{"id":3820,"date":"2025-09-09T17:02:52","date_gmt":"2025-09-09T17:02:52","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3820"},"modified":"2026-03-17T18:05:04","modified_gmt":"2026-03-17T18:05:04","slug":"exponential-functions-apply-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/exponential-functions-apply-it-3\/","title":{"raw":"Exponential Functions: Apply It 3","rendered":"Exponential Functions: Apply It 3"},"content":{"raw":"<h2>Investigating Continuous Growth<\/h2>\r\nSo far we have worked with rational bases for exponential functions. For most real-world phenomena, however, [latex]e[\/latex] is used as the base for exponential functions. Exponential models that use [latex]e[\/latex] as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>the continuous growth\/decay formula<\/h3>\r\nFor all real numbers [latex]t[\/latex], and all positive numbers [latex]a[\/latex] and [latex]r[\/latex], continuous growth or decay is represented by the formula\r\n<p style=\"text-align: center;\">[latex]A(t) = Pe^{rt}[\/latex]<\/p>\r\nwhere\r\n<ul>\r\n \t<li>[latex]P[\/latex] is the initial value or the principal<\/li>\r\n \t<li>[latex]r[\/latex] is the growth or interest rate per unit time<\/li>\r\n \t<li>[latex]t[\/latex] is the period or term of the investment<\/li>\r\n<\/ul>\r\nIf [latex]r \\gt 0[\/latex] <span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">then the formula represents continuous growth. <\/span>\r\n\r\n<span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">If [latex]r \\lt 0[\/latex] then the formula represents continuous decay.<\/span>\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the initial value, rate of growth or decay, and time <em>[latex]t[\/latex]<\/em>, solve a continuous growth or decay function<\/strong>\r\n<ol>\r\n \t<li>Use the information in the problem to determine [latex]P[\/latex], the initial value of the function.<\/li>\r\n \t<li>Use the information in the problem to determine the growth rate [latex]r[\/latex].\r\n<ul>\r\n \t<li>If the problem refers to continuous growth, then [latex]r\u00a0&gt; 0[\/latex].<\/li>\r\n \t<li>If the problem refers to continuous decay, then [latex]r\u00a0&lt; 0[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Use the information in the problem to determine the time [latex]t[\/latex].<\/li>\r\n \t<li>Substitute the given information into the continuous growth formula and solve for [latex]A(t)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">A person invested [latex]$1,000[\/latex] in an account earning a nominal [latex]10\\%[\/latex] per year compounded continuously. How much was in the account at the end of one year?[reveal-answer q=\"864251\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"864251\"]Since the account is growing in value, this is a continuous compounding problem with growth rate [latex]r\u00a0= 0.10[\/latex]. The initial investment was [latex]$1,000[\/latex], so [latex]P\u00a0= 1000[\/latex]. We use the continuous compounding formula to find the value after [latex]t\u00a0= 1[\/latex] year:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill &amp; =P{e}^{rt}\\hfill &amp; \\text{Use the continuous compounding formula}.\\hfill \\\\ \\hfill &amp; =1000{\\left(e\\right)}^{0.1} &amp; \\text{Substitute known values for }P, r,\\text{ and }t.\\hfill \\\\ \\hfill &amp; \\approx 1105.17\\hfill &amp; \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/p>\r\nThe account is worth [latex]$1,105.17[\/latex] after one year.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321421[\/ohm_question]<\/section><section class=\"textbox example\" aria-label=\"Example\">Radon-222 decays at a continuous rate of [latex]17.3 \\%[\/latex] per day. How much will [latex]100[\/latex] mg of Radon-[latex]222[\/latex] decay to in [latex]3[\/latex] days?[reveal-answer q=\"660828\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"660828\"]Since the substance is decaying, the rate, [latex]17.3%[\/latex], is negative. So, [latex]r=-0.173[\/latex]. The initial amount of radon-[latex]222[\/latex] was [latex]100[\/latex] mg, so [latex]a= 100[\/latex]. We use the continuous decay formula to find the value after [latex]t= 3[\/latex] days:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill &amp; =a{e}^{rt}\\hfill &amp; \\text{Use the continuous growth formula}.\\hfill \\\\ \\hfill &amp; =100{e}^{-0.173\\left(3\\right)} &amp; \\text{Substitute known values for }a, r,\\text{ and }t.\\hfill \\\\ \\hfill &amp; \\approx 59.5115\\hfill &amp; \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/p>\r\nSo [latex]59.5115[\/latex] mg of radon-[latex]222[\/latex] will remain.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321422[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]321424[\/ohm_question]<\/section>","rendered":"<h2>Investigating Continuous Growth<\/h2>\n<p>So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, [latex]e[\/latex] is used as the base for exponential functions. Exponential models that use [latex]e[\/latex] as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>the continuous growth\/decay formula<\/h3>\n<p>For all real numbers [latex]t[\/latex], and all positive numbers [latex]a[\/latex] and [latex]r[\/latex], continuous growth or decay is represented by the formula<\/p>\n<p style=\"text-align: center;\">[latex]A(t) = Pe^{rt}[\/latex]<\/p>\n<p>where<\/p>\n<ul>\n<li>[latex]P[\/latex] is the initial value or the principal<\/li>\n<li>[latex]r[\/latex] is the growth or interest rate per unit time<\/li>\n<li>[latex]t[\/latex] is the period or term of the investment<\/li>\n<\/ul>\n<p>If [latex]r \\gt 0[\/latex] <span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">then the formula represents continuous growth. <\/span><\/p>\n<p><span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">If [latex]r \\lt 0[\/latex] then the formula represents continuous decay.<\/span><\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given the initial value, rate of growth or decay, and time <em>[latex]t[\/latex]<\/em>, solve a continuous growth or decay function<\/strong><\/p>\n<ol>\n<li>Use the information in the problem to determine [latex]P[\/latex], the initial value of the function.<\/li>\n<li>Use the information in the problem to determine the growth rate [latex]r[\/latex].\n<ul>\n<li>If the problem refers to continuous growth, then [latex]r\u00a0> 0[\/latex].<\/li>\n<li>If the problem refers to continuous decay, then [latex]r\u00a0< 0[\/latex].<\/li>\n<\/ul>\n<\/li>\n<li>Use the information in the problem to determine the time [latex]t[\/latex].<\/li>\n<li>Substitute the given information into the continuous growth formula and solve for [latex]A(t)[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">A person invested [latex]$1,000[\/latex] in an account earning a nominal [latex]10\\%[\/latex] per year compounded continuously. How much was in the account at the end of one year?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q864251\">Show Solution<\/button><\/p>\n<div id=\"q864251\" class=\"hidden-answer\" style=\"display: none\">Since the account is growing in value, this is a continuous compounding problem with growth rate [latex]r\u00a0= 0.10[\/latex]. The initial investment was [latex]$1,000[\/latex], so [latex]P\u00a0= 1000[\/latex]. We use the continuous compounding formula to find the value after [latex]t\u00a0= 1[\/latex] year:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =P{e}^{rt}\\hfill & \\text{Use the continuous compounding formula}.\\hfill \\\\ \\hfill & =1000{\\left(e\\right)}^{0.1} & \\text{Substitute known values for }P, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 1105.17\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/p>\n<p>The account is worth [latex]$1,105.17[\/latex] after one year.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321421\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321421&theme=lumen&iframe_resize_id=ohm321421&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Radon-222 decays at a continuous rate of [latex]17.3 \\%[\/latex] per day. How much will [latex]100[\/latex] mg of Radon-[latex]222[\/latex] decay to in [latex]3[\/latex] days?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q660828\">Show Solution<\/button><\/p>\n<div id=\"q660828\" class=\"hidden-answer\" style=\"display: none\">Since the substance is decaying, the rate, [latex]17.3%[\/latex], is negative. So, [latex]r=-0.173[\/latex]. The initial amount of radon-[latex]222[\/latex] was [latex]100[\/latex] mg, so [latex]a= 100[\/latex]. We use the continuous decay formula to find the value after [latex]t= 3[\/latex] days:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =a{e}^{rt}\\hfill & \\text{Use the continuous growth formula}.\\hfill \\\\ \\hfill & =100{e}^{-0.173\\left(3\\right)} & \\text{Substitute known values for }a, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 59.5115\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/p>\n<p>So [latex]59.5115[\/latex] mg of radon-[latex]222[\/latex] will remain.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321422\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321422&theme=lumen&iframe_resize_id=ohm321422&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm321424\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=321424&theme=lumen&iframe_resize_id=ohm321424&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":13,"menu_order":16,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":105,"module-header":"- Select 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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3820"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3820\/revisions"}],"predecessor-version":[{"id":5883,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3820\/revisions\/5883"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/105"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3820\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3820"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3820"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3820"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3820"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}