{"id":2143,"date":"2024-07-10T00:57:22","date_gmt":"2024-07-10T00:57:22","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=2143"},"modified":"2024-11-27T20:11:52","modified_gmt":"2024-11-27T20:11:52","slug":"applications-of-exponential-functions-apply-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/applications-of-exponential-functions-apply-it-1\/","title":{"raw":"Applications of Exponential Functions: Apply It 1","rendered":"Applications of Exponential Functions: Apply It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate the values of exponential functions, especially those using the base \ud835\udc52, and understand their equations<\/li>\r\n \t<li>Use compound interest formulas to work out how investments or loans grow over time in real-life financial situations<\/li>\r\n \t<li>Find an exponential function that models continuous growth or decay<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Writing an Exponential Model<\/h2>\r\n<section class=\"textbox example\" aria-label=\"Example\">In 2006, [latex]80[\/latex] deer were introduced into a wildlife refuge. By 2012, the population had grown to [latex]180[\/latex] deer. The population was growing exponentially. Write an algebraic function [latex]N(t) [\/latex] representing the population [latex]N[\/latex]\u00a0of deer over time [latex]t[\/latex].[reveal-answer q=\"910377\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"910377\"]We let our independent variable [latex]t[\/latex]\u00a0be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: [latex](0, 80)[\/latex] and [latex](6, 180)[\/latex]. Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, [latex]a\u00a0= 80[\/latex]. We can now substitute the second point into the equation [latex]N\\left(t\\right)=80{b}^{t}[\/latex] to find [latex]b[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}N\\left(t\\right)\\hfill &amp; =80{b}^{t}\\hfill &amp; \\hfill \\\\ 180\\hfill &amp; =80{b}^{6}\\hfill &amp; \\text{Substitute using point }\\left(6, 180\\right).\\hfill \\\\ \\frac{9}{4}\\hfill &amp; ={b}^{6}\\hfill &amp; \\text{Divide and write in lowest terms}.\\hfill \\\\ b\\hfill &amp; ={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}}\\hfill &amp; \\text{Isolate }b\\text{ using properties of exponents}.\\hfill \\\\ b\\hfill &amp; \\approx 1.1447 &amp; \\text{Round to 4 decimal places}.\\hfill \\end{array}[\/latex]<\/p>\r\n<strong>NOTE:<\/strong> <em>Unless otherwise stated, do not round any intermediate calculations. Round the final answer to four places for the remainder of this section.<\/em>\r\n\r\nThe exponential model for the population of deer is [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex]. Note that this exponential function models short-term growth. As the inputs get larger, the outputs will get increasingly larger resulting in the model not being useful in the long term due to extremely large output values.\r\n\r\nWe can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(0,\\text{ 8}0\\right)[\/latex] and [latex]\\left(\\text{6},\\text{ 18}0\\right)[\/latex]. We can also see that the domain for the function is [latex]\\left[0,\\infty \\right)[\/latex] and the range for the function is [latex]\\left[80,\\infty \\right)[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225444\/CNX_Precalc_Figure_04_01_0022.jpg\" alt=\"Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).\" width=\"487\" height=\"700\" \/> Graph showing the population of deer over time, [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex], t\u00a0years after 2006[\/caption]\r\n[\/hidden-answer]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24721[\/ohm2_question]<\/section><section class=\"textbox example\" aria-label=\"Example\">Find an exponential function that passes through the points [latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(2,1\\right)[\/latex].[reveal-answer q=\"904458\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"904458\"]Because we don\u2019t have the initial value, we substitute both points into an equation of the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex] and then solve the system for [latex]a[\/latex]\u00a0and [latex]b[\/latex].\r\n<ul>\r\n \t<li>Substituting [latex]\\left(-2,6\\right)[\/latex] gives [latex]6=a{b}^{-2}[\/latex]<\/li>\r\n \t<li>Substituting [latex]\\left(2,1\\right)[\/latex] gives [latex]1=a{b}^{2}[\/latex]<\/li>\r\n<\/ul>\r\nUse the first equation to solve for [latex]a[\/latex]\u00a0in terms of [latex]b[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}6=ab^{-2}\\\\\\frac{6}{b^{-2}}=a\\,\\,\\,\\,\\,\\,\\,\\,\\text{Divide.}\\\\a=6b^{2}\\,\\,\\,\\,\\,\\,\\,\\,\\text{Use properties of exponents to rewrite the denominator.}\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]a[\/latex]\u00a0in the second equation and solve for [latex]b:[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}1=ab^{2}\\\\1=6b^{2}b^{2}=6b^{4}\\,\\,\\,\\,\\,\\text{Substitute }a.\\\\b=\\left(\\frac{1}{6}\\right)^{\\frac{1}{4}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Use properties of exponents to isolate }b.\\\\b\\approx0.6389\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Round 4 decimal places.}\\end{array}[\/latex]<\/p>\r\nUse the value of [latex]b[\/latex]\u00a0in the first equation to solve for the value of [latex]a[\/latex]:\r\n<p style=\"text-align: center;\">[latex]a=6b^{2}\\approx6\\left(0.6389\\right)^{2}\\approx2.4492[\/latex]<\/p>\r\nThus, the equation is [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex].\r\n\r\nWe can graph our model to check our work. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(-2,\\text{ 6}\\right)[\/latex] and [latex]\\left(2,\\text{ 1}\\right)[\/latex]. The graph is an example of an <strong>exponential decay<\/strong> function.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225453\/CNX_Precalc_Figure_04_01_0032.jpg\" alt=\"Graph of the exponential function, f(x)=2.4492(0.6389)^x, with labeled points at (-2, 6) and (2, 1).\" width=\"487\" height=\"445\" \/> The graph of [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex] models exponential decay.[\/caption][\/hidden-answer]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24722[\/ohm2_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24723[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate the values of exponential functions, especially those using the base \ud835\udc52, and understand their equations<\/li>\n<li>Use compound interest formulas to work out how investments or loans grow over time in real-life financial situations<\/li>\n<li>Find an exponential function that models continuous growth or decay<\/li>\n<\/ul>\n<\/section>\n<h2>Writing an Exponential Model<\/h2>\n<section class=\"textbox example\" aria-label=\"Example\">In 2006, [latex]80[\/latex] deer were introduced into a wildlife refuge. By 2012, the population had grown to [latex]180[\/latex] deer. The population was growing exponentially. Write an algebraic function [latex]N(t)[\/latex] representing the population [latex]N[\/latex]\u00a0of deer over time [latex]t[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q910377\">Show Solution<\/button><\/p>\n<div id=\"q910377\" class=\"hidden-answer\" style=\"display: none\">We let our independent variable [latex]t[\/latex]\u00a0be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: [latex](0, 80)[\/latex] and [latex](6, 180)[\/latex]. Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, [latex]a\u00a0= 80[\/latex]. We can now substitute the second point into the equation [latex]N\\left(t\\right)=80{b}^{t}[\/latex] to find [latex]b[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}N\\left(t\\right)\\hfill & =80{b}^{t}\\hfill & \\hfill \\\\ 180\\hfill & =80{b}^{6}\\hfill & \\text{Substitute using point }\\left(6, 180\\right).\\hfill \\\\ \\frac{9}{4}\\hfill & ={b}^{6}\\hfill & \\text{Divide and write in lowest terms}.\\hfill \\\\ b\\hfill & ={\\left(\\frac{9}{4}\\right)}^{\\frac{1}{6}}\\hfill & \\text{Isolate }b\\text{ using properties of exponents}.\\hfill \\\\ b\\hfill & \\approx 1.1447 & \\text{Round to 4 decimal places}.\\hfill \\end{array}[\/latex]<\/p>\n<p><strong>NOTE:<\/strong> <em>Unless otherwise stated, do not round any intermediate calculations. Round the final answer to four places for the remainder of this section.<\/em><\/p>\n<p>The exponential model for the population of deer is [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex]. Note that this exponential function models short-term growth. As the inputs get larger, the outputs will get increasingly larger resulting in the model not being useful in the long term due to extremely large output values.<\/p>\n<p>We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(0,\\text{ 8}0\\right)[\/latex] and [latex]\\left(\\text{6},\\text{ 18}0\\right)[\/latex]. We can also see that the domain for the function is [latex]\\left[0,\\infty \\right)[\/latex] and the range for the function is [latex]\\left[80,\\infty \\right)[\/latex].<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225444\/CNX_Precalc_Figure_04_01_0022.jpg\" alt=\"Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).\" width=\"487\" height=\"700\" \/><figcaption class=\"wp-caption-text\">Graph showing the population of deer over time, [latex]N\\left(t\\right)=80{\\left(1.1447\\right)}^{t}[\/latex], t\u00a0years after 2006<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24721\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24721&theme=lumen&iframe_resize_id=ohm24721&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find an exponential function that passes through the points [latex]\\left(-2,6\\right)[\/latex] and [latex]\\left(2,1\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q904458\">Show Solution<\/button><\/p>\n<div id=\"q904458\" class=\"hidden-answer\" style=\"display: none\">Because we don\u2019t have the initial value, we substitute both points into an equation of the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex] and then solve the system for [latex]a[\/latex]\u00a0and [latex]b[\/latex].<\/p>\n<ul>\n<li>Substituting [latex]\\left(-2,6\\right)[\/latex] gives [latex]6=a{b}^{-2}[\/latex]<\/li>\n<li>Substituting [latex]\\left(2,1\\right)[\/latex] gives [latex]1=a{b}^{2}[\/latex]<\/li>\n<\/ul>\n<p>Use the first equation to solve for [latex]a[\/latex]\u00a0in terms of [latex]b[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}6=ab^{-2}\\\\\\frac{6}{b^{-2}}=a\\,\\,\\,\\,\\,\\,\\,\\,\\text{Divide.}\\\\a=6b^{2}\\,\\,\\,\\,\\,\\,\\,\\,\\text{Use properties of exponents to rewrite the denominator.}\\end{array}[\/latex]<\/p>\n<p>Substitute [latex]a[\/latex]\u00a0in the second equation and solve for [latex]b:[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}1=ab^{2}\\\\1=6b^{2}b^{2}=6b^{4}\\,\\,\\,\\,\\,\\text{Substitute }a.\\\\b=\\left(\\frac{1}{6}\\right)^{\\frac{1}{4}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Use properties of exponents to isolate }b.\\\\b\\approx0.6389\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{Round 4 decimal places.}\\end{array}[\/latex]<\/p>\n<p>Use the value of [latex]b[\/latex]\u00a0in the first equation to solve for the value of [latex]a[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]a=6b^{2}\\approx6\\left(0.6389\\right)^{2}\\approx2.4492[\/latex]<\/p>\n<p>Thus, the equation is [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex].<\/p>\n<p>We can graph our model to check our work. Notice that the graph below\u00a0passes through the initial points given in the problem, [latex]\\left(-2,\\text{ 6}\\right)[\/latex] and [latex]\\left(2,\\text{ 1}\\right)[\/latex]. The graph is an example of an <strong>exponential decay<\/strong> function.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02225453\/CNX_Precalc_Figure_04_01_0032.jpg\" alt=\"Graph of the exponential function, f(x)=2.4492(0.6389)^x, with labeled points at (-2, 6) and (2, 1).\" width=\"487\" height=\"445\" \/><figcaption class=\"wp-caption-text\">The graph of [latex]f\\left(x\\right)=2.4492{\\left(0.6389\\right)}^{x}[\/latex] models exponential decay.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24722\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24722&theme=lumen&iframe_resize_id=ohm24722&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24723\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24723&theme=lumen&iframe_resize_id=ohm24723&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":255,"module-header":"apply_it","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\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2143"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2143\/revisions"}],"predecessor-version":[{"id":6503,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2143\/revisions\/6503"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/255"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2143\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=2143"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2143"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=2143"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=2143"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}