{"id":86389,"date":"2021-07-20T09:12:14","date_gmt":"2021-07-20T14:12:14","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=86389"},"modified":"2026-03-28T10:47:54","modified_gmt":"2026-03-28T15:47:54","slug":"population-growth","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/population-growth\/","title":{"rendered":"Population Growth"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_IU2IQHzjrA0\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_IU2IQHzjrA0\" data-source-videoID=\"IU2IQHzjrA0\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Population Growth Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Population Growth\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_IU2IQHzjrA0:hover {cursor:pointer;} img#videoThumbnailImage_IU2IQHzjrA0 {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/1721-population-growth-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_IU2IQHzjrA0\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_IU2IQHzjrA0\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Population Growth\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_IU2IQHzjrA0\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_IU2IQHzjrA0\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_IU2IQHzjrA0\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction Nw6_Function() {\n  var x = document.getElementById(\"Nw6\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"Nw6_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"Nw6\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#The_Basics_of_Functions\" class=\"smooth-scroll\">The Basics of Functions<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Population_Growth\" class=\"smooth-scroll\">Population Growth<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Population_Growth_Examples\" class=\"smooth-scroll\">Population Growth Examples<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Population_Growth_Practice_Questions\" class=\"smooth-scroll\">Population Growth Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this video about population growth! <\/p>\n<p>Today, we\u2019re going to cover the unique features of exponential functions and how they are used to model population growth. I\u2019m going to give you a few practice problems later on, so go ahead and make sure you\u2019ve got a calculator handy if you want to try to solve them yourself.<\/p>\n<p>Let\u2019s get started!<\/p>\n<h2><span id=\"The_Basics_of_Functions\" class=\"m-toc-anchor\"><\/span>The Basics of Functions<\/h2>\n<p>\nBefore we dive in, let\u2019s remind ourselves of the basics of functions. In simple terms, functions provide the rules on how to mathematically adjust \u201cinput\u201d to get \u201coutput.\u201d <\/p>\n<p>Here\u2019s a simple <strong>linear equation<\/strong>:<\/p>\n<div class=\"examplesentence\">\\(f(x)=mx+b\\)<\/div>\n<p>\n&nbsp;<br \/>\nThese rules arrive at the \u201coutput\u201d by multiplying the \u201cinput\u201d by the constant, <span style=\"font-style:normal; font-size:90%\">\\(m\\)<\/span>, and adding the value of <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span>.  As you know, the graph of any linear function is a line that increases at a constant rate when the slope, <span style=\"font-style:normal; font-size:90%\">\\(m\\)<\/span>, is positive.  These functions are helpful to visualize data that is linear in nature. <\/p>\n<p>In contrast, <strong>exponential functions<\/strong>, which are the focus of this video, are used as models for data that increase rapidly over time. This is known as a pattern of exponential growth. An exponential function is written in the form, <span style=\"font-style:normal; font-size:90%\">\\(f(x)=ab^{x}\\)<\/span>, where:<\/p>\n<div class=\"examplesentence\">\\(a\\) is the beginning value, and \\(a\\neq 0\\).<br \/>\n\\(b\\) is considered the base. It reflects the growth factor, \\(b>0\\) and \\(b\\neq 1\\).<br \/>\nAnd \\(x\\) is the exponent on the base.<\/div>\n<p>\n&nbsp;<br \/>\nBecause the growth factor, <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span>, is being raised to a power that is a variable, <span style=\"font-style:normal; font-size:90%\">\\(x\\)<\/span>, the rate of change in an exponential function is not constant.<\/p>\n<p>The graph here shows the difference between the steady increase of a linear function and the extreme, rapid increase of a function showing exponential growth over time:<\/p>\n<p>Now that we\u2019ve seen the shape of a function that shows exponential growth, let\u2019s look at a few ways this relates to population growth.<\/p>\n<h2><span id=\"Population_Growth\" class=\"m-toc-anchor\"><\/span>Population Growth<\/h2>\n<p>\nPopulation growth is an interesting concept from many different perspectives.  For example, biologists can study the growth patterns of living organisms based on reproductive facts and environmental needs of the species. Another application would be global health professionals and conservationists who investigate the exponential growth that our world population has experienced.<\/p>\n<p> <img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-87643\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Population-Growth-1.png\" alt=\"graph of world population growth since 1700\" width=\"777\" height=\"437\" style=\"box-shadow: 1.5px 1.5px 3px grey;\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Population-Growth-1.png 1917w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Population-Growth-1-300x169.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Population-Growth-1-1024x576.png 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Population-Growth-1-768x432.png 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Population-Growth-1-1536x865.png 1536w\" sizes=\"auto, (max-width: 777px) 100vw, 777px\" \/><br \/>\nAs you can see, our planet has sustained an exponential increase of approximately 6.6 billion people since 1800, with most of that growth occurring during the 20th century!<\/p>\n<p>When dealing with \u201cnatural\u201d population growth models, you will notice that the notation is slightly different from a standard math exponential function.<\/p>\n<p>A population model requires the use of the irrational number <span style=\"font-style:normal; font-size:90%\">\\(e\\)<\/span>, which has an approximate value of 2.71828, as the base.  This base is then raised to a power that reflects the growth rate, <span style=\"font-style:normal; font-size:90%\">\\(r\\)<\/span>, and a unit of time, <span style=\"font-style:normal; font-size:90%\">\\(t\\)<\/span>. This is the general form of an exponential population growth model.<\/p>\n<p>It is also important to note that this type of growth model is only effective to use for short cycles of time, as true populations are typically limited by external factors that impact growth such as food supply and mortality. Logistic population growth models are more appropriate to use when there are known limitations on growth capacity.<\/p>\n<p>Let\u2019s apply the exponential population growth formula with a few practice problems:<\/p>\n<h2><span id=\"Population_Growth_Examples\" class=\"m-toc-anchor\"><\/span>Population Growth Examples<\/h2>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nLet\u2019s assume that we\u2019re inspecting a certain bacteria that grows exponentially.  If there were 100 bacteria present in a lab sample at 1pm and the growth rate is known to be 2.5% per hour, calculate the number of bacteria that was observed at 4pm.<\/p>\n<p>Our first step is to identify the components of the formula. Here\u2019s what we know:<\/p>\n<div class=\"examplesentence\">\\(P=P_{0}e^{rt}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe initial population of the bacteria is 100. <\/p>\n<div class=\"examplesentence\">\\(P=100e^{rt}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe rate of growth is 2.5% per hour, which converts to a decimal value of 0.025. <\/p>\n<div class=\"examplesentence\">\\(P=100e^{0.025t}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe amount of time that elapsed in the lab is 3 hours. <\/p>\n<div class=\"examplesentence\">\\( P=100e^{(0.025\\cdot 3)}\\)<\/div>\n<p>\n&nbsp;<br \/>\nFor <span style=\"font-style:normal; font-size:90%\">\\(e\\)<\/span>, you can use the approximate value of 2.71828 if you do not have a scientific calculator.<\/p>\n<p>We need to use this information to determine the final population, <span style=\"font-style:normal; font-size:90%\">\\(P\\)<\/span>.<\/p>\n<p>Using a calculator to compute the final population, <span style=\"font-style:normal; font-size:90%\">\\( P=100e^{(0.025\\cdot 3)}\\)<\/span>, we get that<\/p>\n<div class=\"examplesentence\">\\(P= 107.79 \\text{ bacteria}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn words, the bacteria in the sample grew from 100 to approximately 108 in 3 hours.<\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nOkay, let\u2019s look at another example. This time, I want you to try to work it out on your own before I tell you the answer.<\/p>\n<p>Our world population totaled 7,631,091,040 at the end of 2018. This reflects a 1.1%  increase over the prior year.  To practice using the population growth formula, use this information to estimate the world population in the year 2019.<\/p>\n<p>Pause the video if you want to try it out on your own.<\/p>\n<p>So, let\u2019s look at the information we know and plug it all into our equation.<\/p>\n<div class=\"examplesentence\">\\(P=P_{0}e^{rt}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo we have our initial world population<\/p>\n<div class=\"examplesentence\">\\(P=7,631,091,040e^{rt}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe rate of growth is 1.1%, which converts to a decimal value of 0.011.<\/p>\n<div class=\"examplesentence\">\\(P=7,631,091,040e^{0.011t}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe number of years from 2018 to 2019 is 1.<\/p>\n<div class=\"examplesentence\">\\(P=7,631,091.040e^{(0.011\\cdot 1)}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAn approximate value of <span style=\"font-style:normal; font-size:90%\">\\(e=2.71828\\)<\/span> can be used.<\/p>\n<p>Using this information and a calculator, it is estimated that the world population was approximately 7.716 billion at the end of 2019.<\/p>\n<p>The actual population at the end of 2019 was about 7.714 billion people, which reflects roughly 2 million fewer people from what your estimate should have been!<\/p>\n<h3><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h3>\n<p>\nSo, we\u2019ve looked at a couple of examples that require you to find the final population. But what if you were already given the final population and asked to find the initial population?<\/p>\n<p>Let\u2019s look at an example:<\/p>\n<p>As of 2020, the population of Africa was approximately 1.341 billion. The annual growth rate is calculated at 2.49%. Use the population growth formula to determine the African population at this time in 2019.<\/p>\n<div class=\"examplesentence\">\\(P=P_{0}e^{rt}\\)<\/div>\n<p>\n&nbsp;<br \/>\nHere\u2019s what we know. The final population of Africa is approximately 1.341 billion.<\/p>\n<div class=\"examplesentence\">\\(1,340,598,000=P_{0}e^{rt}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe rate of growth is 2.49%, which converts to a decimal value of 0.0249.<\/p>\n<div class=\"examplesentence\">\\(1,340,598,000=P_{0}e^{0.0249t}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe number of years from 2019 to 2020 is 1.<\/p>\n<div class=\"examplesentence\">\\(1,340,598,000=P_{0}e^{(0.249\\cdot 1)}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAn approximate value of <span style=\"font-style:normal; font-size:90%\">\\(e=2.71828\\)<\/span> can be used.<\/p>\n<p>We are asked to determine the initial population, <span style=\"font-style:normal; font-size:90%\">\\(P_{0}\\)<\/span>. First, use a calculator to determine the constant value that is multiplied by <span style=\"font-style:normal; font-size:90%\">\\(P_{0}\\)<\/span>, <span style=\"font-style:normal; font-size:90%\">\\(e^{0.0249}\\)<\/span>. So now we know that:<\/p>\n<div class=\"examplesentence\">\\(1,340,598,000=P_{0}\\cdot 1.0252\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, solve for <span style=\"font-style:normal; font-size:90%\">\\(P_{0}\\)<\/span> by dividing both sides of the equation by 1.0252.<\/p>\n<div class=\"examplesentence\">\\(\\frac{1,340,598,000}{1.0252}=P_{0}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn this case, the \u201coriginal\u201d population represents the year 2019. Our calculation shows that value to be about 1.308 billion people.<\/p>\n<h3><span id=\"Example_4\" class=\"m-toc-anchor\"><\/span>Example #4<\/h3>\n<p>\nHere is another example:<\/p>\n<p>The United Nations estimated that the United States population will be 331,002,651 by mid-year 2020, and the expected growth rate is estimated at 0.59%.  Use this information to determine the US population at mid-year 2022. (Assume that the growth rate is the same between 2020-2021 and 2021-2022.)<\/p>\n<p>Pause the video now if you\u2019d like to try this one on your own.<\/p>\n<p>Ok, so here\u2019s what we know:<\/p>\n<div class=\"examplesentence\">\\(P=P_{0}e^{rt}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe US population is estimated to be about 331 million at mid-year, 2020.<\/p>\n<div class=\"examplesentence\">\\(P=331,002,651e^{rt}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe rate of growth is 0.59%, which converts to a decimal value of 0.0059.<\/p>\n<div class=\"examplesentence\">\\(P=331,002,651e^{0.0059t}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe number of years from 2020 to 2022 is 2.<\/p>\n<div class=\"examplesentence\">\\(P=331,002,651e^{(0.0059\\cdot 2)}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAn approximate value of <span style=\"font-style:normal; font-size:90%\">\\(e=2.71828\\)<\/span> can be used.<\/p>\n<p>We are asked to determine population, <span style=\"font-style:normal; font-size:90%\">\\(P\\)<\/span>, in mid-year 2022. Since we\u2019ve substituted the known values in our equation, we can use our calculator to find our final answer, which is approximately 335 million.<\/p>\n<p>We can also use this formula to solve for <span style=\"font-style:normal; font-size:90%\">\\(t\\)<\/span>, the number of years it takes for the original population to grow to the final population. When we solve for any variable, we use algebraic steps to isolate it in order to determine its value. However, in this case, the variable for time is in the exponent of the formula. How can we get to it in order to find its value?<\/p>\n<p>This can be done by using the <strong>inverse function<\/strong> that \u201cundoes\u201d the work of the exponential function. It is called the natural log, and is notated as <span style=\"font-style:normal; font-size:90%\">\\(ln\\)<\/span>. Consider that the exponential function, <span style=\"font-style:normal; font-size:90%\">\\(f(x)=e^{x}\\)<\/span>, takes in the input, <span style=\"font-style:normal; font-size:90%\">\\(x\\)<\/span>, and results in the output, <span style=\"font-style:normal; font-size:90%\">\\(e^{x}\\)<\/span>.  Applying the natural log, <span style=\"font-style:normal; font-size:90%\">\\(ln\\)<\/span>, to this output results in the original input, <span style=\"font-style:normal; font-size:90%\">\\(x\\)<\/span>. This identity is shown as <span style=\"font-style:normal; font-size:90%\">\\(ln(e^{x})=x\\)<\/span>.<\/p>\n<p>If we want to solve for the variable <span style=\"font-style:normal; font-size:90%\">\\(t\\)<\/span> in a population formula, it is as simple as hitting the natural log key on your calculator and continuing with a few additional algebraic steps.<\/p>\n<h3><span id=\"Example_5\" class=\"m-toc-anchor\"><\/span>Example #5<\/h3>\n<p>\nLet\u2019s try one last example to practice using the natural log.<\/p>\n<p>The current population of Iceland is approximately 341,400 people. The annual growth rate of the population is approximately 0.65%. Assume that this rate of growth will continue in the future. How many years will it take for the population to reach 400,000?<\/p>\n<p>Well, we could simply multiply the current population by the annual growth rate for as many times as it takes to get to 400,000, but solving the exponential equation for <span style=\"font-style:normal; font-size:90%\">\\(t\\)<\/span> is a lot more efficient and precise.. <\/p>\n<p>As before, let\u2019s list what we know:<\/p>\n<div class=\"examplesentence\">\\(P=P_{0}e^{rt}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIceland\u2019s population is  estimated to be 341,400.<\/p>\n<div class=\"examplesentence\">\\(P=341,400e^{rt}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe rate of growth is 0.65%, which converts to a decimal value of 0.0065.<\/p>\n<div class=\"examplesentence\">\\(P=341,400e^{0.0065t}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe population that we are targeting is 400,000.<\/p>\n<div class=\"examplesentence\">\\(400,000=341,400e^{0.0065t}\\)<\/div>\n<p>\n&nbsp;<br \/>\nAn approximate value of <span style=\"font-style:normal; font-size:90%\">\\(e=2.71828 \\)<\/span> can be used.<\/p>\n<p>As you can see, the unknown in this equation is time. We have to use algebraic steps and the natural log function to solve for the amount of time it will take for the population to be equal to 400,000.<\/p>\n<p>Step 1: Divide both sides of the equation by <span style=\"font-style:normal; font-size:90%\">\\(P_{0}\\)<\/span>, which will leave an exponential expression on the right.<\/p>\n<div class=\"examplesentence\">\\(\\frac{400,000}{341,400}=\\frac{341,400\\cdot e^{0.0065\\cdot t}}{341,400}\\)<\/p>\n\\(1.1716=e^{0.0065\\cdot t}\\)\n<\/div>\n<p>\n&nbsp;<br \/>\nNote that we will be rounding our answers from the calculator to four decimal places to illustrate this example, but leaving the values in the calculator unrounded gives a more accurate answer.<\/p>\n<p>We are now ready to apply the natural log to the exponential expression on the right side of the equation so we can do the work to solve for <span style=\"font-style:normal; font-size:90%\">\\(t\\)<\/span>. However, as we know from solving equations, if we perform an operation on one side of the equal sign, you must do the same on the other side to maintain balance.<\/p>\n<p>Here is what taking the natural log of both sides looks like. Step 2:<\/p>\n<div class=\"examplesentence\">\\(ln(1.1716)=ln(e^{0.0065\\cdot t})\\)<\/div>\n<p>\n&nbsp;<br \/>\nAs mentioned, there is a key on your calculator to determine the value of <span style=\"font-style:normal; font-size:90%\">\\(ln(1.1716)\\)<\/span>. Simply hit the <span style=\"font-style:normal; font-size:90%\">\\(ln\\)<\/span> key, and a parenthesis will appear. Enter 1.1716, and close the parentheses. Next, hit enter for a value of 0.1584. On the right side, the natural log \u201cundoes\u201d the exponential function, leaving the expression in the exponent, <span style=\"font-style:normal; font-size:90%\">\\(0.0065\\cdot t\\)<\/span>.<\/p>\n<div class=\"examplesentence\">\\(0.1584=0.0065\\cdot t\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, the <span style=\"font-style:normal; font-size:90%\">\\(t\\)<\/span>-variable can be solved for by dividing both sides of the equation by 0.0065, as follows. Step 3:<\/p>\n<div class=\"examplesentence\">\\(\\frac{0.1584}{0.0065}=\\frac{0.0065\\cdot t}{0.0065}\\)<br \/>\n&nbsp;<br \/>\n\\(24.4=t\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe now know that it will take about 24 and a half years for the population of Iceland to reach 400,000 people.<\/p>\n<p>That\u2019s all for this review! Now that we\u2019ve worked through some examples, you should better understand the features of exponential functions and how they are used to model population growth.<\/p>\n<p>Thanks for watching, and happy studying!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Population_Growth_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Population Growth Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nCity A had a population of 160,000 in 2010. The city is increasing at a steady exponential rate of 3.6%. What was the population of the city in 2022? Round your answer to the nearest person.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">346,454<\/div><div class=\"PQ\"  id=\"PQ-1-2\">446,454<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-3\">246,454<\/div><div class=\"PQ\"  id=\"PQ-1-4\">146,454<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Start by identifying the components of the population growth formula \\(P=P_oe^{rt}\\), where \\(P\\) represents the final population, \\(P_o\\) represents the initial population, \\(e\\) is the base, which is approximated as 2.71828, \\(r\\) represents the rate of change (as a decimal), and \\(t\\) represents time. <\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\(P_o=160{,}000\\)<br \/>\n\\(e=2.71828\\)<br \/>\n\\(r=3.6\\%=0.036\\)<br \/>\n\\(t=12\\)<\/p>\n<p>Substitute this information into the equation. <\/p>\n<p style=\"text-align: center; line-height: 40px\">\\(P=(160{,}000)(2.71828)^{(0.036)(12)}\\)\\(\\:=246{,}454\\)<\/p>\n<p>In 2022, the population of the city was 246,454.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nA biologist counted 100 black bears in a state park in 2019. The population of the bears grows at a steady exponential rate of 5%. If the population growth continues at the same rate, how many black bears will be in the park in 2043? Round your answer to the nearest bear.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-2-1\">332<\/div><div class=\"PQ\"  id=\"PQ-2-2\">292<\/div><div class=\"PQ\"  id=\"PQ-2-3\">322<\/div><div class=\"PQ\"  id=\"PQ-2-4\">272<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Start by identifying the components of the population growth formula \\(P=P_oe^{rt}\\), where \\(P\\) represents the final population, \\(P_o\\) represents the initial population, \\(e\\) is the base, which is approximated as 2.71828, \\(r\\) represents the rate of change (as a decimal), and \\(t\\) represents time.<\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\(P_o=100\\)<br \/>\n\\(e=2.71828\\)<br \/>\n\\(r=5\\%=0.05\\)<br \/>\n\\(t=24\\)<\/p>\n<p>Substitute this information into the original equation. <\/p>\n<p style=\"text-align: center;\">\\(P=(100)(2.71828)^{(0.05)(24)}=332\\)<\/p>\n<p>There will be 332 black bears in the park in 2043. <\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nA school\u2019s population is growing at an exponential rate of 6.7%. The school is considering expanding and building new classrooms due to overcrowding. The school has classrooms for 2,250 students, and right now there are 1,600 students. If the growth rate stays the same, how long does the school have until they run out of space? <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">Approximately 4 years<\/div><div class=\"PQ\"  id=\"PQ-3-2\">Approximately 6 years<\/div><div class=\"PQ\"  id=\"PQ-3-3\">Approximately 3 years<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-4\">Approximately 5 years<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Start by identifying the components of the population growth formula \\(P=P_oe^{rt}\\), where \\(P\\) represents the final population, \\(P_o\\) represents the initial population, \\(e\\) is the base, which is approximated as 2.71828, \\(r\\) represents the rate of change (as a decimal), and \\(t\\) represents time. <\/p>\n<p>Substitute all values into the population growth formula.<\/p>\n<p style=\"text-align: center\">\\(2{,}250=1{,}600e^{(0.067)t}\\)<\/p>\n<p>We are trying to isolate the variable \\(t\\), so start by dividing both sides of the equation by 1,600. <\/p>\n<p style=\"text-align: center; line-height: 60px;\">\\(\\dfrac{2{,}250}{1{,}600}=\\dfrac{1{,}600e^{(0.067)t}}{1,600}\\)<br \/>\n\\(1.40625=e^{(0.067)t}\\)<\/p>\n<p>Now use the natural log function to solve for \\(t\\). Take the natural log of both sides of the equation.<\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\(\\ln(1.40625)=\\ln(e^{0.067t})\\)<br \/>\n\\(0.3409=0.067t\\)<\/p>\n<p>Now solve for t by dividing both sides by 0.067.<\/p>\n<p style=\"text-align: center; line-height: 55px;\">\\(\\dfrac{0.3409}{0.067}=\\dfrac{0.067t}{0.067}\\)<br \/>\n\\(t=5.0888\\)<\/p>\n<p>Therefore, the school will run out of space in approximately 5 years.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nA biologist is studying a cluster of bacteria cells that have a steady growth rate of 60% each day. There are 8,000 bacteria cells in the cluster now. How many cells will there be in the same cluster 7 days from now? Round your answer to the nearest cell.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">456,491<\/div><div class=\"PQ\"  id=\"PQ-4-2\">673,491<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-3\">533,491<\/div><div class=\"PQ\"  id=\"PQ-4-4\">533,811<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Start by identifying the components of the population growth formula \\(P=P_oe^{rt}\\), where \\(P\\) represents the final population, \\(P_o\\) represents the initial population, \\(e\\) is the base, which is approximated as 2.71828, \\(r\\) represents the rate of change (as a decimal), and \\(t\\) represents time. <\/p>\n<p>In this scenario: <\/p>\n<p style=\"text-align: center; line-height: 35px;\">\\(P_o=8{,}000\\)<br \/>\n\\(e=2.71828\\)<br \/>\n\\(r=60\\%=0.6\\)<br \/>\n\\(t=7\\)<\/p>\n<p>Substitute this information into the original equation.<\/p>\n<p style=\"text-align: center;\">\\(P=8{,}000e^{(0.6)(7)}=533{,}490.648\\)<\/p>\n<p>Therefore, after 7 days there will be approximately 533,491 bacterial cells.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nA city currently has a population of 500,000 residents. The city\u2019s population is growing exponentially at a rate of 1.5% each year. Assuming the growth rate stays the same, what will the population of the city be in 3 years? Round your answer to the nearest person.<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">823,314<\/div><div class=\"PQ\"  id=\"PQ-5-2\">593,814<\/div><div class=\"PQ\"  id=\"PQ-5-3\">623,314<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-4\">523,014<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>Start by identifying the components of the population growth formula \\(P=P_oe^{rt}\\), where \\(P\\) represents the final population, \\(P_o\\) represents the initial population, \\(e\\) is the base, which is approximated as 2.71828, \\(r\\) represents the rate of change (as a decimal), and \\(t\\) represents time.<\/p>\n<p style=\"text-align: center; line-height: 35px;\">\n\\(P_o=500{,}000\\)<br \/>\n\\(e=2.71828\\)<br \/>\n\\(r=1.5\\%=0.015\\)<br \/>\n\\(t=3\\)<\/p>\n<p>Substitute this information into the original equation.<\/p>\n<p style=\"text-align: center;\">\\(P=500{,}000e^{(0.015)(3)}=523{,}014\\)<\/p>\n<p>After 3 years, the population of the city will be approximately 523,014.<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/algebra-i\/\">Return to Algebra I Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Algebra I Videos<\/p>\n","protected":false},"author":1,"featured_media":100711,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-86389","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-algebra-i-videos","7":"page_category-functions-videos","8":"page_type-video","9":"content_type-practice-questions","10":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/86389","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=86389"}],"version-history":[{"count":5,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/86389\/revisions"}],"predecessor-version":[{"id":280601,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/86389\/revisions\/280601"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/100711"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=86389"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}