{"id":362,"date":"2016-10-11T20:20:18","date_gmt":"2016-10-11T20:20:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/math4libarts\/?post_type=chapter&#038;p=362"},"modified":"2021-02-05T23:55:57","modified_gmt":"2021-02-05T23:55:57","slug":"exponential-geometric-growth","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/chapter\/exponential-geometric-growth\/","title":{"raw":"Exponential (Geometric) Growth","rendered":"Exponential (Geometric) Growth"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine whether data or a scenario describe linear or geometric growth<\/li>\r\n \t<li>Identify growth rates, initial values, or point values expressed verbally, graphically, or numerically, and translate them into a format usable in calculation<\/li>\r\n \t<li>Calculate recursive and explicit equations for exponential growth and use those equations to make predictions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Population Growth<\/h2>\r\nSuppose that every year, only 10% of the fish in a lake have surviving offspring. If there were 100 fish in the lake last year, there would now be 110 fish. If there were 1000 fish in the lake last year, there would now be 1100 fish. Absent any inhibiting factors, populations of people and animals tend to grow by a percent of the existing population each year.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21172235\/calf-1869984_1280.jpg\"><img class=\"aligncenter size-large wp-image-906\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21172235\/calf-1869984_1280-1024x714.jpg\" alt=\"Two orange koi at the top of a pond, mouths open. Others are swimming beneath them.\" width=\"1024\" height=\"714\" \/><\/a>\r\nSuppose our lake began with 1000 fish, and 10% of the fish have surviving offspring each year. Since we start with 1000 fish, <em>P\u00ad<sub>0<\/sub><\/em> = 1000. How do we calculate <em>P\u00ad<sub>1<\/sub><\/em>? The new population will be the old population, plus an additional 10%. Symbolically:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P<sub>\u00ad0<\/sub><\/em><\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>representing percent as a decimal<\/h3>\r\nThe above statement can be read as \"the number of fish in the pond after the first year is equivalent to the initial number of fish plus 10%,\" where 10% has been expressed in decimal form at [latex]0.10[\/latex].\r\n\r\nTo rewrite a percent as a decimal, drop the % symbol and move the decimal point two places to the left.\r\n\r\n<\/div>\r\nNotice this could be condensed to a shorter form by factoring:\r\n<p style=\"text-align: center;\"><em>P<sub>\u00ad1<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P\u00ad<sub>0<\/sub><\/em> = 1<em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P<sub>\u00ad0 <\/sub><\/em>= (1+ 0.10)<em>P\u00ad<sub>0 <\/sub><\/em>= 1.10<em>P\u00ad<sub>0<\/sub><\/em><\/p>\r\nWhile 10% is the <strong>growth rate<\/strong>, 1.10 is the <strong>growth multiplier<\/strong>. Notice that 1.10 can be thought of as \u201cthe original 100% plus an additional 10%.\u201d\r\n\r\nFor our fish population,\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub><\/em> = 1.10(1000) = 1100<\/p>\r\nWe could then calculate the population in later years:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub><\/em> = 1.10<em>P\u00ad<sub>1<\/sub><\/em> = 1.10(1100) = 1210<\/p>\r\n<p style=\"text-align: center;\"><em>P<sub>\u00ad3<\/sub><\/em> = 1.10<em>P<sub>2<\/sub><\/em> = 1.10(1210) = 1331<\/p>\r\nNotice that in the first year, the population grew by 100 fish; in the second year, the population grew by 110 fish; and in the third year the population grew by 121 fish.\r\n\r\nWhile there is a constant <em>percentage<\/em> growth, the actual increase in number of fish is increasing each year.\r\n\r\nGraphing these values we see that this growth doesn\u2019t quite appear linear.\r\n\r\n<img class=\"aligncenter wp-image-363\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201507\/yearsfromnow1.png\" alt=\"Line graph. Measured vertically: Fish, in increments of 200, from 800 to 1800. Measured horizontally: Years from now, measured in units of 1, from 0 to 5. Year 0 is at 1000; Year 1 is at roughly 1100; and subsequent years move in an increasing rate so that the overall line is curved to the upper right.\" width=\"350\" height=\"283\" \/>\r\n\r\nA walk-through of this fish scenario can be viewed here:\r\n\r\nhttps:\/\/youtu.be\/3BiU7Ihxvxg\r\n\r\nTo get a better picture of how this percentage-based growth affects things, we need an explicit form, so we can quickly calculate values further out in the future.\r\n\r\nLike we did for the linear model, we will start building from the recursive equation:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub><\/em> = 1.10<em>P\u00ad<sub>0<\/sub><\/em> = 1.10(1000)<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub><\/em> = 1.10<em>P\u00ad<sub>1<\/sub><\/em> = 1.10(1.10(1000)) = 1.102(1000)<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>3<\/sub><\/em> = 1.10<em>P\u00ad<sub>2<\/sub><\/em> = 1.10(1.102(1000)) = 1.103(1000)<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>4<\/sub><\/em> = 1.10<em>P\u00ad<sub>3<\/sub><\/em> = 1.10(1.103(1000)) = 1.104(1000)<\/p>\r\nObserving a pattern, we can generalize the explicit form to be:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = 1.10<sub><em>n<\/em><\/sub>(1000), or equivalently, <em>P\u00ad<sub>n<\/sub><\/em> = 1000(1.10<sub><em>n<\/em><\/sub>)<\/p>\r\nFrom this, we can quickly calculate the number of fish in 10, 20, or 30 years:\r\n\r\n<img class=\"aligncenter wp-image-364\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201611\/yearsfromnow2.png\" alt=\"Line graph. Measured vertically: Fish, in increments of 3000, from 0 to 18,000. Measured horizontally: Years from now, measured in units of 5, from 0 to 30. Year 0 is at 1000, with a tight cluster of dots in the lower left quadrant. The exponential increase becomes more dramatic as time advances, so that year 30 is at 18,000 fish with more spacing between dots.\" width=\"350\" height=\"281\" \/>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = 1.10<sub><em>10<\/em><\/sub>(1000) = 2594<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>20<\/sub><\/em> = 1.10<sub><em>20<\/em><\/sub>(1000) = 6727<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>30<\/sub><\/em> = 1.10<sub><em>30<\/em><\/sub>(1000) = 17449<\/p>\r\nAdding these values to our graph reveals a shape that is definitely not linear. If our fish population had been growing linearly, by 100 fish each year, the population would have only reached 4000 in 30 years, compared to almost 18,000 with this percent-based growth, called <strong>exponential growth.<\/strong>\r\n\r\nA video demonstrating the explicit model of this fish story can be viewed here:\r\n\r\nhttps:\/\/youtu.be\/tg2ysaZ8agY\r\n\r\nIn exponential growth, the population grows proportional to the size of the population, so as the population gets larger, the same percent growth will yield a larger numeric growth.\r\n\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>Exponential Growth<\/h3>\r\nIf a quantity starts at size <em>P\u00ad<sub>0<\/sub><\/em> and grows by <em>R%<\/em> (written as a decimal, <em>r<\/em>) every time period, then the quantity after <em>n<\/em> time periods can be determined using either of these relations:\r\n<h4>Recursive form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = (1+<em>r<\/em>) <em>P\u00ad<sub>n-1<\/sub><\/em><\/p>\r\n\r\n<h4>Explicit form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = (1+<em>r<\/em>)<sup><sub><em>n<\/em><\/sub><\/sup> <em>P\u00ad<sub>0<\/sub> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em>or equivalently, <em>P<sub>\u00adn<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> (1+<em>r<\/em>)<sup><sub><em>n<\/em><\/sub><\/sup><\/p>\r\nWe call <em>r<\/em> the <strong>growth rate<\/strong>.\r\n\r\nThe term (1+<em>r<\/em>) is called the <strong>growth multiplier<\/strong>, or common ratio.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nBetween 2007 and 2008, Olympia, WA grew almost 3% to a population of 245 thousand people. If this growth rate was to continue, what would the population of Olympia be in 2014?\r\n[reveal-answer q=\"54756\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"54756\"]\r\n\r\nAs we did before, we first need to define what year will correspond to <em>n<\/em> = 0. Since we know the population in 2008, it would make sense to have 2008 correspond to <em>n <\/em>= 0, so <em>P\u00ad<sub>0<\/sub><\/em> = 245,000.\u00a0\u00a0 The year 2014 would then be <em>n<\/em> = 6.\r\n\r\nWe know the growth rate is 3%, giving <em>r <\/em>= 0.03.\r\n\r\nUsing the explicit form:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>6<\/sub><\/em> = (1+0.03)<sup>6<\/sup> (245,000) = 1.19405(245,000) = 292,542.25<\/p>\r\nThe model predicts that in 2014, Olympia would have a population of about 293 thousand people.\r\n\r\n[\/hidden-answer]\r\n\r\nThe following video explains this example in detail.\r\n\r\nhttps:\/\/youtu.be\/CDI4xS65rxY\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>Evaluating exponents on the calculator<\/h3>\r\nTo evaluate expressions like (1.03)<sup>6<\/sup>, it will be easier to use a calculator than multiply 1.03 by itself six times. Most scientific calculators have a button for exponents.\u00a0 It is typically either labeled like:\r\n<p style=\"text-align: center;\">^ ,\u00a0\u00a0 y<sup>x<\/sup> ,\u00a0\u00a0 or x<sup>y<\/sup> .<\/p>\r\nTo evaluate 1.03<sup>6<\/sup>\u00a0we\u2019d type 1.03 ^ 6, or 1.03 y<sup>x<\/sup> 6.\u00a0 Try it out - you should get an answer around 1.1940523.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIndia is the second most populous country in the world, with a population in 2008 of about 1.14 billion people. The population is growing by about 1.34% each year. If this trend continues, what will India\u2019s population grow to by 2020?\r\n[reveal-answer q=\"265475\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"265475\"]\r\n<div>Using n = 0 corresponding with 2008, [latex]P_12= (1+0.0134)12(1.14)[\/latex] = about 1.337 billion people in 2020<\/div>\r\n[\/hidden-answer]\r\n\r\n[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=6673&amp;theme=oea&amp;iframe_resize_id=mom3[\/embed]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nA friend is using the equation <em>P\u00ad<sub>n<\/sub><\/em> = 4600(1.072)<sup><em>n <\/em><\/sup>to predict the annual tuition at a local college. She says the formula is based on years after 2010. What does this equation tell us?\r\n\r\n[reveal-answer q=\"224261\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"224261\"]\r\n\r\nIn the equation, <em>P<sub>\u00ad0<\/sub><\/em> = 4600, which is the starting value of the tuition when <em>n<\/em> = 0. This tells us that the tuition in 2010 was $4,600.\r\n\r\nThe growth multiplier is 1.072, so the growth rate is 0.072, or 7.2%. This tells us that the tuition is expected to grow by 7.2% each year.\r\n\r\nPutting this together, we could say that the tuition in 2010 was $4,600, and is expected to grow by 7.2% each year.\r\n\r\n[\/hidden-answer]\r\n\r\nView the following to see this example worked out.\r\n\r\nhttps:\/\/youtu.be\/T8Yz94De5UM\r\n\r\n<hr \/>\r\n\r\nIn 1990, residential energy use in the US was responsible for 962 million metric tons of carbon dioxide emissions. By the year 2000, that number had risen to 1182 million metric tons[footnote]<a href=\"http:\/\/www.eia.doe.gov\/oiaf\/1605\/ggrpt\/carbon.html\" target=\"_blank\" rel=\"noopener\">http:\/\/www.eia.doe.gov\/oiaf\/1605\/ggrpt\/carbon.html<\/a>[\/footnote]. If the emissions grow exponentially and continue at the same rate, what will the emissions grow to by 2050?\r\n[reveal-answer q=\"755963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"755963\"]\r\n\r\nSimilar to before, we will correspond <em>n<\/em> = 0 with 1990, as that is the year for the first piece of data we have. That will make <em>P\u00ad<sub>0<\/sub><\/em> = 962 (million metric tons of CO<sub>2<\/sub>). In this problem, we are not given the growth rate, but instead are given that <em>P\u00ad<sub>10<\/sub><\/em> = 1182.\r\n\r\nWhen <em>n<\/em> = 10, the explicit equation looks like:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = (1+<em>r<\/em>)<sup>10<\/sup> <em>P\u00ad<sub>0<\/sub><\/em><\/p>\r\nWe know the value for <em>P\u00ad<sub>0<\/sub><\/em>, so we can put that into the equation:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = (1+<em>r<\/em>)<sup>10<\/sup> 962<\/p>\r\nWe also know that <em>P\u00ad<sub>10<\/sub><\/em> = 1182, so substituting that in, we get\r\n<p style=\"text-align: center;\">1182 = (1+<em>r<\/em>)<sup>10<\/sup> 962<\/p>\r\nWe can now solve this equation for the growth rate, <em>r<\/em>. Start by dividing by 962.\r\n<p style=\"padding-left: 120px;\">[latex]\\frac{1182}{962}={{(1+r)}^{10}}[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Take the 10th root of both sides<\/p>\r\n<p style=\"padding-left: 120px;\">[latex]\\sqrt[10]{\\frac{1182}{962}}=1+r[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Subtract 1 from both sides<\/p>\r\n<p style=\"padding-left: 120px;\">[latex]r=\\sqrt[10]{\\frac{1182}{962}}-1=0.0208[\/latex] = 2.08%<\/p>\r\nSo if the emissions are growing exponentially, they are growing by about 2.08% per year. We can now predict the emissions in 2050 by finding <em>P<sub>\u00ad60<\/sub><\/em>\r\n<p style=\"text-align: center;\"><em>P<sub>\u00ad60<\/sub><\/em> = (1+0.0208)<sup>60<\/sup> 962 = 3308.4 million metric tons of CO<sub>2<\/sub> in 2050<\/p>\r\n[\/hidden-answer]\r\n\r\nView more about this example here.\r\n\r\nhttps:\/\/youtu.be\/9Zu2uONfLkQ\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Rounding<\/h3>\r\nAs a note on rounding, notice that if we had rounded the growth rate to 2.1%, our calculation for the emissions in 2050 would have been 3347.\u00a0\u00a0 Rounding to 2% would have changed our result to 3156. A very small difference in the growth rates gets magnified greatly in exponential growth. For this reason, it is recommended to round the growth rate as little as possible.\r\n\r\nIf you need to round, <strong>keep at least three significant digits<\/strong> - numbers after any leading zeros.\u00a0\u00a0 So 0.4162 could be reasonably rounded to 0.416. A growth rate of 0.001027 could be reasonably rounded to 0.00103.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Evaluating roots on the calculator<\/h3>\r\nIn the previous example, we had to calculate the 10th root of a number. This is different than taking the basic square root, \u221a. Many scientific calculators have a button for general roots.\u00a0 It is typically labeled like:\r\n<p style=\"text-align: center;\">[latex]\\sqrt[y]{x}[\/latex]<\/p>\r\nTo evaluate the 3rd root of 8, for example, we\u2019d either type 3 [latex]\\sqrt[x]{{}}[\/latex] 8, or 8 [latex]\\sqrt[x]{{}}[\/latex] 3, depending on the calculator. Try it on yours to see which to use \u2013 you should get an answer of 2.\r\n\r\nIf your calculator does not have a general root button, all is not lost. You can instead use the property of exponents which states that:\r\n<p style=\"text-align: center;\"><span style=\"color: #000000;\">[latex]\\sqrt[n]{a}={a}^{\\frac{1}{2}}[\/latex]<\/span>.<\/p>\r\nSo, to compute the 3rd root of 8, you could use your calculator\u2019s exponent key to evaluate 8<sup>1\/3<\/sup>. To do this, type:\r\n<p style=\"text-align: center;\">8 y<sup>x<\/sup> ( 1 \u00f7 3 )<\/p>\r\nThe parentheses tell the calculator to divide 1\/3 before doing the exponent.\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>a note about parentheses<\/h3>\r\nDon't overlook the use of parentheses in the above example. They are necessary to force the desired order of operations when using your calculator.\r\n\r\nIf you were to leave them out and type [latex]8y^x1\/3[\/latex] or 8^1\/3, it would tell the calculator to first raise 8 to the power of 1, then divide the result by 3.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe number of users on a social networking site was 45 thousand in February when they officially went public, and grew to 60 thousand by October. If the site is growing exponentially, and growth continues at the same rate, how many users should they expect two years after they went public?\r\n\r\n[reveal-answer q=\"774576\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"774576\"]\r\n\r\nHere we will measure n in months rather than years, with n = 0 corresponding to the February when they went public. This gives [latex]P_0= 45[\/latex] thousand. October is 8 months later, so [latex]P_8= 60[\/latex].\r\n[latex]P_8=(1+r)^{8}P_0[\/latex]\r\n[latex]60=(1+r)^{8}45[\/latex]\r\n[latex]\\frac{60}{45}=(1+r)^8[\/latex]\r\n[latex]\\sqrt[8]{\\frac{60}{45}}=1+r[\/latex]\r\n[latex]r=\\sqrt[8]{\\frac{60}{45}}-1=0.0366\\text{ or }3.66%[\/latex]\r\nThe general explicit equation is [latex]P_n =(1.0366)^{n}45[\/latex]. Predicting 24 months after they went public gives [latex]P_{24}=(1.0366)^{24}45=106.63[\/latex] thousand users.\r\n[\/hidden-answer]\r\n\r\n[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=6598&amp;theme=oea&amp;iframe_resize_id=mom4[\/embed]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLooking back at the last example, for the sake of comparison, what would the carbon emissions be in 2050 if emissions grow linearly at the same rate?\r\n[reveal-answer q=\"355767\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"355767\"]\r\n\r\nAgain we will get <em>n<\/em> = 0 correspond with 1990, giving <em>P<sub>\u00ad0<\/sub><\/em> = 962. To find <em>d<\/em>, we could take the same approach as earlier, noting that the emissions increased by 220 million metric tons in 10 years, giving a common difference of 22 million metric tons each year.\r\n\r\nAlternatively, we could use an approach similar to that which we used to find the exponential equation. When <em>n<\/em> = 10, the explicit linear equation looks like:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = <em>P<sub>\u00ad0<\/sub><\/em> + 10<em>d<\/em><em>\u00a0<\/em><\/p>\r\nWe know the value for <em>P\u00ad0<\/em>, so we can put that into the equation:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = 962 + 10<em>d<\/em><\/p>\r\nSince we know that <em>P\u00ad<sub>10<\/sub><\/em> = 1182, substituting that in we get\r\n<p style=\"text-align: center;\">1182 = 962 + 10<em>d<\/em><\/p>\r\nWe can now solve this equation for the common difference, <em>d<\/em>.\r\n<p style=\"text-align: center;\">1182 \u2013 962 = 10<em>d<\/em><\/p>\r\n<p style=\"text-align: center;\">220 = 10<em>d<\/em><\/p>\r\n<p style=\"text-align: center;\"><em>d<\/em> = 22<\/p>\r\nThis tells us that if the emissions are changing linearly, they are growing by 22 million metric tons each year. Predicting the emissions in 2050,\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>60<\/sub><\/em> = 962 + 22(60) = 2282 million metric tons.<\/p>\r\nYou will notice that this number is substantially smaller than the prediction from the exponential growth model. Calculating and plotting more values helps illustrate the differences.\r\n\r\n<img class=\"aligncenter wp-image-365\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201942\/yearsafter1990.png\" alt=\"Line graph. Vertical measures Emissions (Millions Metric Tons) in increments of 500, from 0 to 3500. Horizontal measures Years after 1990, in increments of 10, from 0 to 60. A blue line shows a linear growth from 1000 in year 0 to over 2000 in year 60. A pink line shows an exponential growth from 1000 in year 0 to 3500 in year 60.\" width=\"350\" height=\"283\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\nA demonstration of this example can be seen in the following video.\r\n\r\nhttps:\/\/youtu.be\/yiuZoiRMtYM\r\n\r\n<\/div>\r\nSo how do we know which growth model to use when working with data? There are two approaches which should be used together whenever possible:\r\n<ol>\r\n \t<li>Find more than two pieces of data. Plot the values, and look for a trend. Does the data appear to be changing like a line, or do the values appear to be curving upwards?<\/li>\r\n \t<li>Consider the factors contributing to the data. Are they things you would expect to change linearly or exponentially? For example, in the case of carbon emissions, we could expect that, absent other factors, they would be tied closely to population values, which tend to change exponentially.<\/li>\r\n<\/ol>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine whether data or a scenario describe linear or geometric growth<\/li>\n<li>Identify growth rates, initial values, or point values expressed verbally, graphically, or numerically, and translate them into a format usable in calculation<\/li>\n<li>Calculate recursive and explicit equations for exponential growth and use those equations to make predictions<\/li>\n<\/ul>\n<\/div>\n<h2>Population Growth<\/h2>\n<p>Suppose that every year, only 10% of the fish in a lake have surviving offspring. If there were 100 fish in the lake last year, there would now be 110 fish. If there were 1000 fish in the lake last year, there would now be 1100 fish. Absent any inhibiting factors, populations of people and animals tend to grow by a percent of the existing population each year.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21172235\/calf-1869984_1280.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-large wp-image-906\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21172235\/calf-1869984_1280-1024x714.jpg\" alt=\"Two orange koi at the top of a pond, mouths open. Others are swimming beneath them.\" width=\"1024\" height=\"714\" \/><\/a><br \/>\nSuppose our lake began with 1000 fish, and 10% of the fish have surviving offspring each year. Since we start with 1000 fish, <em>P\u00ad<sub>0<\/sub><\/em> = 1000. How do we calculate <em>P\u00ad<sub>1<\/sub><\/em>? The new population will be the old population, plus an additional 10%. Symbolically:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P<sub>\u00ad0<\/sub><\/em><\/p>\n<div class=\"textbox examples\">\n<h3>representing percent as a decimal<\/h3>\n<p>The above statement can be read as &#8220;the number of fish in the pond after the first year is equivalent to the initial number of fish plus 10%,&#8221; where 10% has been expressed in decimal form at [latex]0.10[\/latex].<\/p>\n<p>To rewrite a percent as a decimal, drop the % symbol and move the decimal point two places to the left.<\/p>\n<\/div>\n<p>Notice this could be condensed to a shorter form by factoring:<\/p>\n<p style=\"text-align: center;\"><em>P<sub>\u00ad1<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P\u00ad<sub>0<\/sub><\/em> = 1<em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P<sub>\u00ad0 <\/sub><\/em>= (1+ 0.10)<em>P\u00ad<sub>0 <\/sub><\/em>= 1.10<em>P\u00ad<sub>0<\/sub><\/em><\/p>\n<p>While 10% is the <strong>growth rate<\/strong>, 1.10 is the <strong>growth multiplier<\/strong>. Notice that 1.10 can be thought of as \u201cthe original 100% plus an additional 10%.\u201d<\/p>\n<p>For our fish population,<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub><\/em> = 1.10(1000) = 1100<\/p>\n<p>We could then calculate the population in later years:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub><\/em> = 1.10<em>P\u00ad<sub>1<\/sub><\/em> = 1.10(1100) = 1210<\/p>\n<p style=\"text-align: center;\"><em>P<sub>\u00ad3<\/sub><\/em> = 1.10<em>P<sub>2<\/sub><\/em> = 1.10(1210) = 1331<\/p>\n<p>Notice that in the first year, the population grew by 100 fish; in the second year, the population grew by 110 fish; and in the third year the population grew by 121 fish.<\/p>\n<p>While there is a constant <em>percentage<\/em> growth, the actual increase in number of fish is increasing each year.<\/p>\n<p>Graphing these values we see that this growth doesn\u2019t quite appear linear.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-363\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201507\/yearsfromnow1.png\" alt=\"Line graph. Measured vertically: Fish, in increments of 200, from 800 to 1800. Measured horizontally: Years from now, measured in units of 1, from 0 to 5. Year 0 is at 1000; Year 1 is at roughly 1100; and subsequent years move in an increasing rate so that the overall line is curved to the upper right.\" width=\"350\" height=\"283\" \/><\/p>\n<p>A walk-through of this fish scenario can be viewed here:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Exponential Growth Model Part 1\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/3BiU7Ihxvxg?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>To get a better picture of how this percentage-based growth affects things, we need an explicit form, so we can quickly calculate values further out in the future.<\/p>\n<p>Like we did for the linear model, we will start building from the recursive equation:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub><\/em> = 1.10<em>P\u00ad<sub>0<\/sub><\/em> = 1.10(1000)<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub><\/em> = 1.10<em>P\u00ad<sub>1<\/sub><\/em> = 1.10(1.10(1000)) = 1.102(1000)<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>3<\/sub><\/em> = 1.10<em>P\u00ad<sub>2<\/sub><\/em> = 1.10(1.102(1000)) = 1.103(1000)<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>4<\/sub><\/em> = 1.10<em>P\u00ad<sub>3<\/sub><\/em> = 1.10(1.103(1000)) = 1.104(1000)<\/p>\n<p>Observing a pattern, we can generalize the explicit form to be:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = 1.10<sub><em>n<\/em><\/sub>(1000), or equivalently, <em>P\u00ad<sub>n<\/sub><\/em> = 1000(1.10<sub><em>n<\/em><\/sub>)<\/p>\n<p>From this, we can quickly calculate the number of fish in 10, 20, or 30 years:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-364\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201611\/yearsfromnow2.png\" alt=\"Line graph. Measured vertically: Fish, in increments of 3000, from 0 to 18,000. Measured horizontally: Years from now, measured in units of 5, from 0 to 30. Year 0 is at 1000, with a tight cluster of dots in the lower left quadrant. The exponential increase becomes more dramatic as time advances, so that year 30 is at 18,000 fish with more spacing between dots.\" width=\"350\" height=\"281\" \/><\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = 1.10<sub><em>10<\/em><\/sub>(1000) = 2594<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>20<\/sub><\/em> = 1.10<sub><em>20<\/em><\/sub>(1000) = 6727<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>30<\/sub><\/em> = 1.10<sub><em>30<\/em><\/sub>(1000) = 17449<\/p>\n<p>Adding these values to our graph reveals a shape that is definitely not linear. If our fish population had been growing linearly, by 100 fish each year, the population would have only reached 4000 in 30 years, compared to almost 18,000 with this percent-based growth, called <strong>exponential growth.<\/strong><\/p>\n<p>A video demonstrating the explicit model of this fish story can be viewed here:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Exponential Growth Model Part 2\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tg2ysaZ8agY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In exponential growth, the population grows proportional to the size of the population, so as the population gets larger, the same percent growth will yield a larger numeric growth.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>Exponential Growth<\/h3>\n<p>If a quantity starts at size <em>P\u00ad<sub>0<\/sub><\/em> and grows by <em>R%<\/em> (written as a decimal, <em>r<\/em>) every time period, then the quantity after <em>n<\/em> time periods can be determined using either of these relations:<\/p>\n<h4>Recursive form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = (1+<em>r<\/em>) <em>P\u00ad<sub>n-1<\/sub><\/em><\/p>\n<h4>Explicit form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = (1+<em>r<\/em>)<sup><sub><em>n<\/em><\/sub><\/sup> <em>P\u00ad<sub>0<\/sub> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em>or equivalently, <em>P<sub>\u00adn<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> (1+<em>r<\/em>)<sup><sub><em>n<\/em><\/sub><\/sup><\/p>\n<p>We call <em>r<\/em> the <strong>growth rate<\/strong>.<\/p>\n<p>The term (1+<em>r<\/em>) is called the <strong>growth multiplier<\/strong>, or common ratio.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Between 2007 and 2008, Olympia, WA grew almost 3% to a population of 245 thousand people. If this growth rate was to continue, what would the population of Olympia be in 2014?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q54756\">Show Solution<\/span><\/p>\n<div id=\"q54756\" class=\"hidden-answer\" style=\"display: none\">\n<p>As we did before, we first need to define what year will correspond to <em>n<\/em> = 0. Since we know the population in 2008, it would make sense to have 2008 correspond to <em>n <\/em>= 0, so <em>P\u00ad<sub>0<\/sub><\/em> = 245,000.\u00a0\u00a0 The year 2014 would then be <em>n<\/em> = 6.<\/p>\n<p>We know the growth rate is 3%, giving <em>r <\/em>= 0.03.<\/p>\n<p>Using the explicit form:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>6<\/sub><\/em> = (1+0.03)<sup>6<\/sup> (245,000) = 1.19405(245,000) = 292,542.25<\/p>\n<p>The model predicts that in 2014, Olympia would have a population of about 293 thousand people.<\/p>\n<\/div>\n<\/div>\n<p>The following video explains this example in detail.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Predicting future population using an exponential model\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/CDI4xS65rxY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>Evaluating exponents on the calculator<\/h3>\n<p>To evaluate expressions like (1.03)<sup>6<\/sup>, it will be easier to use a calculator than multiply 1.03 by itself six times. Most scientific calculators have a button for exponents.\u00a0 It is typically either labeled like:<\/p>\n<p style=\"text-align: center;\">^ ,\u00a0\u00a0 y<sup>x<\/sup> ,\u00a0\u00a0 or x<sup>y<\/sup> .<\/p>\n<p>To evaluate 1.03<sup>6<\/sup>\u00a0we\u2019d type 1.03 ^ 6, or 1.03 y<sup>x<\/sup> 6.\u00a0 Try it out &#8211; you should get an answer around 1.1940523.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>India is the second most populous country in the world, with a population in 2008 of about 1.14 billion people. The population is growing by about 1.34% each year. If this trend continues, what will India\u2019s population grow to by 2020?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q265475\">Show Solution<\/span><\/p>\n<div id=\"q265475\" class=\"hidden-answer\" style=\"display: none\">\n<div>Using n = 0 corresponding with 2008, [latex]P_12= (1+0.0134)12(1.14)[\/latex] = about 1.337 billion people in 2020<\/div>\n<\/div>\n<\/div>\n<p><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6673&#38;theme=oea&#38;iframe_resize_id=mom3\">https:\/\/www.myopenmath.com\/multiembedq.php?id=6673&amp;theme=oea&amp;iframe_resize_id=mom3<\/a><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>A friend is using the equation <em>P\u00ad<sub>n<\/sub><\/em> = 4600(1.072)<sup><em>n <\/em><\/sup>to predict the annual tuition at a local college. She says the formula is based on years after 2010. What does this equation tell us?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q224261\">Show Solution<\/span><\/p>\n<div id=\"q224261\" class=\"hidden-answer\" style=\"display: none\">\n<p>In the equation, <em>P<sub>\u00ad0<\/sub><\/em> = 4600, which is the starting value of the tuition when <em>n<\/em> = 0. This tells us that the tuition in 2010 was $4,600.<\/p>\n<p>The growth multiplier is 1.072, so the growth rate is 0.072, or 7.2%. This tells us that the tuition is expected to grow by 7.2% each year.<\/p>\n<p>Putting this together, we could say that the tuition in 2010 was $4,600, and is expected to grow by 7.2% each year.<\/p>\n<\/div>\n<\/div>\n<p>View the following to see this example worked out.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Interpreting an Exponential Equation\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/T8Yz94De5UM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<hr \/>\n<p>In 1990, residential energy use in the US was responsible for 962 million metric tons of carbon dioxide emissions. By the year 2000, that number had risen to 1182 million metric tons<a class=\"footnote\" title=\"http:\/\/www.eia.doe.gov\/oiaf\/1605\/ggrpt\/carbon.html\" id=\"return-footnote-362-1\" href=\"#footnote-362-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a>. If the emissions grow exponentially and continue at the same rate, what will the emissions grow to by 2050?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q755963\">Show Solution<\/span><\/p>\n<div id=\"q755963\" class=\"hidden-answer\" style=\"display: none\">\n<p>Similar to before, we will correspond <em>n<\/em> = 0 with 1990, as that is the year for the first piece of data we have. That will make <em>P\u00ad<sub>0<\/sub><\/em> = 962 (million metric tons of CO<sub>2<\/sub>). In this problem, we are not given the growth rate, but instead are given that <em>P\u00ad<sub>10<\/sub><\/em> = 1182.<\/p>\n<p>When <em>n<\/em> = 10, the explicit equation looks like:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = (1+<em>r<\/em>)<sup>10<\/sup> <em>P\u00ad<sub>0<\/sub><\/em><\/p>\n<p>We know the value for <em>P\u00ad<sub>0<\/sub><\/em>, so we can put that into the equation:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = (1+<em>r<\/em>)<sup>10<\/sup> 962<\/p>\n<p>We also know that <em>P\u00ad<sub>10<\/sub><\/em> = 1182, so substituting that in, we get<\/p>\n<p style=\"text-align: center;\">1182 = (1+<em>r<\/em>)<sup>10<\/sup> 962<\/p>\n<p>We can now solve this equation for the growth rate, <em>r<\/em>. Start by dividing by 962.<\/p>\n<p style=\"padding-left: 120px;\">[latex]\\frac{1182}{962}={{(1+r)}^{10}}[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Take the 10th root of both sides<\/p>\n<p style=\"padding-left: 120px;\">[latex]\\sqrt[10]{\\frac{1182}{962}}=1+r[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Subtract 1 from both sides<\/p>\n<p style=\"padding-left: 120px;\">[latex]r=\\sqrt[10]{\\frac{1182}{962}}-1=0.0208[\/latex] = 2.08%<\/p>\n<p>So if the emissions are growing exponentially, they are growing by about 2.08% per year. We can now predict the emissions in 2050 by finding <em>P<sub>\u00ad60<\/sub><\/em><\/p>\n<p style=\"text-align: center;\"><em>P<sub>\u00ad60<\/sub><\/em> = (1+0.0208)<sup>60<\/sup> 962 = 3308.4 million metric tons of CO<sub>2<\/sub> in 2050<\/p>\n<\/div>\n<\/div>\n<p>View more about this example here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Finding an Exponential Model\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/9Zu2uONfLkQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Rounding<\/h3>\n<p>As a note on rounding, notice that if we had rounded the growth rate to 2.1%, our calculation for the emissions in 2050 would have been 3347.\u00a0\u00a0 Rounding to 2% would have changed our result to 3156. A very small difference in the growth rates gets magnified greatly in exponential growth. For this reason, it is recommended to round the growth rate as little as possible.<\/p>\n<p>If you need to round, <strong>keep at least three significant digits<\/strong> &#8211; numbers after any leading zeros.\u00a0\u00a0 So 0.4162 could be reasonably rounded to 0.416. A growth rate of 0.001027 could be reasonably rounded to 0.00103.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Evaluating roots on the calculator<\/h3>\n<p>In the previous example, we had to calculate the 10th root of a number. This is different than taking the basic square root, \u221a. Many scientific calculators have a button for general roots.\u00a0 It is typically labeled like:<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt[y]{x}[\/latex]<\/p>\n<p>To evaluate the 3rd root of 8, for example, we\u2019d either type 3 [latex]\\sqrt[x]{{}}[\/latex] 8, or 8 [latex]\\sqrt[x]{{}}[\/latex] 3, depending on the calculator. Try it on yours to see which to use \u2013 you should get an answer of 2.<\/p>\n<p>If your calculator does not have a general root button, all is not lost. You can instead use the property of exponents which states that:<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #000000;\">[latex]\\sqrt[n]{a}={a}^{\\frac{1}{2}}[\/latex]<\/span>.<\/p>\n<p>So, to compute the 3rd root of 8, you could use your calculator\u2019s exponent key to evaluate 8<sup>1\/3<\/sup>. To do this, type:<\/p>\n<p style=\"text-align: center;\">8 y<sup>x<\/sup> ( 1 \u00f7 3 )<\/p>\n<p>The parentheses tell the calculator to divide 1\/3 before doing the exponent.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>a note about parentheses<\/h3>\n<p>Don&#8217;t overlook the use of parentheses in the above example. They are necessary to force the desired order of operations when using your calculator.<\/p>\n<p>If you were to leave them out and type [latex]8y^x1\/3[\/latex] or 8^1\/3, it would tell the calculator to first raise 8 to the power of 1, then divide the result by 3.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The number of users on a social networking site was 45 thousand in February when they officially went public, and grew to 60 thousand by October. If the site is growing exponentially, and growth continues at the same rate, how many users should they expect two years after they went public?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q774576\">Show Solution<\/span><\/p>\n<div id=\"q774576\" class=\"hidden-answer\" style=\"display: none\">\n<p>Here we will measure n in months rather than years, with n = 0 corresponding to the February when they went public. This gives [latex]P_0= 45[\/latex] thousand. October is 8 months later, so [latex]P_8= 60[\/latex].<br \/>\n[latex]P_8=(1+r)^{8}P_0[\/latex]<br \/>\n[latex]60=(1+r)^{8}45[\/latex]<br \/>\n[latex]\\frac{60}{45}=(1+r)^8[\/latex]<br \/>\n[latex]\\sqrt[8]{\\frac{60}{45}}=1+r[\/latex]<br \/>\n[latex]r=\\sqrt[8]{\\frac{60}{45}}-1=0.0366\\text{ or }3.66%[\/latex]<br \/>\nThe general explicit equation is [latex]P_n =(1.0366)^{n}45[\/latex]. Predicting 24 months after they went public gives [latex]P_{24}=(1.0366)^{24}45=106.63[\/latex] thousand users.\n<\/p><\/div>\n<\/div>\n<p><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6598&#38;theme=oea&#38;iframe_resize_id=mom4\">https:\/\/www.myopenmath.com\/multiembedq.php?id=6598&amp;theme=oea&amp;iframe_resize_id=mom4<\/a><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Looking back at the last example, for the sake of comparison, what would the carbon emissions be in 2050 if emissions grow linearly at the same rate?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q355767\">Show Solution<\/span><\/p>\n<div id=\"q355767\" class=\"hidden-answer\" style=\"display: none\">\n<p>Again we will get <em>n<\/em> = 0 correspond with 1990, giving <em>P<sub>\u00ad0<\/sub><\/em> = 962. To find <em>d<\/em>, we could take the same approach as earlier, noting that the emissions increased by 220 million metric tons in 10 years, giving a common difference of 22 million metric tons each year.<\/p>\n<p>Alternatively, we could use an approach similar to that which we used to find the exponential equation. When <em>n<\/em> = 10, the explicit linear equation looks like:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = <em>P<sub>\u00ad0<\/sub><\/em> + 10<em>d<\/em><em>\u00a0<\/em><\/p>\n<p>We know the value for <em>P\u00ad0<\/em>, so we can put that into the equation:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = 962 + 10<em>d<\/em><\/p>\n<p>Since we know that <em>P\u00ad<sub>10<\/sub><\/em> = 1182, substituting that in we get<\/p>\n<p style=\"text-align: center;\">1182 = 962 + 10<em>d<\/em><\/p>\n<p>We can now solve this equation for the common difference, <em>d<\/em>.<\/p>\n<p style=\"text-align: center;\">1182 \u2013 962 = 10<em>d<\/em><\/p>\n<p style=\"text-align: center;\">220 = 10<em>d<\/em><\/p>\n<p style=\"text-align: center;\"><em>d<\/em> = 22<\/p>\n<p>This tells us that if the emissions are changing linearly, they are growing by 22 million metric tons each year. Predicting the emissions in 2050,<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>60<\/sub><\/em> = 962 + 22(60) = 2282 million metric tons.<\/p>\n<p>You will notice that this number is substantially smaller than the prediction from the exponential growth model. Calculating and plotting more values helps illustrate the differences.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-365\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201942\/yearsafter1990.png\" alt=\"Line graph. Vertical measures Emissions (Millions Metric Tons) in increments of 500, from 0 to 3500. Horizontal measures Years after 1990, in increments of 10, from 0 to 60. A blue line shows a linear growth from 1000 in year 0 to over 2000 in year 60. A pink line shows an exponential growth from 1000 in year 0 to 3500 in year 60.\" width=\"350\" height=\"283\" \/><\/p>\n<\/div>\n<\/div>\n<p>A demonstration of this example can be seen in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Comparing exponential to linear growth\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/yiuZoiRMtYM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>So how do we know which growth model to use when working with data? There are two approaches which should be used together whenever possible:<\/p>\n<ol>\n<li>Find more than two pieces of data. Plot the values, and look for a trend. Does the data appear to be changing like a line, or do the values appear to be curving upwards?<\/li>\n<li>Consider the factors contributing to the data. Are they things you would expect to change linearly or exponentially? For example, in the case of carbon emissions, we could expect that, absent other factors, they would be tied closely to population values, which tend to change exponentially.<\/li>\n<\/ol>\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-362\">\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>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>Exponential (Geometric) Growth. <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>Project<\/strong>: Math in Society. <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>calf-fish-lake-pond-river. <strong>Authored by<\/strong>: Pexels. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/pixabay.com\/en\/calf-fish-lake-pond-river-1869984\/\">https:\/\/pixabay.com\/en\/calf-fish-lake-pond-river-1869984\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/cc0\">CC0: No Rights Reserved<\/a><\/em><\/li><li>Exponential Growth Model Part 1. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/3BiU7Ihxvxg\">https:\/\/youtu.be\/3BiU7Ihxvxg<\/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>Exponential Growth Model Part 2. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tg2ysaZ8agY\">https:\/\/youtu.be\/tg2ysaZ8agY<\/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>Predicting future population using an exponential model. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/CDI4xS65rxY\">https:\/\/youtu.be\/CDI4xS65rxY<\/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>Interpreting an Exponential Equation. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/T8Yz94De5UM\">https:\/\/youtu.be\/T8Yz94De5UM<\/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>Finding an Exponential Model. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/9Zu2uONfLkQ\">https:\/\/youtu.be\/9Zu2uONfLkQ<\/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>Comparing exponential to linear growth. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/yiuZoiRMtYM\">https:\/\/youtu.be\/yiuZoiRMtYM<\/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 6598. <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-362-1\"><a href=\"http:\/\/www.eia.doe.gov\/oiaf\/1605\/ggrpt\/carbon.html\" target=\"_blank\" rel=\"noopener\">http:\/\/www.eia.doe.gov\/oiaf\/1605\/ggrpt\/carbon.html<\/a> <a href=\"#return-footnote-362-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":20,"menu_order":18,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Exponential (Geometric) Growth\",\"author\":\"David Lippman\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"Math in Society\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"calf-fish-lake-pond-river\",\"author\":\"Pexels\",\"organization\":\"\",\"url\":\"https:\/\/pixabay.com\/en\/calf-fish-lake-pond-river-1869984\/\",\"project\":\"\",\"license\":\"cc0\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Exponential Growth Model Part 1\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/3BiU7Ihxvxg\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Exponential Growth Model Part 2\",\"author\":\"OCLPhase2\\'s 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