{"id":3169,"date":"2025-08-15T22:27:01","date_gmt":"2025-08-15T22:27:01","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=3169"},"modified":"2026-02-24T21:36:08","modified_gmt":"2026-02-24T21:36:08","slug":"exponential-and-logarithmic-models-apply-it","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/exponential-and-logarithmic-models-apply-it\/","title":{"raw":"Exponential and Logarithmic Models: Apply It","rendered":"Exponential and Logarithmic Models: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Model exponential growth and decay.<\/li>\r\n \t<li>Use Newton\u2019s Law of Cooling.<\/li>\r\n \t<li>Use logistic-growth models.<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox recall\" aria-label=\"Recall\">The general form for an exponential growth model is [latex]A=A_0e^{rt}[\/latex]. (Growth if [latex]r&gt;0[\/latex], decay if [latex]r&lt;0[\/latex].)<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p data-start=\"731\" data-end=\"778\">A population is measured in regular time steps:<\/p>\r\n\r\n<div class=\"contain-inline-size rounded-2xl relative bg-token-sidebar-surface-primary\">\r\n<div class=\"sticky top-9\">\r\n<div class=\"absolute end-0 bottom-0 flex h-9 items-center pe-2\">\r\n<div class=\"bg-token-bg-elevated-secondary text-token-text-secondary flex items-center gap-4 rounded-sm px-2 font-sans text-xs\">\r\n<table style=\"border-collapse: collapse; width: 100%;\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 50%;\">[latex]t[\/latex]<\/td>\r\n<td style=\"width: 50%;\">[latex]A(t)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">0<\/td>\r\n<td style=\"width: 50%;\">500<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">5<\/td>\r\n<td style=\"width: 50%;\">402<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">10<\/td>\r\n<td style=\"width: 50%;\">323<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%;\">15<\/td>\r\n<td style=\"width: 50%;\">260<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p data-start=\"940\" data-end=\"968\">Write the exponential model.<\/p>\r\n<p data-start=\"940\" data-end=\"968\">[reveal-answer q=\"639255\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"639255\"]From the y-intercept: [latex]A_0=500[\/latex]. Use the second point to solve for [latex]r[\/latex]: [latex]500e^{5r}=402 ;\\Rightarrow; e^{5r}=\\frac{402}{500} ;\\Rightarrow; r=\\frac{1}{5}\\ln!\\Big(\\frac{402}{500}\\Big)\\approx -0.095.[\/latex] Model: [latex]A(t)=500e^{-0.095t}[\/latex]. (Equivalently, [latex]A(t)=500,(e^{-0.095})^{t}\\approx 500(0.909)^{t}[\/latex].)[\/hidden-answer]<\/p>\r\n\r\n<\/section><section class=\"textbox recall\" aria-label=\"Recall\">The general form for Newton's Law of Cooling:[latex]T(t)=A e^{kt}+T_s[\/latex], where [latex]T_s[\/latex] is the ambient temperature (horizontal asymptote), [latex]A=T_0-T_s[\/latex], and [latex]k[\/latex] is found from any additional point.<\/section><section class=\"textbox example\" aria-label=\"Example\">On a cool morning, a mug of tea is set on a patio where the air temperature is 10\u00b0C. The temperature of the tea is shown in the graph:<img class=\"alignnone wp-image-4001 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/08\/15213545\/Screenshot-2025-09-15-at-2.35.21%E2%80%AFPM.png\" alt=\"Coordinate plane with a horizontal red dotted line that cross the y-axis at 10, and a black curved arc the cross the y-axis at 80 and continues down towards the red line as the slope gets less steep. Green dot at the point (20,30) on the black curve.\" width=\"1242\" height=\"948\" \/>Write the cooling model.[reveal-answer q=\"90008\"]Show Solutions[\/reveal-answer]\r\n[hidden-answer a=\"90008\"]\r\n<ol>\r\n \t<li>Read the asymptote (ambient temperature): [latex]T_s=10[\/latex].<\/li>\r\n \t<li>Read the initial temperature: [latex]T_0=80\\Rightarrow A=T_0-T_s=70[\/latex].<\/li>\r\n \t<li>Use the point [latex](20,30)[\/latex] to solve for [latex]k[\/latex]:<\/li>\r\n<\/ol>\r\n<p style=\"padding-left: 40px;\">\\begin{align}\r\n30 &amp;= 10 + 70e^{20k} \\\\\r\n20 &amp;= 70e^{20k} \\\\\r\ne^{20k} &amp;= \\frac{2}{7} \\\\\r\n20k &amp;= \\ln\\left(\\frac{2}{7}\\right) \\\\\r\nk &amp;= \\frac{1}{20}\\ln\\left(\\frac{2}{7}\\right) \\\\\r\nk &amp;\\approx -0.0627\r\n\\end{align}<\/p>\r\n<p style=\"padding-left: 40px;\">Model: [latex]\\displaystyle T(t)=70e^{-0.0627t}+10.[\/latex]<\/p>\r\n<p style=\"padding-left: 40px;\">[\/hidden-answer]<\/p>\r\n\r\n<\/section>&nbsp;\r\n<p data-start=\"940\" data-end=\"968\"><\/p>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Model exponential growth and decay.<\/li>\n<li>Use Newton\u2019s Law of Cooling.<\/li>\n<li>Use logistic-growth models.<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox recall\" aria-label=\"Recall\">The general form for an exponential growth model is [latex]A=A_0e^{rt}[\/latex]. (Growth if [latex]r>0[\/latex], decay if [latex]r<0[\/latex].)<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p data-start=\"731\" data-end=\"778\">A population is measured in regular time steps:<\/p>\n<div class=\"contain-inline-size rounded-2xl relative bg-token-sidebar-surface-primary\">\n<div class=\"sticky top-9\">\n<div class=\"absolute end-0 bottom-0 flex h-9 items-center pe-2\">\n<div class=\"bg-token-bg-elevated-secondary text-token-text-secondary flex items-center gap-4 rounded-sm px-2 font-sans text-xs\">\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 50%;\">[latex]t[\/latex]<\/td>\n<td style=\"width: 50%;\">[latex]A(t)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">0<\/td>\n<td style=\"width: 50%;\">500<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">5<\/td>\n<td style=\"width: 50%;\">402<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">10<\/td>\n<td style=\"width: 50%;\">323<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%;\">15<\/td>\n<td style=\"width: 50%;\">260<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p data-start=\"940\" data-end=\"968\">Write the exponential model.<\/p>\n<p data-start=\"940\" data-end=\"968\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q639255\">Show Solution<\/button><\/p>\n<div id=\"q639255\" class=\"hidden-answer\" style=\"display: none\">From the y-intercept: [latex]A_0=500[\/latex]. Use the second point to solve for [latex]r[\/latex]: [latex]500e^{5r}=402 ;\\Rightarrow; e^{5r}=\\frac{402}{500} ;\\Rightarrow; r=\\frac{1}{5}\\ln!\\Big(\\frac{402}{500}\\Big)\\approx -0.095.[\/latex] Model: [latex]A(t)=500e^{-0.095t}[\/latex]. (Equivalently, [latex]A(t)=500,(e^{-0.095})^{t}\\approx 500(0.909)^{t}[\/latex].)<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox recall\" aria-label=\"Recall\">The general form for Newton&#8217;s Law of Cooling:[latex]T(t)=A e^{kt}+T_s[\/latex], where [latex]T_s[\/latex] is the ambient temperature (horizontal asymptote), [latex]A=T_0-T_s[\/latex], and [latex]k[\/latex] is found from any additional point.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">On a cool morning, a mug of tea is set on a patio where the air temperature is 10\u00b0C. The temperature of the tea is shown in the graph:<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4001 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/08\/15213545\/Screenshot-2025-09-15-at-2.35.21%E2%80%AFPM.png\" alt=\"Coordinate plane with a horizontal red dotted line that cross the y-axis at 10, and a black curved arc the cross the y-axis at 80 and continues down towards the red line as the slope gets less steep. Green dot at the point (20,30) on the black curve.\" width=\"1242\" height=\"948\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/08\/15213545\/Screenshot-2025-09-15-at-2.35.21%E2%80%AFPM.png 1242w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/08\/15213545\/Screenshot-2025-09-15-at-2.35.21%E2%80%AFPM-300x229.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/08\/15213545\/Screenshot-2025-09-15-at-2.35.21%E2%80%AFPM-1024x782.png 1024w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/08\/15213545\/Screenshot-2025-09-15-at-2.35.21%E2%80%AFPM-768x586.png 768w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/08\/15213545\/Screenshot-2025-09-15-at-2.35.21%E2%80%AFPM-65x50.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/08\/15213545\/Screenshot-2025-09-15-at-2.35.21%E2%80%AFPM-225x172.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/61\/2025\/08\/15213545\/Screenshot-2025-09-15-at-2.35.21%E2%80%AFPM-350x267.png 350w\" sizes=\"(max-width: 1242px) 100vw, 1242px\" \/>Write the cooling model.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q90008\">Show Solutions<\/button><\/p>\n<div id=\"q90008\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Read the asymptote (ambient temperature): [latex]T_s=10[\/latex].<\/li>\n<li>Read the initial temperature: [latex]T_0=80\\Rightarrow A=T_0-T_s=70[\/latex].<\/li>\n<li>Use the point [latex](20,30)[\/latex] to solve for [latex]k[\/latex]:<\/li>\n<\/ol>\n<p style=\"padding-left: 40px;\">\\begin{align}<br \/>\n30 &amp;= 10 + 70e^{20k} \\\\<br \/>\n20 &amp;= 70e^{20k} \\\\<br \/>\ne^{20k} &amp;= \\frac{2}{7} \\\\<br \/>\n20k &amp;= \\ln\\left(\\frac{2}{7}\\right) \\\\<br \/>\nk &amp;= \\frac{1}{20}\\ln\\left(\\frac{2}{7}\\right) \\\\<br \/>\nk &amp;\\approx -0.0627<br \/>\n\\end{align}<\/p>\n<p style=\"padding-left: 40px;\">Model: [latex]\\displaystyle T(t)=70e^{-0.0627t}+10.[\/latex]<\/p>\n<p style=\"padding-left: 40px;\"><\/div>\n<\/div>\n<\/section>\n<p>&nbsp;<\/p>\n<p data-start=\"940\" data-end=\"968\">\n","protected":false},"author":67,"menu_order":23,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":510,"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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3169"}],"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\/67"}],"version-history":[{"count":18,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3169\/revisions"}],"predecessor-version":[{"id":5733,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3169\/revisions\/5733"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/510"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/3169\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=3169"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=3169"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=3169"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=3169"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}