{"id":271,"date":"2024-10-18T21:19:52","date_gmt":"2024-10-18T21:19:52","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/chapter\/logarithms-and-logistic-growth-learn-it-2\/"},"modified":"2024-10-18T21:19:52","modified_gmt":"2024-10-18T21:19:52","slug":"logarithms-and-logistic-growth-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/chapter\/logarithms-and-logistic-growth-learn-it-2\/","title":{"raw":"Logarithms and Logistic Growth: Learn It 2","rendered":"Logarithms and Logistic Growth: Learn It 2"},"content":{"raw":"\n<h2>Reversing an Exponent Cont.<\/h2>\n<p>The exponential property of logarithms will finally allow us to answer the original question posed on the last page.<\/p>\n<section class=\"textbox questionHelp\">\n<p><b>How To: Solve Exponential Equations with Logarithms<\/b><\/p>\n<ol>\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>\n\t<li>Take the log of both sides of the equation.<\/li>\n\t<li>Use the exponent property of logs to rewrite the exponential with the variable exponent multiplying the logarithm.<\/li>\n\t<li>Divide as needed to solve for the variable.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">If Olympia is growing according to the equation,&nbsp;[latex]P_{n}= 245(1.03)^{n}[\/latex], where&nbsp;[latex]n[\/latex] is years after 2008, and the population is measured in thousands. Find when the population will be&nbsp;[latex]400[\/latex] thousand.<br>\n[reveal-answer q=\"770708\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"770708\"]We need to solve the equation\n\n<p style=\"padding-left: 120px;\">[latex]400 = 245(1.03)^n[\/latex]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Begin by dividing both sides by [latex]245[\/latex] to isolate the exponential<\/p>\n<p style=\"padding-left: 120px;\">[latex]1.633 = 1.03^n[\/latex]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Now take the log of both sides<\/p>\n<p style=\"padding-left: 120px;\">[latex]log(1.633) = log(1.03^n)[\/latex]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Use the exponent property of logs on the right side<\/p>\n<p style=\"padding-left: 120px;\">[latex]log(1.633)= n log(1.03)[\/latex]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Now we can divide by [latex]log(1.03)[\/latex]<\/p>\n<p style=\"padding-left: 120px;\">[latex]\\frac{\\log(1.633)}{\\log(1.03)}=n[\/latex]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;We can approximate this value on a calculator<\/p>\n<p style=\"padding-left: 120px;\">[latex]n \u2248 16.591[\/latex]<\/p>\n<p>A full walk-through of this problem is available here.<\/p>\n<p><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/liNffAACIUs?si=wExYZoVqH9lBAwio\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Using+logs+to+solve+for+time.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cUsing logs to solve for time\u201d here (opens in new window).<\/a><\/p>\n<p><br>\n[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]6928[\/ohm2_question]<\/section>\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<section class=\"textbox example\">Polluted water is passed through a series of filters. Each filter removes&nbsp;[latex]90\\%[\/latex] of the remaining impurities from the water. If you have&nbsp;[latex]10[\/latex] million particles of pollutant per gallon originally, how many filters would the water need to be passed through to reduce the pollutant to&nbsp;[latex]500[\/latex] particles per gallon?<br>\n[reveal-answer q=\"878881\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"878881\"]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&nbsp;[latex]P_0 = 10,000,000[\/latex]. Instead of changing with time, the pollutant changes with the number of filters, so&nbsp;[latex]n[\/latex] will represent the number of filters the water passes through.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&nbsp;[latex]r = -0.90[\/latex].We can then write the explicit equation for the pollutant:\n\n<p style=\"text-align: center;\">[latex]P_n = 10,000,000(1 \u2013 0.90)^{n} = 10,000,000(0.10)^{n}[\/latex]<\/p>\n<p>To solve the question of how many filters are needed to lower the pollutant to&nbsp;[latex]500[\/latex] particles per gallon, we can set&nbsp;[latex]P_n[\/latex] equal to&nbsp;[latex]500[\/latex], and solve for&nbsp;[latex]n[\/latex].<\/p>\n<p style=\"padding-left: 150px;\">[latex]500 = 10,000,000(0.10)n[\/latex]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Divide both sides by [latex]10,000,000[\/latex]<\/p>\n<p style=\"padding-left: 150px;\">[latex]0.00005 = 0.10n[\/latex]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Take the log of both sides<\/p>\n<p style=\"padding-left: 150px;\">[latex]log(0.00005) = log(0.10n)[\/latex]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Use the exponent property of logs on the right side<\/p>\n<p style=\"padding-left: 150px;\">[latex]log(0.00005) = n log(0.10)[\/latex] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Evaluate the logarithms to a decimal approximation<\/p>\n<p style=\"padding-left: 150px;\">[latex]-4.301 = n (-1)[\/latex] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Divide by&nbsp;[latex]-1[\/latex], the value multiplying&nbsp;[latex]n[\/latex]<\/p>\n<p style=\"padding-left: 150px;\">[latex]4.301 = n[\/latex]<\/p>\n<p>It would take about&nbsp;[latex]4.301[\/latex] filters. Of course, since we probably can\u2019t install&nbsp;[latex]0.3[\/latex] filters, we would need to use&nbsp;[latex]5[\/latex] filters to bring the pollutant below the desired level.<\/p>\n<p>More details about solving this scenario are available in this video.<\/p>\n<p><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/sLLu0u1YgM0?si=cHd0iBWpPQ0IJrfC\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Using+logs+to+solve+an+exponential.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cUsing logs to solve an exponential\u201d here (opens in new window).<\/a><\/p>\n\n[\/hidden-answer]<\/section>\n<section class=\"textbox proTip\">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 \"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&nbsp;problems is in the exponent.<\/section>\n","rendered":"<h2>Reversing an Exponent Cont.<\/h2>\n<p>The exponential property of logarithms will finally allow us to answer the original question posed on the last page.<\/p>\n<section class=\"textbox questionHelp\">\n<p><b>How To: Solve Exponential Equations with Logarithms<\/b><\/p>\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<\/section>\n<section class=\"textbox example\">If Olympia is growing according to the equation,&nbsp;[latex]P_{n}= 245(1.03)^{n}[\/latex], where&nbsp;[latex]n[\/latex] is years after 2008, and the population is measured in thousands. Find when the population will be&nbsp;[latex]400[\/latex] thousand.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q770708\">Show Solution<\/button><\/p>\n<div id=\"q770708\" class=\"hidden-answer\" style=\"display: none\">We need to solve the equation<\/p>\n<p style=\"padding-left: 120px;\">[latex]400 = 245(1.03)^n[\/latex]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Begin by dividing both sides by [latex]245[\/latex] to isolate the exponential<\/p>\n<p style=\"padding-left: 120px;\">[latex]1.633 = 1.03^n[\/latex]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Now take the log of both sides<\/p>\n<p style=\"padding-left: 120px;\">[latex]log(1.633) = log(1.03^n)[\/latex]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Use the exponent property of logs on the right side<\/p>\n<p style=\"padding-left: 120px;\">[latex]log(1.633)= n log(1.03)[\/latex]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Now we can divide by [latex]log(1.03)[\/latex]<\/p>\n<p style=\"padding-left: 120px;\">[latex]\\frac{\\log(1.633)}{\\log(1.03)}=n[\/latex]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;We can approximate this value on a calculator<\/p>\n<p style=\"padding-left: 120px;\">[latex]n \u2248 16.591[\/latex]<\/p>\n<p>A full walk-through of this problem is available here.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/liNffAACIUs?si=wExYZoVqH9lBAwio\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Using+logs+to+solve+for+time.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cUsing logs to solve for time\u201d here (opens in new window).<\/a><\/p>\n<p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm6928\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=6928&theme=lumen&iframe_resize_id=ohm6928&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\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<section class=\"textbox example\">Polluted water is passed through a series of filters. Each filter removes&nbsp;[latex]90\\%[\/latex] of the remaining impurities from the water. If you have&nbsp;[latex]10[\/latex] million particles of pollutant per gallon originally, how many filters would the water need to be passed through to reduce the pollutant to&nbsp;[latex]500[\/latex] particles per gallon?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q878881\">Show Solution<\/button><\/p>\n<div id=\"q878881\" class=\"hidden-answer\" style=\"display: none\">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&nbsp;[latex]P_0 = 10,000,000[\/latex]. Instead of changing with time, the pollutant changes with the number of filters, so&nbsp;[latex]n[\/latex] will represent the number of filters the water passes through.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&nbsp;[latex]r = -0.90[\/latex].We can then write the explicit equation for the pollutant:<\/p>\n<p style=\"text-align: center;\">[latex]P_n = 10,000,000(1 \u2013 0.90)^{n} = 10,000,000(0.10)^{n}[\/latex]<\/p>\n<p>To solve the question of how many filters are needed to lower the pollutant to&nbsp;[latex]500[\/latex] particles per gallon, we can set&nbsp;[latex]P_n[\/latex] equal to&nbsp;[latex]500[\/latex], and solve for&nbsp;[latex]n[\/latex].<\/p>\n<p style=\"padding-left: 150px;\">[latex]500 = 10,000,000(0.10)n[\/latex]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Divide both sides by [latex]10,000,000[\/latex]<\/p>\n<p style=\"padding-left: 150px;\">[latex]0.00005 = 0.10n[\/latex]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Take the log of both sides<\/p>\n<p style=\"padding-left: 150px;\">[latex]log(0.00005) = log(0.10n)[\/latex]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Use the exponent property of logs on the right side<\/p>\n<p style=\"padding-left: 150px;\">[latex]log(0.00005) = n log(0.10)[\/latex] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Evaluate the logarithms to a decimal approximation<\/p>\n<p style=\"padding-left: 150px;\">[latex]-4.301 = n (-1)[\/latex] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Divide by&nbsp;[latex]-1[\/latex], the value multiplying&nbsp;[latex]n[\/latex]<\/p>\n<p style=\"padding-left: 150px;\">[latex]4.301 = n[\/latex]<\/p>\n<p>It would take about&nbsp;[latex]4.301[\/latex] filters. Of course, since we probably can\u2019t install&nbsp;[latex]0.3[\/latex] filters, we would need to use&nbsp;[latex]5[\/latex] filters to bring the pollutant below the desired level.<\/p>\n<p>More details about solving this scenario are available in this video.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/sLLu0u1YgM0?si=cHd0iBWpPQ0IJrfC\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Using+logs+to+solve+an+exponential.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cUsing logs to solve an exponential\u201d here (opens in new window).<\/a><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox proTip\">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&nbsp;problems is in the exponent.<\/section>\n","protected":false},"author":6,"menu_order":5,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":266,"module-header":"","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/chapters\/271"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/chapters\/271\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/parts\/266"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/chapters\/271\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/media?parent=271"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/pressbooks\/v2\/chapter-type?post=271"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/contributor?post=271"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebrademo\/wp-json\/wp\/v2\/license?post=271"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}