{"id":1173,"date":"2025-07-23T18:29:47","date_gmt":"2025-07-23T18:29:47","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1173"},"modified":"2025-09-15T16:53:33","modified_gmt":"2025-09-15T16:53:33","slug":"exponential-and-logarithmic-equations-apply-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/exponential-and-logarithmic-equations-apply-it-1\/","title":{"raw":"Exponential and Logarithmic Equations: Apply It 1","rendered":"Exponential and Logarithmic Equations: Apply It 1"},"content":{"raw":"<section id=\"fs-id1165137755280\"><section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use like bases to solve exponential equations.<\/li>\r\n \t<li>Use logarithms to solve exponential equations.<\/li>\r\n \t<li>Solve logarithmic equations<\/li>\r\n \t<li>Solve applied problems involving exponential and logarithmic equations.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Solve applied problems involving exponential and logarithmic equations<\/h2>\r\n<\/section><section id=\"fs-id1165137828382\">\r\n<p id=\"fs-id1165137828387\">We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/p>\r\n<p id=\"fs-id1165134192326\">One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its <strong>half-life<\/strong>. The table below\u00a0lists the half-life for several of the more common radioactive substances.<\/p>\r\n\r\n<table id=\"Table_04_06_001\" summary=\"Seven rows and three columns. The first column is labeled, \">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center;\">Substance<\/th>\r\n<th style=\"text-align: center;\">Use<\/th>\r\n<th style=\"text-align: center;\">Half-life<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>gallium-67<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>80 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>cobalt-60<\/td>\r\n<td>manufacturing<\/td>\r\n<td>5.3 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>technetium-99m<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>6 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>americium-241<\/td>\r\n<td>construction<\/td>\r\n<td>432 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>carbon-14<\/td>\r\n<td>archeological dating<\/td>\r\n<td>5,715 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>uranium-235<\/td>\r\n<td>atomic power<\/td>\r\n<td>703,800,000 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165135209669\">We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>radioactive decay<\/h3>\r\n<div id=\"eip-247\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&amp;A\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{T}t} \\\\ &amp;A\\left(t\\right)={A}_{0}{e}^{\\mathrm{ln}\\left(0.5\\right)\\frac{t}{T}} \\\\ &amp;A\\left(t\\right)={A}_{0}{\\left({e}^{\\mathrm{ln}\\left(0.5\\right)}\\right)}^{\\frac{t}{T}} \\\\ &amp;A\\left(t\\right)={A}_{0}{\\left(\\frac{1}{2}\\right)}^{\\frac{t}{T}} \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137408740\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137408743\">\r\n \t<li>[latex]{A}_{0}[\/latex] is the amount initially present<\/li>\r\n \t<li><em>T<\/em>\u00a0is the half-life of the substance<\/li>\r\n \t<li><em>t<\/em>\u00a0is the time period over which the substance is studied<\/li>\r\n \t<li><em>y<\/em>\u00a0is the amount of the substance present after time\u00a0<em>t<\/em><\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1165137628659\">How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?<\/p>\r\n[reveal-answer q=\"860588\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"860588\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=\\text{1000}e\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t \\\\ &amp;900=1000{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}&amp;&amp; \\text{After 10% decays, 900 grams are left}. \\\\ &amp;0.9={e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}&amp;&amp; \\text{Divide by 1000}. \\\\ &amp;\\mathrm{ln}\\left(0.9\\right)=\\mathrm{ln}\\left({e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\right)&amp;&amp; \\text{Take ln of both sides}. \\\\ &amp;\\mathrm{ln}\\left(0.9\\right)=\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t&amp;&amp; \\text{ln}\\left({e}^{M}\\right)=M \\\\ &amp;t=\\text{703,800,000}\\times \\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{years}&amp;&amp; \\text{Solve for }t. \\\\ &amp;t\\approx \\text{106,979,777 years}\\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137453460\">Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]312047[\/ohm_question]<\/section><section aria-label=\"Try It\"><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]312049[\/ohm_question]\r\n\r\n<\/section><\/section><\/section>","rendered":"<section id=\"fs-id1165137755280\">\n<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use like bases to solve exponential equations.<\/li>\n<li>Use logarithms to solve exponential equations.<\/li>\n<li>Solve logarithmic equations<\/li>\n<li>Solve applied problems involving exponential and logarithmic equations.<\/li>\n<\/ul>\n<\/section>\n<h2>Solve applied problems involving exponential and logarithmic equations<\/h2>\n<\/section>\n<section id=\"fs-id1165137828382\">\n<p id=\"fs-id1165137828387\">We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/p>\n<p id=\"fs-id1165134192326\">One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its <strong>half-life<\/strong>. The table below\u00a0lists the half-life for several of the more common radioactive substances.<\/p>\n<table id=\"Table_04_06_001\" summary=\"Seven rows and three columns. The first column is labeled,\">\n<thead>\n<tr>\n<th style=\"text-align: center;\">Substance<\/th>\n<th style=\"text-align: center;\">Use<\/th>\n<th style=\"text-align: center;\">Half-life<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>gallium-67<\/td>\n<td>nuclear medicine<\/td>\n<td>80 hours<\/td>\n<\/tr>\n<tr>\n<td>cobalt-60<\/td>\n<td>manufacturing<\/td>\n<td>5.3 years<\/td>\n<\/tr>\n<tr>\n<td>technetium-99m<\/td>\n<td>nuclear medicine<\/td>\n<td>6 hours<\/td>\n<\/tr>\n<tr>\n<td>americium-241<\/td>\n<td>construction<\/td>\n<td>432 years<\/td>\n<\/tr>\n<tr>\n<td>carbon-14<\/td>\n<td>archeological dating<\/td>\n<td>5,715 years<\/td>\n<\/tr>\n<tr>\n<td>uranium-235<\/td>\n<td>atomic power<\/td>\n<td>703,800,000 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165135209669\">We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>radioactive decay<\/h3>\n<div id=\"eip-247\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&A\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{T}t} \\\\ &A\\left(t\\right)={A}_{0}{e}^{\\mathrm{ln}\\left(0.5\\right)\\frac{t}{T}} \\\\ &A\\left(t\\right)={A}_{0}{\\left({e}^{\\mathrm{ln}\\left(0.5\\right)}\\right)}^{\\frac{t}{T}} \\\\ &A\\left(t\\right)={A}_{0}{\\left(\\frac{1}{2}\\right)}^{\\frac{t}{T}} \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137408740\">where<\/p>\n<ul id=\"fs-id1165137408743\">\n<li>[latex]{A}_{0}[\/latex] is the amount initially present<\/li>\n<li><em>T<\/em>\u00a0is the half-life of the substance<\/li>\n<li><em>t<\/em>\u00a0is the time period over which the substance is studied<\/li>\n<li><em>y<\/em>\u00a0is the amount of the substance present after time\u00a0<em>t<\/em><\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1165137628659\">How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q860588\">Show Solution<\/button><\/p>\n<div id=\"q860588\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=\\text{1000}e\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t \\\\ &900=1000{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}&& \\text{After 10% decays, 900 grams are left}. \\\\ &0.9={e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}&& \\text{Divide by 1000}. \\\\ &\\mathrm{ln}\\left(0.9\\right)=\\mathrm{ln}\\left({e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\right)&& \\text{Take ln of both sides}. \\\\ &\\mathrm{ln}\\left(0.9\\right)=\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t&& \\text{ln}\\left({e}^{M}\\right)=M \\\\ &t=\\text{703,800,000}\\times \\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{years}&& \\text{Solve for }t. \\\\ &t\\approx \\text{106,979,777 years}\\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137453460\">Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm312047\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=312047&theme=lumen&iframe_resize_id=ohm312047&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section aria-label=\"Try It\">\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm312049\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=312049&theme=lumen&iframe_resize_id=ohm312049&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/section>\n<\/section>\n<\/section>\n","protected":false},"author":13,"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":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\/1173"}],"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":4,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1173\/revisions"}],"predecessor-version":[{"id":3996,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1173\/revisions\/3996"}],"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\/1173\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1173"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1173"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1173"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1173"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}