUsing this equation, we were able to predict the population in the future.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21154211\/population-1282377_1280.jpg\"><img class=\"aligncenter size-large wp-image-899\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21154211\/population-1282377_1280-1024x426.jpg\" alt=\"A soft-focus landscape photo of a crowd of people. Overlaid on top is a dotted red horizontal line and many vertical blue lines rising above and below the dotted red line, to give the impression of a population growth chart.\" width=\"1024\" height=\"426\" \/><\/a>\r\n\r\nSuppose we wanted to know when the population of Olympia would reach 400 thousand. Since we are looking for the year <em>n<\/em> when the population will be 400 thousand, we would need to solve the equation\r\n<p style=\"text-align: center;\">400,000 = 245,000(1.03)<sup><em>n<\/em><\/sup><\/p>\r\n<p style=\"text-align: left;\">Dividing both sides by 245,000 gives<\/p>\r\n<p style=\"text-align: center;\">1.6327 = 1.03<sup><em>n<\/em><\/sup><\/p>\r\n&nbsp;\r\n\r\nOne approach to this problem would be to create a table of values, or to use technology to draw a graph to estimate the solution.\r\n\r\n<img class=\"aligncenter wp-image-368\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11202224\/estimatesolution-300x164.png\" alt=\"Line graph. Vertical measures increments of 50 from 0 to 450. Horizontal measures increments of 1 from 0 to 20. The line starts at 250\/0, and moves in a sloping up trend towards 450\/20.\" width=\"350\" height=\"191\" \/>\r\n\r\nFrom the graph, we can estimate that the solution will be around 16 to 17 years after 2008 (2024 to 2025). This is pretty good, but we\u2019d really like to have an algebraic tool to answer this question. To do that, we need to introduce a new function that will undo exponentials, similar to how a square root undoes a square. For exponentials, the function we need is called a logarithm. It is the inverse of the exponential, meaning it undoes the exponential. While there is a whole family of logarithms with different bases, we will focus on the common log, which is based on the exponential 10<sup><em>x<\/em><\/sup>.\r\n<div>\r\n<div class=\"textbox\">\r\n<div>\r\n<h3>Common Logarithm<\/h3>\r\n<\/div>\r\n<div>\r\n\r\nThe common logarithm, written log(<em>x<\/em>), undoes the exponential 10<sup><em>x<\/em><\/sup>\r\n\r\nThis means that log(10<sup><em>x<\/em><\/sup>) = <em>x<\/em>, and likewise 10<sup>log(<em>x<\/em>)<\/sup> = <em>x<\/em>\r\n\r\nThis also means the statement 10<sup><em>a<\/em><\/sup> = <em>b<\/em> is equivalent to the statement log(<em>b<\/em>) = <em>a<\/em>\r\n\r\nlog(<em>x<\/em>) is read as \u201clog of <em>x<\/em>\u201d, and means \u201cthe logarithm of the value <em>x<\/em>\u201d. It is important to note that this is <em>not<\/em> multiplication \u2013 the log doesn\u2019t mean anything by itself, just like \u221a doesn\u2019t mean anything by itself; it has to be applied to a number.\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<div>\r\n\r\nEvaluate each of the following\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>log(100)<\/li>\r\n \t<li>log(1000)<\/li>\r\n \t<li>log(10000)<\/li>\r\n \t<li>log(1\/100)<\/li>\r\n \t<li>log(1)<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"121444\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"121444\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>log(100) can be written as log(10<sup>2<\/sup>). Since the log undoes the exponential, log(10<sup>2<\/sup>) = 2<\/li>\r\n \t<li>log(1000) = log(10<sup>3<\/sup>) = 3<\/li>\r\n \t<li>log(10000) = log(10<sup>4<\/sup>) = 4<\/li>\r\n \t<li>Recall that [latex]{{x}^{-n}}=\\frac{1}{{{x}^{n}}}[\/latex].\u00a0\u00a0 [latex]\\log\\left(\\frac{1}{100}\\right)=\\log\\left({{10}^{-2}}\\right)=-2[\/latex]<\/li>\r\n \t<li>Recall that <em>x<\/em><sup>0<\/sup> = 1.\u00a0\u00a0 log(1) = log(10<sup>0<\/sup>) = 0<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=100163&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"100%\"><\/iframe>\r\n\r\n<\/div>\r\n<\/div>\r\nIt is helpful to note that from the first three parts of the previous example that the number we\u2019re taking the log of has to get <em>10 times bigger<\/em> for the log to increase in value by 1.\r\n\r\nOf course, most numbers cannot be written as a nice simple power of 10. For those numbers, we can evaluate the log using a scientific calculator with a log button.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<div>\r\n\r\nEvaluate log(300)\r\n[reveal-answer q=\"517989\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"517989\"]Using a calculator, log(300) is approximately 2.477121[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWith an equation, just like we can add a number to both sides, multiply both sides by a number, or square both sides, we can also take the logarithm of both sides of the equation and end up with an equivalent equation. This will allow us to solve some simple equations.\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\n<div>\r\n<ol>\r\n \t<li>Solve 10<sup><em>x<\/em><\/sup> = 1000<\/li>\r\n \t<li>Solve 10<sup><em>x<\/em><\/sup> = 3<\/li>\r\n \t<li>Solve 2(10<sup><em>x<\/em><\/sup>) = 8<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"588505\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"588505\"]\r\n<ol>\r\n \t<li>Taking the log of both sides gives log(10<sup><em>x<\/em><\/sup>) = log(1000)\r\n<ul>\r\n \t<li>Since the log undoes the exponential, log(10<sup><em>x<\/em><\/sup>) = <em>x<\/em>. Similarly log(1000) = log(10<sup>3<\/sup>) = 3.<\/li>\r\n \t<li>The equation simplifies then to <em>x<\/em> = 3.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Taking the log of both sides gives log(10<sup><em>x<\/em><\/sup>) = log(3).\r\n<ul>\r\n \t<li>On the left side, log(10<sup><em>x<\/em><\/sup>) = <em>x<\/em>, so <em>x <\/em>= log(3).<\/li>\r\n \t<li>We can approximate this value with a calculator. <em>x<\/em> \u2248 0.477<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Here we would first want to isolate the exponential by dividing both sides of the equation by 2, giving 10<sup><em>x<\/em><\/sup>\u00a0= 4.\r\n<ul>\r\n \t<li>Now we can take the log of both sides, giving log(10<sup><em>x<\/em><\/sup>) = log(4), which simplifies to\u00a0<em>x<\/em> = log(4) \u2248 0.602<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n\r\n&nbsp;\r\n\r\nThis approach allows us to solve exponential equations with powers of 10, but what about problems like 2 = 1.03<sup><em>n<\/em><\/sup> from earlier, which have a base of 1.03? For that, we need the exponent property for logs.\r\n\r\n<\/div>\r\n<div>\r\n<div class=\"textbox\">\r\n<div>\r\n<h3>Properties of Logs: Exponent Property<\/h3>\r\n<\/div>\r\n<div>[latex]\\log\\left({{A}^{r}}\\right)=r\\log\\left(A\\right)[\/latex]<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\nTo show why this is true, we offer a proof.\r\n\r\nSince the logarithm and exponential undo each other, [latex]{{10}^{\\log{A}}}=A[\/latex].\r\n\r\nSo [latex]{{A}^{r}}={{\\left({{10}^{\\log{A}}}\\right)}^{r}}[\/latex]\r\n\r\nUtilizing the exponential rule that states [latex]{{\\left({{x}^{a}}\\right)}^{b}}={{x}^{ab}}[\/latex],\r\n\r\n[latex]{{A}^{r}}={{\\left({{10}^{\\log{A}}}\\right)}^{r}}={{10}^{r\\log{A}}}[\/latex]\r\n\r\n&nbsp;\r\n\r\nSo then [latex]\\log\\left({{A}^{r}}\\right)=\\log\\left({{10}^{r\\log{A}}}\\right)[\/latex]\r\n\r\nAgain utilizing the property that the log undoes the exponential on the right side yields the result\r\n\r\n[latex]\\log\\left({{A}^{r}}\\right)=r\\log{A}[\/latex]\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<div>\r\n\r\nRewrite log(25) using the exponent property for logs.\r\n[reveal-answer q=\"925419\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"925419\"]log(25) = log(5<sup>2<\/sup>) = 2log(5)[\/hidden-answer]\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1529&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"100%\"><\/iframe>\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\nThis property will finally allow us to answer our original question.\r\n<div>\r\n<div class=\"textbox\">\r\n<div>\r\n<h3>Solving exponential equations with logarithms<\/h3>\r\n<\/div>\r\n<div>\r\n<ol>\r\n \t<li>Isolate the exponential. In other words, get it by itself on one side of the equation. This usually involves dividing by a number multiplying it.<\/li>\r\n \t<li>Take the log of both sides of the equation.<\/li>\r\n \t<li>Use the exponent property of logs to rewrite the exponential with the variable exponent multiplying the logarithm.<\/li>\r\n \t<li>Divide as needed to solve for the variable.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf Olympia is growing according to the equation, <em>P<sub>\u00adn<\/sub><\/em> = 245(1.03)<sup><em>n<\/em><\/sup>, where <em>n<\/em> is years after 2008, and the population is measured in thousands. Find when the population will be 400 thousand.\r\n[reveal-answer q=\"770708\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"770708\"]\r\n\r\nWe need to solve the equation\r\n<p style=\"padding-left: 120px;\">400 = 245(1.03)<em>n<\/em> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Begin by dividing both sides by 245 to isolate the exponential<\/p>\r\n<p style=\"padding-left: 120px;\">1.633 = 1.03<em>n<\/em>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Now take the log of both sides<\/p>\r\n<p style=\"padding-left: 120px;\">log(1.633) = log(1.03<em>n<\/em>)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Use the exponent property of logs on the right side<\/p>\r\n<p style=\"padding-left: 120px;\">log(1.633)= <em>n<\/em> log(1.03) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Now we can divide by log(1.03)<\/p>\r\n<p style=\"padding-left: 120px;\">[latex]\\frac{\\log(1.633)}{\\log(1.03)}=n[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 We can approximate this value on a calculator<\/p>\r\n<p style=\"padding-left: 120px;\"><em>n<\/em> \u2248 16.591<\/p>\r\n[\/hidden-answer]\r\n\r\nA full walkthrough of this problem is available here.\r\n\r\nhttps:\/\/youtu.be\/liNffAACIUs\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1764&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"100%\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\nAlternatively, after applying the exponent property of logs on the right side, we could have evaluated the logarithms to decimal approximations and completed our calculations using those approximations, as you\u2019ll see in the next example. While the final answer may come out slightly differently, as long as we keep enough significant values during calculation, our answer will be close enough for most purposes.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nPolluted water is passed through a series of filters. Each filter removes 90% of the remaining impurities from the water. If you have 10 million particles of pollutant per gallon originally, how many filters would the water need to be passed through to reduce the pollutant to 500 particles per gallon?\r\n[reveal-answer q=\"878881\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"878881\"]\r\n\r\nIn this problem, our \u201cpopulation\u201d is the number of particles of pollutant per gallon. The initial pollutant is 10 million particles per gallon, so <em>P<sub>0<\/sub><\/em> = 10,000,000. Instead of changing with time, the pollutant changes with the number of filters, so <em>n<\/em> will represent the number of filters the water passes through.\r\n\r\nAlso, since the amount of pollutant is <em>decreasing<\/em> with each filter instead of increasing, our \u201cgrowth\u201d rate will be negative, indicating that the population is decreasing instead of increasing, so <em>r<\/em> = -0.90.\r\n\r\nWe can then write the explicit equation for the pollutant:\r\n<p style=\"text-align: center;\"><em>P<sub>n<\/sub><\/em> = 10,000,000(1 \u2013 0.90)<sup><em>n<\/em><\/sup> = 10,000,000(0.10)<sup><em>n<\/em><\/sup><\/p>\r\nTo solve the question of how many filters are needed to lower the pollutant to 500 particles per gallon, we can set <em>P<sub>n<\/sub><\/em> equal to 500, and solve for <em>n<\/em>.\r\n<p style=\"padding-left: 150px;\">500 = 10,000,000(0.10)<em>n<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divide both sides by 10,000,000<\/p>\r\n<p style=\"padding-left: 150px;\">0.00005 = 0.10<em>n<\/em>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Take the log of both sides<\/p>\r\n<p style=\"padding-left: 150px;\">log(0.00005) = log(0.10<em>n<\/em>)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Use the exponent property of logs on the right side<\/p>\r\n<p style=\"padding-left: 150px;\">log(0.00005) = <em>n <\/em>log(0.10) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Evaluate the logarithms to a decimal approximation<\/p>\r\n<p style=\"padding-left: 150px;\">-4.301 = <em>n <\/em>(-1) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Divide by -1, the value multiplying <em>n<\/em><\/p>\r\n<p style=\"padding-left: 150px;\">4.301 = <em>n<\/em><\/p>\r\nIt would take about 4.301 filters. Of course, since we probably can\u2019t install 0.3 filters, we would need to use 5 filters to bring the pollutant below the desired level.\r\n\r\n[\/hidden-answer]\r\n\r\nMore details about solving this scenario are available in this video.\r\n\r\nhttps:\/\/youtu.be\/sLLu0u1YgM0\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIndia had a population in 2008 of about 1.14 billion people. The population is growing by about 1.34% each year. If this trend continues, when will India\u2019s population reach 1.2 billion?\r\n<iframe id=\"mom7\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=114634&amp;theme=oea&amp;iframe_resize_id=mom7\" width=\"100%\" height=\"100%\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>TIP<\/h3>\r\nWhen you are solving growth problems, use the language in the question to determine whether you are solving for time, future value, present value or growth rate. Questions that uses words like \"when\", \"what year\", or \"how long\" are asking you to solve for time and you will need to use logarithms to solve them because the time variable in growth\u00a0problems is in the exponent.\r\n\r\n<\/div>\r\n<\/div>\r\n<h2>Logistic Growth<\/h2>\r\n<h3>Limits on Exponential Growth<\/h3>\r\nIn our basic exponential growth scenario, we had a recursive equation of the form\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n <\/sub><\/em>= <em>P<sub>\u00adn-1 <\/sub><\/em>+ <em>r P<sub>\u00adn-1<\/sub><\/em><\/p>\r\nIn a confined environment, however, the growth rate may not remain constant. In a lake, for example, there is some <em>maximum sustainable population<\/em> of fish, also called a <strong>carrying capacity<\/strong>.\r\n<div class=\"textbox\">\r\n<h3>Carrying Capacity<\/h3>\r\nThe <strong>carrying capacity<\/strong>, or <strong>maximum sustainable population<\/strong>, is the largest population that an environment can support.\r\n\r\n<\/div>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/20205537\/fishes-1711002_1280.jpg\"><img class=\"aligncenter size-large wp-image-892\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/20205537\/fishes-1711002_1280-1024x682.jpg\" alt=\"Dense koi fish population in water\" width=\"1024\" height=\"682\" \/><\/a>\r\n\r\nFor our fish, the carrying capacity is the largest population that the resources in the lake can sustain. If the population in the lake is far below the carrying capacity, then we would expect the population to grow essentially exponentially. However, as the population approaches the carrying capacity, there will be a scarcity of food and space available, and the growth rate will decrease. If the population exceeds the carrying capacity, there won\u2019t be enough resources to sustain all the fish and there will be a negative growth rate, causing the population to decrease back to the carrying capacity.\r\n\r\nIf the carrying capacity was 5000, the growth rate might vary something like that in the graph shown.<img class=\"aligncenter wp-image-371\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11203520\/populationgrowthrate.png\" alt=\"Graph. Vertical measures Growth Rate, in increments of .1, from -0.1 to 0.1. Horizontal measures Population, in increments of 5000, from 0 to 10000. There's a diagonal line sloping down from 0.1 at 0 to -0.1 at 10000.\" width=\"350\" height=\"246\" \/>\r\n\r\nNote that this is a linear equation with intercept at 0.1 and slope [latex]-\\frac{0.1}{5000}[\/latex], so we could write an equation for this adjusted growth rate as:\r\n<p style=\"text-align: center;\"><em>r<sub>adjusted <\/sub><\/em>= [latex]0.1-\\frac{0.1}{5000}P=0.1\\left(1-\\frac{P}{5000}\\right)[\/latex]<\/p>\r\nSubstituting this in to our original exponential growth model for <em>r<\/em> gives\r\n<p style=\"text-align: center;\">[latex]{{P}_{n}}={{P}_{n-1}}+0.1\\left(1-\\frac{{{P}_{n-1}}}{5000}\\right){{P}_{n-1}}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">View the following for a detailed explanation of the concept.<\/p>\r\nhttps:\/\/youtu.be\/-6VLXCTkP_c\r\n<div class=\"textbox\">\r\n<h3>Logistic Growth<\/h3>\r\nIf a population is growing in a constrained environment with carrying capacity <em>K<\/em>, and absent constraint would grow exponentially with growth rate <em>r<\/em>, then the population behavior can be described by the logistic growth model:\r\n\r\n[latex]{{P}_{n}}={{P}_{n-1}}+r\\left(1-\\frac{{{P}_{n-1}}}{K}\\right){{P}_{n-1}}[\/latex]\r\n\r\n<\/div>\r\nUnlike linear and exponential growth, logistic growth behaves differently if the populations grow steadily throughout the year or if they have one breeding time per year. The recursive formula provided above models generational growth, where there is one breeding time per year (or, at least a finite number); there is no explicit formula for this type of logistic growth.\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nA forest is currently home to a population of 200 rabbits. The forest is estimated to be able to sustain a population of 2000 rabbits. Absent any restrictions, the rabbits would grow by 50% per year. Predict the future population using the logistic growth model.\r\n[reveal-answer q=\"543594\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"543594\"]\r\n\r\nModeling this with a logistic growth model, <em>r <\/em> = 0.50, <em>K<\/em> = 2000, and <em>P<sub>\u00ad0 <\/sub><\/em>= 200. Calculating the next year:\r\n<p style=\"text-align: center;\">[latex]{{P}_{1}}={{P}_{0}}+0.50\\left(1-\\frac{{{P}_{0}}}{2000}\\right){{P}_{0}}=200+0.50\\left(1-\\frac{200}{2000}\\right)200=290[\/latex]<\/p>\r\nWe can use this to calculate the following year:\r\n<p style=\"text-align: center;\">[latex]{{P}_{2}}={{P}_{1}}+0.50\\left(1-\\frac{{{P}_{1}}}{2000}\\right){{P}_{1}}=290+0.50\\left(1-\\frac{290}{2000}\\right)290\\approx414[\/latex]<\/p>\r\nA calculator was used to compute several more values:\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><em>n<\/em><\/td>\r\n<td>0<\/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<td>9<\/td>\r\n<td>10<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em>P<sub>n<\/sub><\/em><\/td>\r\n<td>200<\/td>\r\n<td>290<\/td>\r\n<td>414<\/td>\r\n<td>578<\/td>\r\n<td>784<\/td>\r\n<td>1022<\/td>\r\n<td>1272<\/td>\r\n<td>1503<\/td>\r\n<td>1690<\/td>\r\n<td>1821<\/td>\r\n<td>1902<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlotting these values, we can see that the population starts to increase faster and the graph curves upwards during the first few years, like exponential growth, but then the growth slows down as the population approaches the carrying capacity.\r\n\r\n<img class=\"aligncenter wp-image-372\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11203921\/populationyears.png\" alt=\"Graph. Vertical measures Population, in increments of 500, from 0 to 2000. Horizontal measures Years, in increments of 1, from 0 to 10. The line increases quickly and then tapers, similar to the first half of a bell curve.\" width=\"350\" height=\"283\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\nView more about this example below.\r\n\r\nhttps:\/\/youtu.be\/dPOlEgJ2QX0\r\n\r\n<hr \/>\r\n\r\nOn an island that can support a population of 1000 lizards, there is currently a population of 600. These lizards have a lot of offspring and not a lot of natural predators, so have very high growth rate, around 150%. Calculating out the next couple generations:\r\n<p style=\"text-align: center;\">[latex]{{P}_{1}}={{P}_{0}}+1.50\\left(1-\\frac{{{P}_{0}}}{1000}\\right){{P}_{0}}=600+1.50\\left(1-\\frac{600}{1000}\\right)600=960[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{{P}_{2}}={{P}_{1}}+1.50\\left(1-\\frac{{{P}_{1}}}{1000}\\right){{P}_{1}}=960+1.50\\left(1-\\frac{960}{1000}\\right)960=1018[\/latex]<\/p>\r\nInterestingly, even though the factor that limits the growth rate slowed the growth a lot, the population still overshot the carrying capacity. We would expect the population to decline the next year.\r\n<p style=\"text-align: center;\">[latex]{{P}_{3}}={{P}_{2}}+1.50\\left(1-\\frac{{{P}_{3}}}{1000}\\right){{P}_{3}}=1018+1.50\\left(1-\\frac{1018}{1000}\\right)1018=991[\/latex]<\/p>\r\nCalculating out a few more years and plotting the results, we see the population wavers above and below the carrying capacity, but eventually settles down, leaving a steady population near the carrying capacity.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21175945\/Screen-Shot-2016-12-21-at-12.59.24-PM.png\"><img class=\"aligncenter wp-image-910\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21175945\/Screen-Shot-2016-12-21-at-12.59.24-PM.png\" alt=\"Graph. Vertical measures Population, in increments of 200 from 0 to 1200. Horizontal measures Years, in increments of 1 from 0 to 10. Year 0 shows population of 600, jumping to ~1000 in year 1, a little over 1000 in year 2, and staying close to 1000 in every subsequent year. \" width=\"350\" height=\"277\" \/><\/a>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA field currently contains 20 mint plants. Absent constraints, the number of plants would increase by 70% each year, but the field can only support a maximum population of 300 plants. Use the logistic model to predict the population in the next three years.\r\n[reveal-answer q=\"726473\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"726473\"][latex]P_1=P_0+0.70(1-\\frac{P_0}{300})P_0=20+0.70(1-\\frac{20}{300})20=33[\/latex]\r\n\r\n[latex]P_2=54[\/latex]\r\n\r\n[latex]P_3=85[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nOn a neighboring island to the one from the previous example, there is another population of lizards, but the growth rate is even higher \u2013 about 205%.\r\n\r\nCalculating out several generations and plotting the results, we get a surprise: the population seems to be oscillating between two values, a pattern called a 2-cycle.<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21184924\/Screen-Shot-2016-12-21-at-1.47.24-PM.png\"><img class=\"aligncenter wp-image-911\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21184924\/Screen-Shot-2016-12-21-at-1.47.24-PM.png\" alt=\"Graph. Vertical measures Population, in increments of 200 from 0 to 1200. Horizontal measures Years, in increments of 1 from 0 to 10. Year 0 shows population of 600, jumping to ~1100 in year 1, down to ~900 in year 2, and vacillating from 1100 to 800 in alternating years through the rest of the graph.\" width=\"350\" height=\"276\" \/><\/a>\r\n\r\nWhile it would be tempting to treat this only as a strange side effect of mathematics, this has actually been observed in nature. Researchers from the University of California observed a stable 2-cycle in a lizard population in California.[footnote]<a href=\"http:\/\/users.rcn.com\/jkimball.ma.ultranet\/BiologyPages\/P\/Populations2.html\" target=\"_blank\" rel=\"noopener\">http:\/\/users.rcn.com\/jkimball.ma.ultranet\/BiologyPages\/P\/Populations2.html<\/a>[\/footnote]\r\n\r\nTaking this even further, we get more and more extreme behaviors as the growth rate increases higher. It is possible to get stable 4-cycles, 8-cycles, and higher. Quickly, though, the behavior approaches chaos (remember the movie <em>Jurassic Park<\/em>?).\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21185438\/Screen-Shot-2016-12-21-at-1.50.01-PM.png\"><img class=\"aligncenter wp-image-913 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21185438\/Screen-Shot-2016-12-21-at-1.50.01-PM.png\" alt=\"Two graphs. Left graph, labeled \u201cr=2.46 A 4-cycle.\u201d Vertical measures Population, in increments of 200 from 0 to 1400. Horizontal measures Years, in increments of 1 from 0 to 10. The plotted line moves in sharp up and down cycles, from ~700 in year 0, 1200 in year 1, 600 in year 2, 1200 in year 3, ~600 in year 4, and so forth. Right graph, labeled \u201cr=2.9 Chaos!\u201d Vertical measures Population, in increments of 200 from 0 to 1400. Horizontal measures Years, in increments of 5 from 0 to 30. The plotted line moves in sharp up and down cycles of varying lengths, forming an erratic back and forth between highs and lows.\" width=\"613\" height=\"279\" \/><\/a>\r\n\r\nAll of the lizard island examples are discussed in this video.\r\n\r\nhttps:\/\/youtu.be\/fuJF_uZGoFc\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6589&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"100%\"><\/iframe>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Evaluate and rewrite logarithms using the properties of logarithms<\/li>\n<li>Use the properties of logarithms to solve exponential models\u00a0for time<\/li>\n<li>Identify the carrying capacity in a logistic growth model<\/li>\n<li>Use a logistic growth model to predict growth<\/li>\n<\/ul>\n<\/div>\n<p>In a confined environment the growth rate of a population may not remain constant. In a lake, for example, there is some <em>maximum sustainable population<\/em> of fish, also called a <strong>carrying capacity<\/strong>. In this section, we will develop a model that contains a carrying capacity term, and use it to predict growth under constraints. \u00a0Because resources are typically limited in systems, these types of models are much more common than linear or geometric growth.<\/p>\n<div id=\"attachment_1279\" style=\"width: 544px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/02000229\/Mandel_zoom_08_satellite_antenna.jpg\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1279\" class=\"wp-image-1279\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/02000229\/Mandel_zoom_08_satellite_antenna-300x225.jpg\" alt=\"The famous Mandelbrot set, a fractal whose growth is constrained.\" width=\"534\" height=\"402\" \/><\/a><\/p>\n<p id=\"caption-attachment-1279\" class=\"wp-caption-text\">The famous Mandelbrot set, a fractal whose growth is constrained.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Solve Exponentials for Time: Logarithms<\/h2>\n<h3>Reversing an Exponent<\/h3>\n<p>Earlier, we found that since Olympia, WA had a population of 245 thousand in 2008 and had been growing at 3% per year, the population could be modeled by the equation<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = (1+0.03)<sup><em>n <\/em><\/sup>(245,000), or equivalently, <em>P<sub>\u00adn<\/sub><\/em> = 245,000(1.03)<sup><em>n<\/em><\/sup>.<\/p>\n<p>Using this equation, we were able to predict the population in the future.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21154211\/population-1282377_1280.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-large wp-image-899\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21154211\/population-1282377_1280-1024x426.jpg\" alt=\"A soft-focus landscape photo of a crowd of people. Overlaid on top is a dotted red horizontal line and many vertical blue lines rising above and below the dotted red line, to give the impression of a population growth chart.\" width=\"1024\" height=\"426\" \/><\/a><\/p>\n<p>Suppose we wanted to know when the population of Olympia would reach 400 thousand. Since we are looking for the year <em>n<\/em> when the population will be 400 thousand, we would need to solve the equation<\/p>\n<p style=\"text-align: center;\">400,000 = 245,000(1.03)<sup><em>n<\/em><\/sup><\/p>\n<p style=\"text-align: left;\">Dividing both sides by 245,000 gives<\/p>\n<p style=\"text-align: center;\">1.6327 = 1.03<sup><em>n<\/em><\/sup><\/p>\n<p>&nbsp;<\/p>\n<p>One approach to this problem would be to create a table of values, or to use technology to draw a graph to estimate the solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-368\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11202224\/estimatesolution-300x164.png\" alt=\"Line graph. Vertical measures increments of 50 from 0 to 450. Horizontal measures increments of 1 from 0 to 20. The line starts at 250\/0, and moves in a sloping up trend towards 450\/20.\" width=\"350\" height=\"191\" \/><\/p>\n<p>From the graph, we can estimate that the solution will be around 16 to 17 years after 2008 (2024 to 2025). This is pretty good, but we\u2019d really like to have an algebraic tool to answer this question. To do that, we need to introduce a new function that will undo exponentials, similar to how a square root undoes a square. For exponentials, the function we need is called a logarithm. It is the inverse of the exponential, meaning it undoes the exponential. While there is a whole family of logarithms with different bases, we will focus on the common log, which is based on the exponential 10<sup><em>x<\/em><\/sup>.<\/p>\n<div>\n<div class=\"textbox\">\n<div>\n<h3>Common Logarithm<\/h3>\n<\/div>\n<div>\n<p>The common logarithm, written log(<em>x<\/em>), undoes the exponential 10<sup><em>x<\/em><\/sup><\/p>\n<p>This means that log(10<sup><em>x<\/em><\/sup>) = <em>x<\/em>, and likewise 10<sup>log(<em>x<\/em>)<\/sup> = <em>x<\/em><\/p>\n<p>This also means the statement 10<sup><em>a<\/em><\/sup> = <em>b<\/em> is equivalent to the statement log(<em>b<\/em>) = <em>a<\/em><\/p>\n<p>log(<em>x<\/em>) is read as \u201clog of <em>x<\/em>\u201d, and means \u201cthe logarithm of the value <em>x<\/em>\u201d. It is important to note that this is <em>not<\/em> multiplication \u2013 the log doesn\u2019t mean anything by itself, just like \u221a doesn\u2019t mean anything by itself; it has to be applied to a number.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<div>\n<p>Evaluate each of the following<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>log(100)<\/li>\n<li>log(1000)<\/li>\n<li>log(10000)<\/li>\n<li>log(1\/100)<\/li>\n<li>log(1)<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q121444\">Show Solution<\/span><\/p>\n<div id=\"q121444\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>log(100) can be written as log(10<sup>2<\/sup>). Since the log undoes the exponential, log(10<sup>2<\/sup>) = 2<\/li>\n<li>log(1000) = log(10<sup>3<\/sup>) = 3<\/li>\n<li>log(10000) = log(10<sup>4<\/sup>) = 4<\/li>\n<li>Recall that [latex]{{x}^{-n}}=\\frac{1}{{{x}^{n}}}[\/latex].\u00a0\u00a0 [latex]\\log\\left(\\frac{1}{100}\\right)=\\log\\left({{10}^{-2}}\\right)=-2[\/latex]<\/li>\n<li>Recall that <em>x<\/em><sup>0<\/sup> = 1.\u00a0\u00a0 log(1) = log(10<sup>0<\/sup>) = 0<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=100163&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"100%\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<p>It is helpful to note that from the first three parts of the previous example that the number we\u2019re taking the log of has to get <em>10 times bigger<\/em> for the log to increase in value by 1.<\/p>\n<p>Of course, most numbers cannot be written as a nice simple power of 10. For those numbers, we can evaluate the log using a scientific calculator with a log button.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<div>\n<p>Evaluate log(300)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q517989\">Show Solution<\/span><\/p>\n<div id=\"q517989\" class=\"hidden-answer\" style=\"display: none\">Using a calculator, log(300) is approximately 2.477121<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>With an equation, just like we can add a number to both sides, multiply both sides by a number, or square both sides, we can also take the logarithm of both sides of the equation and end up with an equivalent equation. This will allow us to solve some simple equations.<\/p>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<div>\n<ol>\n<li>Solve 10<sup><em>x<\/em><\/sup> = 1000<\/li>\n<li>Solve 10<sup><em>x<\/em><\/sup> = 3<\/li>\n<li>Solve 2(10<sup><em>x<\/em><\/sup>) = 8<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q588505\">Show Solution<\/span><\/p>\n<div id=\"q588505\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Taking the log of both sides gives log(10<sup><em>x<\/em><\/sup>) = log(1000)\n<ul>\n<li>Since the log undoes the exponential, log(10<sup><em>x<\/em><\/sup>) = <em>x<\/em>. Similarly log(1000) = log(10<sup>3<\/sup>) = 3.<\/li>\n<li>The equation simplifies then to <em>x<\/em> = 3.<\/li>\n<\/ul>\n<\/li>\n<li>Taking the log of both sides gives log(10<sup><em>x<\/em><\/sup>) = log(3).\n<ul>\n<li>On the left side, log(10<sup><em>x<\/em><\/sup>) = <em>x<\/em>, so <em>x <\/em>= log(3).<\/li>\n<li>We can approximate this value with a calculator. <em>x<\/em> \u2248 0.477<\/li>\n<\/ul>\n<\/li>\n<li>Here we would first want to isolate the exponential by dividing both sides of the equation by 2, giving 10<sup><em>x<\/em><\/sup>\u00a0= 4.\n<ul>\n<li>Now we can take the log of both sides, giving log(10<sup><em>x<\/em><\/sup>) = log(4), which simplifies to\u00a0<em>x<\/em> = log(4) \u2248 0.602<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<p>&nbsp;<\/p>\n<p>This approach allows us to solve exponential equations with powers of 10, but what about problems like 2 = 1.03<sup><em>n<\/em><\/sup> from earlier, which have a base of 1.03? For that, we need the exponent property for logs.<\/p>\n<\/div>\n<div>\n<div class=\"textbox\">\n<div>\n<h3>Properties of Logs: Exponent Property<\/h3>\n<\/div>\n<div>[latex]\\log\\left({{A}^{r}}\\right)=r\\log\\left(A\\right)[\/latex]<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p>To show why this is true, we offer a proof.<\/p>\n<p>Since the logarithm and exponential undo each other, [latex]{{10}^{\\log{A}}}=A[\/latex].<\/p>\n<p>So [latex]{{A}^{r}}={{\\left({{10}^{\\log{A}}}\\right)}^{r}}[\/latex]<\/p>\n<p>Utilizing the exponential rule that states [latex]{{\\left({{x}^{a}}\\right)}^{b}}={{x}^{ab}}[\/latex],<\/p>\n<p>[latex]{{A}^{r}}={{\\left({{10}^{\\log{A}}}\\right)}^{r}}={{10}^{r\\log{A}}}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>So then [latex]\\log\\left({{A}^{r}}\\right)=\\log\\left({{10}^{r\\log{A}}}\\right)[\/latex]<\/p>\n<p>Again utilizing the property that the log undoes the exponential on the right side yields the result<\/p>\n<p>[latex]\\log\\left({{A}^{r}}\\right)=r\\log{A}[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<div>\n<p>Rewrite log(25) using the exponent property for logs.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q925419\">Show Solution<\/span><\/p>\n<div id=\"q925419\" class=\"hidden-answer\" style=\"display: none\">log(25) = log(5<sup>2<\/sup>) = 2log(5)<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1529&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"100%\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>This property will finally allow us to answer our original question.<\/p>\n<div>\n<div class=\"textbox\">\n<div>\n<h3>Solving exponential equations with logarithms<\/h3>\n<\/div>\n<div>\n<ol>\n<li>Isolate the exponential. In other words, get it by itself on one side of the equation. This usually involves dividing by a number multiplying it.<\/li>\n<li>Take the log of both sides of the equation.<\/li>\n<li>Use the exponent property of logs to rewrite the exponential with the variable exponent multiplying the logarithm.<\/li>\n<li>Divide as needed to solve for the variable.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If Olympia is growing according to the equation, <em>P<sub>\u00adn<\/sub><\/em> = 245(1.03)<sup><em>n<\/em><\/sup>, where <em>n<\/em> is years after 2008, and the population is measured in thousands. Find when the population will be 400 thousand.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q770708\">Show Solution<\/span><\/p>\n<div id=\"q770708\" class=\"hidden-answer\" style=\"display: none\">\n<p>We need to solve the equation<\/p>\n<p style=\"padding-left: 120px;\">400 = 245(1.03)<em>n<\/em> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Begin by dividing both sides by 245 to isolate the exponential<\/p>\n<p style=\"padding-left: 120px;\">1.633 = 1.03<em>n<\/em>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Now take the log of both sides<\/p>\n<p style=\"padding-left: 120px;\">log(1.633) = log(1.03<em>n<\/em>)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Use the exponent property of logs on the right side<\/p>\n<p style=\"padding-left: 120px;\">log(1.633)= <em>n<\/em> log(1.03) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Now we can divide by log(1.03)<\/p>\n<p style=\"padding-left: 120px;\">[latex]\\frac{\\log(1.633)}{\\log(1.03)}=n[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 We can approximate this value on a calculator<\/p>\n<p style=\"padding-left: 120px;\"><em>n<\/em> \u2248 16.591<\/p>\n<\/div>\n<\/div>\n<p>A full walkthrough of this problem is available here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Using logs to solve for time\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/liNffAACIUs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1764&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"100%\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<p>Alternatively, after applying the exponent property of logs on the right side, we could have evaluated the logarithms to decimal approximations and completed our calculations using those approximations, as you\u2019ll see in the next example. While the final answer may come out slightly differently, as long as we keep enough significant values during calculation, our answer will be close enough for most purposes.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Polluted water is passed through a series of filters. Each filter removes 90% of the remaining impurities from the water. If you have 10 million particles of pollutant per gallon originally, how many filters would the water need to be passed through to reduce the pollutant to 500 particles per gallon?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q878881\">Show Solution<\/span><\/p>\n<div id=\"q878881\" class=\"hidden-answer\" style=\"display: none\">\n<p>In this problem, our \u201cpopulation\u201d is the number of particles of pollutant per gallon. The initial pollutant is 10 million particles per gallon, so <em>P<sub>0<\/sub><\/em> = 10,000,000. Instead of changing with time, the pollutant changes with the number of filters, so <em>n<\/em> will represent the number of filters the water passes through.<\/p>\n<p>Also, since the amount of pollutant is <em>decreasing<\/em> with each filter instead of increasing, our \u201cgrowth\u201d rate will be negative, indicating that the population is decreasing instead of increasing, so <em>r<\/em> = -0.90.<\/p>\n<p>We can then write the explicit equation for the pollutant:<\/p>\n<p style=\"text-align: center;\"><em>P<sub>n<\/sub><\/em> = 10,000,000(1 \u2013 0.90)<sup><em>n<\/em><\/sup> = 10,000,000(0.10)<sup><em>n<\/em><\/sup><\/p>\n<p>To solve the question of how many filters are needed to lower the pollutant to 500 particles per gallon, we can set <em>P<sub>n<\/sub><\/em> equal to 500, and solve for <em>n<\/em>.<\/p>\n<p style=\"padding-left: 150px;\">500 = 10,000,000(0.10)<em>n<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divide both sides by 10,000,000<\/p>\n<p style=\"padding-left: 150px;\">0.00005 = 0.10<em>n<\/em>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Take the log of both sides<\/p>\n<p style=\"padding-left: 150px;\">log(0.00005) = log(0.10<em>n<\/em>)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Use the exponent property of logs on the right side<\/p>\n<p style=\"padding-left: 150px;\">log(0.00005) = <em>n <\/em>log(0.10) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Evaluate the logarithms to a decimal approximation<\/p>\n<p style=\"padding-left: 150px;\">-4.301 = <em>n <\/em>(-1) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Divide by -1, the value multiplying <em>n<\/em><\/p>\n<p style=\"padding-left: 150px;\">4.301 = <em>n<\/em><\/p>\n<p>It would take about 4.301 filters. Of course, since we probably can\u2019t install 0.3 filters, we would need to use 5 filters to bring the pollutant below the desired level.<\/p>\n<\/div>\n<\/div>\n<p>More details about solving this scenario are available in this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Using logs to solve an exponential\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/sLLu0u1YgM0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>India had a population in 2008 of about 1.14 billion people. The population is growing by about 1.34% each year. If this trend continues, when will India\u2019s population reach 1.2 billion?<br \/>\n<iframe loading=\"lazy\" id=\"mom7\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=114634&amp;theme=oea&amp;iframe_resize_id=mom7\" width=\"100%\" height=\"100%\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>TIP<\/h3>\n<p>When you are solving growth problems, use the language in the question to determine whether you are solving for time, future value, present value or growth rate. Questions that uses words like &#8220;when&#8221;, &#8220;what year&#8221;, or &#8220;how long&#8221; are asking you to solve for time and you will need to use logarithms to solve them because the time variable in growth\u00a0problems is in the exponent.<\/p>\n<\/div>\n<\/div>\n<h2>Logistic Growth<\/h2>\n<h3>Limits on Exponential Growth<\/h3>\n<p>In our basic exponential growth scenario, we had a recursive equation of the form<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n <\/sub><\/em>= <em>P<sub>\u00adn-1 <\/sub><\/em>+ <em>r P<sub>\u00adn-1<\/sub><\/em><\/p>\n<p>In a confined environment, however, the growth rate may not remain constant. In a lake, for example, there is some <em>maximum sustainable population<\/em> of fish, also called a <strong>carrying capacity<\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>Carrying Capacity<\/h3>\n<p>The <strong>carrying capacity<\/strong>, or <strong>maximum sustainable population<\/strong>, is the largest population that an environment can support.<\/p>\n<\/div>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/20205537\/fishes-1711002_1280.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-large wp-image-892\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/20205537\/fishes-1711002_1280-1024x682.jpg\" alt=\"Dense koi fish population in water\" width=\"1024\" height=\"682\" \/><\/a><\/p>\n<p>For our fish, the carrying capacity is the largest population that the resources in the lake can sustain. If the population in the lake is far below the carrying capacity, then we would expect the population to grow essentially exponentially. However, as the population approaches the carrying capacity, there will be a scarcity of food and space available, and the growth rate will decrease. If the population exceeds the carrying capacity, there won\u2019t be enough resources to sustain all the fish and there will be a negative growth rate, causing the population to decrease back to the carrying capacity.<\/p>\n<p>If the carrying capacity was 5000, the growth rate might vary something like that in the graph shown.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-371\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11203520\/populationgrowthrate.png\" alt=\"Graph. Vertical measures Growth Rate, in increments of .1, from -0.1 to 0.1. Horizontal measures Population, in increments of 5000, from 0 to 10000. There's a diagonal line sloping down from 0.1 at 0 to -0.1 at 10000.\" width=\"350\" height=\"246\" \/><\/p>\n<p>Note that this is a linear equation with intercept at 0.1 and slope [latex]-\\frac{0.1}{5000}[\/latex], so we could write an equation for this adjusted growth rate as:<\/p>\n<p style=\"text-align: center;\"><em>r<sub>adjusted <\/sub><\/em>= [latex]0.1-\\frac{0.1}{5000}P=0.1\\left(1-\\frac{P}{5000}\\right)[\/latex]<\/p>\n<p>Substituting this in to our original exponential growth model for <em>r<\/em> gives<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{n}}={{P}_{n-1}}+0.1\\left(1-\\frac{{{P}_{n-1}}}{5000}\\right){{P}_{n-1}}[\/latex]<\/p>\n<p style=\"text-align: left;\">View the following for a detailed explanation of the concept.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Logistic model\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/-6VLXCTkP_c?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox\">\n<h3>Logistic Growth<\/h3>\n<p>If a population is growing in a constrained environment with carrying capacity <em>K<\/em>, and absent constraint would grow exponentially with growth rate <em>r<\/em>, then the population behavior can be described by the logistic growth model:<\/p>\n<p>[latex]{{P}_{n}}={{P}_{n-1}}+r\\left(1-\\frac{{{P}_{n-1}}}{K}\\right){{P}_{n-1}}[\/latex]<\/p>\n<\/div>\n<p>Unlike linear and exponential growth, logistic growth behaves differently if the populations grow steadily throughout the year or if they have one breeding time per year. The recursive formula provided above models generational growth, where there is one breeding time per year (or, at least a finite number); there is no explicit formula for this type of logistic growth.<\/p>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>A forest is currently home to a population of 200 rabbits. The forest is estimated to be able to sustain a population of 2000 rabbits. Absent any restrictions, the rabbits would grow by 50% per year. Predict the future population using the logistic growth model.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q543594\">Show Solution<\/span><\/p>\n<div id=\"q543594\" class=\"hidden-answer\" style=\"display: none\">\n<p>Modeling this with a logistic growth model, <em>r <\/em> = 0.50, <em>K<\/em> = 2000, and <em>P<sub>\u00ad0 <\/sub><\/em>= 200. Calculating the next year:<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{1}}={{P}_{0}}+0.50\\left(1-\\frac{{{P}_{0}}}{2000}\\right){{P}_{0}}=200+0.50\\left(1-\\frac{200}{2000}\\right)200=290[\/latex]<\/p>\n<p>We can use this to calculate the following year:<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{2}}={{P}_{1}}+0.50\\left(1-\\frac{{{P}_{1}}}{2000}\\right){{P}_{1}}=290+0.50\\left(1-\\frac{290}{2000}\\right)290\\approx414[\/latex]<\/p>\n<p>A calculator was used to compute several more values:<\/p>\n<table>\n<tbody>\n<tr>\n<td><em>n<\/em><\/td>\n<td>0<\/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<td>9<\/td>\n<td>10<\/td>\n<\/tr>\n<tr>\n<td><em>P<sub>n<\/sub><\/em><\/td>\n<td>200<\/td>\n<td>290<\/td>\n<td>414<\/td>\n<td>578<\/td>\n<td>784<\/td>\n<td>1022<\/td>\n<td>1272<\/td>\n<td>1503<\/td>\n<td>1690<\/td>\n<td>1821<\/td>\n<td>1902<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plotting these values, we can see that the population starts to increase faster and the graph curves upwards during the first few years, like exponential growth, but then the growth slows down as the population approaches the carrying capacity.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-372\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11203921\/populationyears.png\" alt=\"Graph. Vertical measures Population, in increments of 500, from 0 to 2000. Horizontal measures Years, in increments of 1, from 0 to 10. The line increases quickly and then tapers, similar to the first half of a bell curve.\" width=\"350\" height=\"283\" \/><\/p>\n<\/div>\n<\/div>\n<p>View more about this example below.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Logistic growth of rabbits\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/dPOlEgJ2QX0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<hr \/>\n<p>On an island that can support a population of 1000 lizards, there is currently a population of 600. These lizards have a lot of offspring and not a lot of natural predators, so have very high growth rate, around 150%. Calculating out the next couple generations:<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{1}}={{P}_{0}}+1.50\\left(1-\\frac{{{P}_{0}}}{1000}\\right){{P}_{0}}=600+1.50\\left(1-\\frac{600}{1000}\\right)600=960[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{2}}={{P}_{1}}+1.50\\left(1-\\frac{{{P}_{1}}}{1000}\\right){{P}_{1}}=960+1.50\\left(1-\\frac{960}{1000}\\right)960=1018[\/latex]<\/p>\n<p>Interestingly, even though the factor that limits the growth rate slowed the growth a lot, the population still overshot the carrying capacity. We would expect the population to decline the next year.<\/p>\n<p style=\"text-align: center;\">[latex]{{P}_{3}}={{P}_{2}}+1.50\\left(1-\\frac{{{P}_{3}}}{1000}\\right){{P}_{3}}=1018+1.50\\left(1-\\frac{1018}{1000}\\right)1018=991[\/latex]<\/p>\n<p>Calculating out a few more years and plotting the results, we see the population wavers above and below the carrying capacity, but eventually settles down, leaving a steady population near the carrying capacity.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21175945\/Screen-Shot-2016-12-21-at-12.59.24-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-910\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21175945\/Screen-Shot-2016-12-21-at-12.59.24-PM.png\" alt=\"Graph. Vertical measures Population, in increments of 200 from 0 to 1200. Horizontal measures Years, in increments of 1 from 0 to 10. Year 0 shows population of 600, jumping to ~1000 in year 1, a little over 1000 in year 2, and staying close to 1000 in every subsequent year.\" width=\"350\" height=\"277\" \/><\/a><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A field currently contains 20 mint plants. Absent constraints, the number of plants would increase by 70% each year, but the field can only support a maximum population of 300 plants. Use the logistic model to predict the population in the next three years.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q726473\">Show Solution<\/span><\/p>\n<div id=\"q726473\" class=\"hidden-answer\" style=\"display: none\">[latex]P_1=P_0+0.70(1-\\frac{P_0}{300})P_0=20+0.70(1-\\frac{20}{300})20=33[\/latex]<\/p>\n<p>[latex]P_2=54[\/latex]<\/p>\n<p>[latex]P_3=85[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>On a neighboring island to the one from the previous example, there is another population of lizards, but the growth rate is even higher \u2013 about 205%.<\/p>\n<p>Calculating out several generations and plotting the results, we get a surprise: the population seems to be oscillating between two values, a pattern called a 2-cycle.<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21184924\/Screen-Shot-2016-12-21-at-1.47.24-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-911\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21184924\/Screen-Shot-2016-12-21-at-1.47.24-PM.png\" alt=\"Graph. Vertical measures Population, in increments of 200 from 0 to 1200. Horizontal measures Years, in increments of 1 from 0 to 10. Year 0 shows population of 600, jumping to ~1100 in year 1, down to ~900 in year 2, and vacillating from 1100 to 800 in alternating years through the rest of the graph.\" width=\"350\" height=\"276\" \/><\/a><\/p>\n<p>While it would be tempting to treat this only as a strange side effect of mathematics, this has actually been observed in nature. Researchers from the University of California observed a stable 2-cycle in a lizard population in California.<a class=\"footnote\" title=\"http:\/\/users.rcn.com\/jkimball.ma.ultranet\/BiologyPages\/P\/Populations2.html\" id=\"return-footnote-1238-1\" href=\"#footnote-1238-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a><\/p>\n<p>Taking this even further, we get more and more extreme behaviors as the growth rate increases higher. It is possible to get stable 4-cycles, 8-cycles, and higher. Quickly, though, the behavior approaches chaos (remember the movie <em>Jurassic Park<\/em>?).<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21185438\/Screen-Shot-2016-12-21-at-1.50.01-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-913 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21185438\/Screen-Shot-2016-12-21-at-1.50.01-PM.png\" alt=\"Two graphs. Left graph, labeled \u201cr=2.46 A 4-cycle.\u201d Vertical measures Population, in increments of 200 from 0 to 1400. Horizontal measures Years, in increments of 1 from 0 to 10. The plotted line moves in sharp up and down cycles, from ~700 in year 0, 1200 in year 1, 600 in year 2, 1200 in year 3, ~600 in year 4, and so forth. Right graph, labeled \u201cr=2.9 Chaos!\u201d Vertical measures Population, in increments of 200 from 0 to 1400. Horizontal measures Years, in increments of 5 from 0 to 30. The plotted line moves in sharp up and down cycles of varying lengths, forming an erratic back and forth between highs and lows.\" width=\"613\" height=\"279\" \/><\/a><\/p>\n<p>All of the lizard island examples are discussed in this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Logistic growth of lizards\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/fuJF_uZGoFc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6589&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"100%\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1238\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Introduction and Learning Objectives. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Math in Society. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>population-statistics-human. <strong>Authored by<\/strong>: geralt. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/pixabay.com\/en\/population-statistics-human-1282377\/\">https:\/\/pixabay.com\/en\/population-statistics-human-1282377\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Using logs to solve for time. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/liNffAACIUs\">https:\/\/youtu.be\/liNffAACIUs<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Using logs to solve an exponential. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/sLLu0u1YgM0\">https:\/\/youtu.be\/sLLu0u1YgM0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 100163. <strong>Authored by<\/strong>: Rieman, Rick. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 1764. <strong>Authored by<\/strong>: Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 34625. <strong>Authored by<\/strong>: Smart, Jim. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>fishes-colourful-beautiful-koi. <strong>Authored by<\/strong>: sharonang. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/pixabay.com\/en\/fishes-colourful-beautiful-koi-1711002\/\">https:\/\/pixabay.com\/en\/fishes-colourful-beautiful-koi-1711002\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/cc0\">CC0: No Rights Reserved<\/a><\/em><\/li><li>Logistic model. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/-6VLXCTkP_c\">https:\/\/youtu.be\/-6VLXCTkP_c<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Logistic growth of rabbits. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/dPOlEgJ2QX0\">https:\/\/youtu.be\/dPOlEgJ2QX0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Logistic growth of lizards. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/fuJF_uZGoFc\">https:\/\/youtu.be\/fuJF_uZGoFc<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 54627. <strong>Authored by<\/strong>: Hartley, Josiah. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 6589. <strong>Authored by<\/strong>: Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-1238-1\"><a href=\"http:\/\/users.rcn.com\/jkimball.ma.ultranet\/BiologyPages\/P\/Populations2.html\" target=\"_blank\" rel=\"noopener\">http:\/\/users.rcn.com\/jkimball.ma.ultranet\/BiologyPages\/P\/Populations2.html<\/a> <a href=\"#return-footnote-1238-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Introduction and Learning Objectives\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"David Lippman\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"population-statistics-human\",\"author\":\"geralt\",\"organization\":\"\",\"url\":\"https:\/\/pixabay.com\/en\/population-statistics-human-1282377\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Using logs to solve for time\",\"author\":\"OCLPhase2\\'s 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