{"id":456,"date":"2019-03-06T22:07:05","date_gmt":"2019-03-06T22:07:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/exponential-and-logarithmic-equations\/"},"modified":"2020-01-19T19:06:05","modified_gmt":"2020-01-19T19:06:05","slug":"exponential-and-logarithmic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/exponential-and-logarithmic-equations\/","title":{"raw":"2.5 Exponential and Logarithmic Equations","rendered":"2.5 Exponential and Logarithmic Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nIn this section, you will:\r\n<ul>\r\n \t<li>Use logarithms to solve exponential equations.<\/li>\r\n \t<li>Use the definition of a logarithm to solve logarithmic equations.<\/li>\r\n \t<li>Solve applied problems involving exponential and logarithmic equations.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"CNX_Precalc_Figure_04_06_001\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/06220634\/CNX_Precalc_Figure_04_06_001.jpg\" alt=\"Seven rabbits in front of a brick building.\" width=\"488\" height=\"324\" \/> <strong>Figure 1.\u00a0<\/strong>Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the \u201crabbit plague.\u201d (credit: Richard Taylor, Flickr)[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137871518\">In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.<\/p>\r\n<p id=\"fs-id1165135695212\">Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential equations.<\/p>\r\n\r\n<h3>Solving Exponential Equations Using Logarithms<\/h3>\r\nMany times we need to solve equations of the form [latex]n=ab^x[\/latex] where [latex]n[\/latex] is a real number.\u00a0 To do this we will divide both sides by [latex]a[\/latex] to get\u00a0[latex]\\frac{n}{a}=b^x[\/latex] and then take the logarithm of both sides giving the equation\u00a0[latex]\\mathrm{log}\\left(\\frac{n}{a}\\right)=\\mathrm{log}\\left(b^x\\right).[\/latex] Next, we use the power rule for logarithms to get\u00a0[latex]\\mathrm{log}\\left(\\frac{n}{a}\\right)=x\\mathrm{log}\\left(b\\right)[\/latex].\u00a0 Finally, divide both sides by\u00a0[latex]\\mathrm{log}\\left(b\\right)[\/latex]\u00a0 to get\u00a0[latex]x=\\frac{\\mathrm{log}\\left(\\frac{n}{a}\\right)}{\\mathrm{log}\\left(b\\right)}[\/latex].\u00a0 Note that any base for the logarithm can be used but base 10 and base [latex]e[\/latex] are most commonly used.\r\n<div class=\"textbox examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137673524\"><strong>Given an exponential equation in the form [latex]n=ab^x[\/latex], solve for the unknown.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137784632\" type=\"1\">\r\n \t<li>Divide both sides by [latex]a[\/latex] or the initial condition.<\/li>\r\n \t<li>Apply the logarithm of both sides of the equation.\r\n<ul id=\"fs-id1165137824134\">\r\n \t<li>If one of the terms in the equation has base 10, use the common logarithm.<\/li>\r\n \t<li>If none of the terms in the equation has base 10, use the natural logarithm.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Use the rules of logarithms to solve for the unknown.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 1:\u00a0 Solving a Basic Exponential Equation<\/h3>\r\nSolve [latex]5=3\\left(2\\right)^x.[\/latex]\r\n\r\n[reveal-answer q=\"280679\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"280679\"]\r\n\r\nBegin by dividing both sides by 3 to get\r\n<p style=\"text-align: center\">[latex]5\/3=\\left(2\\right)^x.[\/latex]<\/p>\r\nThen, take the natural logarithm of both sides to get the equation\r\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left(5\/3\\right)=\\mathrm{ln}\\left(2^x\\right).[\/latex]<\/p>\r\nUse the power rule of logarithms [latex]\\mathrm{ln}\\left(b^x\\right)=x\\mathrm{ln}\\left(b\\right)[\/latex] to get\r\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left(5\/3\\right)=x\\mathrm{ln}\\left(2\\right).[\/latex]<\/p>\r\nFinally, divide both sides by [latex]\\mathrm{ln}\\left(2\\right)[\/latex] to get\r\n<p style=\"text-align: center\">[latex]x=\\frac{\\mathrm{ln}\\left(5\/3\\right)}{\\mathrm{ln}\\left(2\\right)}\\approx0.7370.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It #1<\/h3>\r\nSolve [latex]7=15\\left(4\\right)^x.[\/latex]\r\n\r\n[reveal-answer q=\"261872\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"261872\"]\r\n\r\n[latex]x=\\frac{\\mathrm{ln}\\left(7\/15\\right)}{\\mathrm{ln}\\left(4\\right)}\\approx-0.5498.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 2:\u00a0 Solving an Exponential Equation with an Algebraic Expression in the Exponent<\/h3>\r\nSolve [latex]15=3\\left(0.5\\right)^{x+1}.[\/latex]\r\n\r\n[reveal-answer q=\"346617\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"346617\"]\r\n\r\nSince the algebraic expression is in the exponent, this equation will be solved using the same steps as above but the x+1 should be kept in parenthesis until we are ready to simplify at the end.\u00a0 Begin by dividing both sides by 3 to get\r\n<p style=\"text-align: center\">[latex]5=\\left(0.5\\right)^{\\left(x+1\\right)}.[\/latex]<\/p>\r\nThen, take the natural logarithm of both sides to get the equation\r\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left(5\\right)=\\mathrm{ln}\\left(0.5^{\\left(x+1\\right)}\\right).[\/latex]<\/p>\r\nUse the power rule of logarithms [latex]\\mathrm{ln}\\left(b^x\\right)=x\\mathrm{ln}\\left(b\\right)[\/latex] so\r\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left(5\\right)={\\left(x+1\\right)}\\mathrm{ln}\\left(0.5\\right).[\/latex]<\/p>\r\nDivide both sides of the equation by [latex]\\mathrm{ln}\\left(0.5\\right)[\/latex] to get the equation\r\n<p style=\"text-align: center\">[latex]\\left(x+1\\right)=\\frac{\\mathrm{ln}\\left(5\\right)}{\\mathrm{ln}\\left(0.5\\right)}.[\/latex]<\/p>\r\nNote that this is just a linear equation so we subtract 1 from both sides and the solution is\r\n<p style=\"text-align: center\">[latex]x=\\frac{\\mathrm{ln}\\left(5\\right)}{\\mathrm{ln}\\left(0.5\\right)}-1\\approx-3.3219.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It #2<\/h3>\r\nSolve [latex]15=2\\left(7\\right)^{x^2+1}.[\/latex]\r\n\r\n[reveal-answer q=\"653920\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"653920\"]\r\n\r\n[latex]x=\\pm\\sqrt{\\frac{\\mathrm{ln}\\left(7.5\\right)}{\\mathrm{ln}\\left(7\\right)}-1}\\approx\\pm0.1883[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h4>Equations Containing <em><span class=\"e2\">e<\/span><\/em><\/h4>\r\n<p id=\"fs-id1165137606150\">One common type of exponential equations uses base [latex]e.[\/latex] This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base [latex]e[\/latex] on either side, we can use the <span class=\"no-emphasis\">natural logarithm<\/span> to solve it.<\/p>\r\n\r\n<div id=\"fs-id1165137727175\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137656412\"><strong>Given an equation of the form [latex]y=a{e}^{kt}\\text{,}[\/latex] solve for [latex]t.[\/latex]<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165135189859\" type=\"1\">\r\n \t<li>Divide both sides of the equation by [latex]a.[\/latex]<\/li>\r\n \t<li>Apply the natural logarithm of both sides of the equation.<\/li>\r\n \t<li>Divide both sides of the equation by [latex]k.[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_06_06\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137846466\">\r\n<div id=\"fs-id1165137846468\">\r\n<h3>Example 3:\u00a0 Solve an Equation with Continuous Growth<\/h3>\r\n<p id=\"fs-id1165137581969\">Solve [latex]100=20{e}^{2t}.[\/latex]<\/p>\r\n[reveal-answer q=\"538063\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"538063\"]\r\n\r\nFor this problem, use the natural logarithm since the equation contains base [latex]e.[\/latex]\r\n<p style=\"text-align: center\">[latex]\\begin{align*}100&amp;=20{e}^{2t}&amp;&amp; \\hfill \\\\5&amp;={e}^{2t}&amp;&amp;\\textrm{Divide both sides by 20}\\text{.}\\hfill \\\\ \\mathrm{ln}\\left(5\\right)&amp;=2t&amp;&amp;\\textrm{Take the natural logarithm of both sides.}\\\\ \\text{ }&amp;\\text{ } &amp;&amp;\\text{Use the fact that }\\mathrm{ln}\\left(e^{x}\\right)=x\\text{.}\\hfill \\\\t&amp;=\\frac{\\mathrm{ln}\\left(5\\right)}{2}&amp;&amp;\\textrm{Divide by the coefficient of }t\\text{.}\\hfill \\end{align*}[\/latex]<\/p>\r\n\r\n<h3>Analysis<\/h3>\r\nUsing laws of logs, we can also write this answer in the form [latex]t=\\mathrm{ln}\\sqrt[\\leftroot{1}\\uproot{2} ]{5}.[\/latex] If we want a decimal approximation of the answer, we use a calculator.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135186732\" class=\"precalculus tryit\">\r\n<h3>Try it #3<\/h3>\r\n<div id=\"ti_04_06_06\">\r\n<div id=\"fs-id1165137737732\">\r\n<p id=\"fs-id1165137431561\">Solve [latex]3{e}^{0.5t}=11.[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1165137726884\"]Show Solution[\/reveal-answer][hidden-answer a=\"fs-id1165137726884\"]\r\n<p id=\"fs-id1165137698501\">[latex]t=2\\mathrm{ln}\\left(\\frac{11}{3}\\right)[\/latex] or [latex]\\mathrm{ln}\\left({\\left(\\frac{11}{3}\\right)}^{2}\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137748966\" class=\"bc-section section\">\r\n<div id=\"fs-id1165137667260\" class=\"bc-section section\">\r\n<div id=\"fs-id1165135369638\" class=\"precalculus qa key-takeaways\">\r\n<h3>Q&amp;A<\/h3>\r\n<p id=\"fs-id1165137547216\"><strong>Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?<\/strong><\/p>\r\n<p id=\"fs-id1165137435646\"><em>No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined.\u00a0<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_04_06_04\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137405247\">\r\n<div id=\"fs-id1165137635295\">\r\n<h3>Example 4:\u00a0 Solving an Equation with Positive and Negative Powers<\/h3>\r\n<p id=\"fs-id1165137805073\">Solve [latex]{3}^{x+1}=-2.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137723720\">[reveal-answer q=\"fs-id1165137723720\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137723720\"]\r\n<p id=\"fs-id1165137874850\">This equation has no solution. There is no real value of [latex]x[\/latex] that will make the equation a true statement because any power of a positive number is positive.<\/p>\r\n<p id=\"fs-id1165137578263\"><a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_06_002\">Figure 2<\/a> shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_06_002\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"330\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/06220637\/CNX_Precalc_Figure_04_06_002.jpg\" alt=\"Graph of 3^(x+1)=-2 and y=-2. The graph notes that they do not cross.\" width=\"330\" height=\"297\" \/> Figure 2[\/caption]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137849213\">\r\n<div id=\"CNX_Precalc_Figure_04_06_002\" class=\"small\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137920668\" class=\"precalculus tryit\">\r\n<h3>Try it #4<\/h3>\r\n<div id=\"ti_04_06_04\">\r\n<div id=\"fs-id1165137761184\">\r\n<p id=\"fs-id1165137847264\">Solve [latex]{2}^{x}=-100.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137844325\">[reveal-answer q=\"fs-id1165137844325\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137844325\"]\r\n<p id=\"fs-id1165135208714\">The equation has no solution.\u00a0 We know that\u00a0[latex]{2}^{x}[\/latex] is always greater than 0.\u00a0 Therefore, there is no value for the input [latex] x [\/latex] which produces an output of\u00a0 -100.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137641700\" class=\"bc-section section\">\r\n<h3><\/h3>\r\n<p id=\"fs-id1165137939689\">Sometimes there are exponential functions on both sides of the equation such as [latex]ab^x=cd^x[\/latex].\u00a0 The process is similar to solving [latex]n=ab^x[\/latex] but we will need to be careful to use the product rule of logarithms before applying the power rule of logarithms.<\/p>\r\n\r\n<div id=\"fs-id1165137939925\" class=\"precalculus howto examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1165137673524\"><strong>Given an equation with exponential expression on each side, solve for the unknown.<\/strong><\/p>\r\n\r\n<ol id=\"fs-id1165137784632\" type=\"1\">\r\n \t<li>Divide both sides by one of the initial conditions.<\/li>\r\n \t<li>Apply the logarithm of both sides of the equation.\r\n<ul id=\"fs-id1165137824134\">\r\n \t<li>If one of the terms in the equation has base 10, use the common logarithm.<\/li>\r\n \t<li>If none of the terms in the equation has base 10, use the natural logarithm.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Use the product and power rules of logarithms and then solve for the unknown.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_04_06_05\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137653913\">\r\n<div id=\"fs-id1165137715230\">\r\n<h3>Example 5:\u00a0 Solving an Equation Containing Powers of Different Bases<\/h3>\r\n<p id=\"fs-id1165137549801\">Solve [latex]6\\left(5\\right)^{x+2}=2\\left(4\\right)^{x}.[\/latex]<\/p>\r\n[reveal-answer q=\"336145\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"336145\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align*}6\\left(5\\right)^{x+2}&amp;=2\\left(4\\right)^x&amp;&amp;\\text{ }\\\\3\\left(5\\right)^{x+2}&amp;=\\left(4\\right)^x&amp;&amp;\\text{Divide both sides by 2.}\\\\\\mathrm{ln}\\left(3\\left(5\\right)^{x+2}\\right)&amp;=\\mathrm{ln}\\left(4^x\\right)&amp;&amp;\\text{Take the natural logarithm of both sides.}\\\\\\mathrm{ln}\\left(3\\right)+\\mathrm{ln}\\left(5^{x+2}\\right)&amp;=\\mathrm{ln}\\left(4^x\\right)&amp;&amp;\\text{Use the product rule of logarithms.}\\\\\\mathrm{ln}\\left(3\\right)+\\left(x+2\\right)\\mathrm{ln}\\left(5\\right)&amp;=x\\mathrm{ln}\\left(4\\right)&amp;&amp;\\text{Use the power law of logarithms.}\\\\\\mathrm{ln}\\left(3\\right)+x\\mathrm{ln}\\left(5\\right)+2\\mathrm{ln}\\left(5\\right)&amp;=x\\mathrm{ln}\\left(4\\right)&amp;&amp;\\text{Use the distributive law.}\\\\x\\mathrm{ln}\\left(5\\right)-x\\mathrm{ln}\\left(4\\right)&amp;=-2\\mathrm{ln}\\left(5\\right)-\\mathrm{ln}\\left(3\\right)&amp;&amp;\\text{Move terms containing x on one side, }\\\\&amp;\\text{}&amp;&amp;\\text{and terms without x on the other.}\\\\x\\left(\\mathrm{ln}\\left(5\\right)-\\mathrm{ln}\\left(4\\right)\\right)&amp;=-2\\mathrm{ln}\\left(5\\right)-\\mathrm{ln}\\left(3\\right)&amp;&amp; \\text{On the left hand side, factor out an }x.\\\\x\\mathrm{ln}\\left(\\frac{5}{4}\\right)&amp;=\\mathrm{ln}\\left(\\frac{1}{75}\\right)&amp;&amp;\\text{Use the laws of logarithms.}\\\\x&amp;=\\frac{\\mathrm{ln}\\left(\\frac{1}{75}\\right)}{\\mathrm{ln}\\left(\\frac{5}{4}\\right)}&amp;&amp; \\text{Divide by the coefficient of }x.\\\\\\end{align*}[\/latex]<\/p>\r\n\r\n<h3>Analysis<\/h3>\r\nNotice that the product rule of logarithms was used before the power rule because the power rule cannot be applied to [latex]ab^x[\/latex]; [latex]a[\/latex] is not raised to the [latex]x[\/latex] power.\u00a0 Also, note that when the power rule is used on [latex]\\mathrm{ln}\\left(5^{x+2}\\right)[\/latex] the [latex]\\left(x+2\\right)[\/latex] has parenthesis so that the [latex]\\mathrm{ln}\\left(5\\right)[\/latex] gets properly distributed.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137473583\" class=\"precalculus tryit\">\r\n<h3>Try it #5<\/h3>\r\n<div id=\"ti_04_06_05\">\r\n<div id=\"fs-id1165137402173\">\r\n<p id=\"fs-id1165137402175\">Solve [latex]{2}^{x}={3}^{x+1}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137605315\">[reveal-answer q=\"fs-id1165137605315\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137605315\"]\r\n[latex]x=\\frac{\\mathrm{ln}\\left(3\\right)}{\\mathrm{ln}\\left(\\frac{2}{3}\\right)}[\/latex]\r\n[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137469838\" class=\"bc-section section\">\r\n<div id=\"fs-id1165137532335\" class=\"precalculus qa key-takeaways\">\r\n<h3>Q&amp;A<\/h3>\r\n<p id=\"eip-id1165135593037\"><strong>Does every equation of the form\u00a0<\/strong>[latex]y=A{e}^{kt}[\/latex]\u00a0<strong>have a solution?<\/strong><\/p>\r\n<p id=\"fs-id1165137423829\"><em>No. There is a solution when [latex]k\\ne 0,[\/latex] and when [latex]y[\/latex] and [latex]A[\/latex] are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is [latex]2=-3{e}^{t}.[\/latex]<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_04_06_07\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137828228\">\r\n<div id=\"fs-id1165137704438\">\r\n<h3>Example 6:\u00a0 Solving an Equation That Requires Algebra First<\/h3>\r\n<p id=\"fs-id1165134223341\">Solve<\/p>\r\n\r\n<ol>\r\n \t<li>[latex]4{e}^{2x}+5=12.[\/latex]<\/li>\r\n \t<li>[latex]{e}^{2t}-3=-4{e}^{2t}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1165137675352\">[reveal-answer q=\"fs-id1165137675352\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137675352\"]\r\n<ol>\r\n \t<li style=\"text-align: center\"><span style=\"font-size: 1rem;text-align: initial\">[latex]\\\\[\/latex][latex]\\begin{align*}4{e}^{2x}+5&amp;=12&amp;&amp;\\text{ }\\\\4{e}^{2x}&amp;=7&amp;&amp;\\text{Combine like terms.}\\\\e^{2x}&amp;=\\frac{7}{4}&amp;&amp;\\text{Divide by the coefficient.}\\\\2x&amp;=\\mathrm{ln}\\left(\\frac{7}{4}\\right)&amp;&amp;\\text{Take the natural logarithm of both sides.}\\\\x&amp;=\\frac{1}{2}\\mathrm{ln}\\left(\\frac{7}{4}\\right)&amp;&amp;\\text{Solve for }x. \\end{align*}[\/latex]<\/span><\/li>\r\n \t<li style=\"text-align: center\">[latex]\\\\[\/latex][latex]\\begin{align*}{e}^{2t}-3&amp;=-4{e}^{2t}&amp;&amp;\\text{ }\\\\5{e}^{2t}-3&amp;=0&amp;&amp;\\text{Add }4{e}^{2t}\\text{ to both sides.}\\\\5{e}^{2t}&amp;=3&amp;&amp;\\text{Add 3 to both sides.}\\\\{e}^{2t}&amp;=\\frac{3}{5}&amp;&amp;\\text{Divide both sides by 5.}\\\\2t&amp;=\\mathrm{ln}\\left(\\frac{3}{5}\\right)&amp;&amp;\\text{Take the natural logarithm of both sides.}\\\\t&amp;=\\frac{1}{2}\\mathrm{ln}\\left(\\frac{3}{5}\\right)&amp;&amp;\\text{Solve for }x.\\\\\\end{align*}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1165137828228\">\r\n<div id=\"fs-id1165137675352\">\r\n<div class=\"unnumbered\" style=\"text-align: left\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137827163\" class=\"precalculus tryit\">\r\n<h3>Try it #6<\/h3>\r\n<div id=\"ti_04_06_07\">\r\n<div id=\"fs-id1165137936638\">\r\n<p id=\"fs-id1165137936640\">Solve [latex]3+{e}^{2t}=7{e}^{2t}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137819982\">[reveal-answer q=\"fs-id1165137819982\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137819982\"]\r\n<p id=\"fs-id1165137819984\">[latex]t=\\mathrm{ln}\\left(\\frac{1}{\\sqrt[\\leftroot{1}\\uproot{2} ]{2}}\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(2\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137665482\" class=\"bc-section section\">\r\n<h4>Extraneous Solutions<\/h4>\r\n<p id=\"fs-id1165137742403\">Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.<\/p>\r\nThese extraneous solutions frequently occur when the exponential function is a quadratic form. Recall that quadratic equations can be solved by factoring and setting each factor equal to zero, or by the quadratic equation.\u00a0 W will look for the pattern of the quadratic and then choose which technique can most easily be used.\r\n<div id=\"Example_04_06_08\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137828517\">\r\n<div id=\"fs-id1165137828519\">\r\n<h3>Example 7:\u00a0 Solving Exponential Functions in Quadratic Form<\/h3>\r\n<p id=\"fs-id1165137443096\">Solve [latex]{e}^{2x}-{e}^{x}=56.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div>\r\n<div id=\"eip-id1165133221795\" class=\"unnumbered\" style=\"text-align: left\">[reveal-answer q=\"884164\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"884164\"]We first re-write the equation using the fact that [latex]e^{2x}={\\left({e^x}\\right)}^2.[\/latex] Notice that this equation has the form [latex]e^x[\/latex] squared minus\u00a0[latex]e^x[\/latex] equals 56. This is a quadratic equation which can be solved by factoring.\r\n<p style=\"text-align: center\">[latex]\\begin{align*}{e}^{2x}-{e}^{x}&amp;=56&amp;&amp; \\hfill \\\\ {e}^{2x}-{e}^{x}-56&amp;=0&amp;&amp;\\text{Get one side of the equation equal to zero}.\\hfill \\\\ \\left({e}^{x}+7\\right)\\left({e}^{x}-8\\right)&amp;=0&amp;&amp;\\text{Factor by the FOIL method}.\\hfill \\\\{e}^{x}+7&amp;=0\\text{ or }{e}^{x}-8=0&amp;&amp; \\text{If a product is zero, then one factor must be zero}.\\hfill \\\\{e}^{x}&amp;=-7{\\text{ or e}}^{x}=8&amp;&amp; \\text{Isolate the exponentials}.\\hfill \\\\ {e}^{x}&amp;=8&amp;&amp;\\text{Reject the equation that has no solution}.\\hfill \\\\x&amp;=\\mathrm{ln}\\left(8\\right)&amp;&amp;\\text{Write as a logarmithm}.\\hfill \\end{align*}[\/latex]<\/p>\r\n\r\n<h3>Analysis<\/h3>\r\n<div>When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation [latex]{e}^{x}=-7[\/latex] because a positive number never equals a negative number. The solution [latex]\\mathrm{ln}\\left(-7\\right)[\/latex] is not a real number, and in the real number system this solution is rejected as an extraneous solution.<\/div>\r\n<div class=\"unnumbered\" style=\"text-align: left\">[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135560773\" class=\"precalculus tryit\">\r\n<h3>Try it #7<\/h3>\r\n<div id=\"ti_04_06_08\">\r\n<div id=\"fs-id1165134378706\">\r\n<p id=\"fs-id1165134378708\">Solve [latex]{e}^{2x}={e}^{x}+2.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135638513\">[reveal-answer q=\"fs-id1165135638513\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135638513\"]\r\n<p id=\"fs-id1165135638516\">[latex]x=\\mathrm{ln}2.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137862548\" class=\"bc-section section\">\r\n<h3>Using the Definition of a Logarithm to Solve Logarithmic Equations<\/h3>\r\n<p id=\"fs-id1165137862553\">We have already seen that the\u00a0<span class=\"no-emphasis\">logarithmic equation\u00a0<\/span>[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equivalent to the exponential equation [latex]{b}^{y}=x,[\/latex] for [latex]x&gt;0.[\/latex] We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 8:\u00a0 Solving a Logarithmic Equation<\/h3>\r\n<p id=\"fs-id1165134148350\">Solve the equation [latex]{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x-5\\right)=3.[\/latex]<\/p>\r\n[reveal-answer q=\"809039\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"809039\"]\r\n\r\nTo solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for [latex]x:[\/latex]\r\n<div id=\"eip-id2205910\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{log}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x-5\\right)&amp;=3&amp;&amp;\\text{ }\\\\\\mathrm{log}_{2}\\left(2\\left(3x-5\\right)\\right)&amp;=3&amp;&amp;\\text{Apply the product rule of logarithms}.\\\\{\\mathrm{log}}_{2}\\left(6x-10\\right)&amp;=3&amp;&amp;\\text{Distribute}.\\\\{2}^{3}&amp;=6x-10&amp;&amp;\\text{Apply the definition of a logarithm}.\\\\8&amp;=6x-10&amp;&amp;\\text{Calculate }{2}^{3}.\\\\18&amp;=6x&amp;&amp;\\text{Add 10 to both sides}.\\\\x&amp;=3&amp;&amp; \\text{Divide by 6}.\\end{align*}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_06_09\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137841585\">\r\n<div id=\"fs-id1165137725474\">\r\n<h3>Example 9:\u00a0 Using Algebra to Solve a Logarithmic Equation<\/h3>\r\n<p id=\"fs-id1165135152135\">Solve [latex]2\\mathrm{ln}\\left(x\\right)+3=7.[\/latex]<\/p>\r\n[reveal-answer q=\"414654\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"414654\"]\r\n<p style=\"text-align: center\">[latex]\\begin{align*}2\\mathrm{ln}\\left(x\\right)+3&amp;=7&amp;&amp;\\text{ }\\\\2\\mathrm{ln}\\left(x\\right)&amp;=4&amp;&amp;\\text{Subtract 3}.\\\\\\mathrm{ln}\\left(x\\right)&amp;=2&amp;&amp;\\text{Divide by 2}.\\\\x&amp;={e}^{2}&amp;&amp;\\text{Rewrite in exponential form}.\\end{align*}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n<div id=\"eip-id1165135466384\" class=\"unnumbered\" style=\"text-align: center\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137571063\" class=\"precalculus tryit\">\r\n<h3>Try it #8<\/h3>\r\n<div id=\"ti_04_06_09\">\r\n<div id=\"fs-id1165137437378\">\r\n<p id=\"fs-id1165137437380\">Solve [latex]6+\\mathrm{ln}\\left(x\\right)=10.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135209084\">[reveal-answer q=\"fs-id1165135209084\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165135209084\"]\r\n<p id=\"fs-id1165137673564\">[latex]x={e}^{4}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_06_10\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137482839\">\r\n<div id=\"fs-id1165137557104\">\r\n<h3>Example 10:\u00a0 Using Algebra Before and After Using the Definition of the Natural Logarithm<\/h3>\r\n<p id=\"fs-id1165137557109\">Solve [latex]2\\mathrm{ln}\\left(6x\\right)=7.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137501502\">[reveal-answer q=\"fs-id1165137501502\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137501502\"]\r\n<div id=\"eip-id1165135388424\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\mathrm{ln}\\left(6x\\right)&amp;=7&amp;&amp;\\text{ }\\\\\\mathrm{ln}\\left(6x\\right)&amp;=\\frac{7}{2}&amp;&amp; \\text{Divide by 2}.\\\\6x&amp;={e}^{\\left(\\frac{7}{2}\\right)}&amp;&amp;\\text{Use the definition of the natural logarithm}.\\\\x&amp;=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}&amp;&amp;\\text{Divide by 6}.\\end{align*}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"precalculus tryit\">\r\n<h3>Try it #9<\/h3>\r\n<div id=\"ti_04_06_10\">\r\n<div id=\"fs-id1165137862628\">\r\n<p id=\"fs-id1165137862630\">Solve [latex]2\\mathrm{ln}\\left(x+1\\right)=10.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137530181\">[reveal-answer q=\"fs-id1165137530181\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137530181\"]\r\n<p id=\"fs-id1165137530183\">[latex]x={e}^{5}-1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_04_06_11\" class=\"textbox examples\">\r\n<div id=\"fs-id1165137531831\">\r\n<div id=\"fs-id1165137805003\">\r\n<h3>Example 11:\u00a0 Using a Graph to Understand the Solution to a Logarithmic Equation<\/h3>\r\n<p id=\"fs-id1165137805008\">Solve [latex]\\mathrm{ln}\\left(x\\right)=3.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div>\r\n<div class=\"wp-caption-text\">[reveal-answer q=\"409852\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"409852\"]\r\n<div class=\"wp-caption-text\">\r\n\r\nTo solve [latex]\\mathrm{ln}\\left(x\\right)=3,[\/latex] write the equation in exponential form using the definition of the natural logarithm to get [latex]x={e}^{3}.[\/latex][latex]\\\\[\/latex]\r\n<p id=\"fs-id1165137443165\"><a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_06_003\">Figure 3<\/a>\u00a0represents the graph of the equation. On the graph, the <em>x<\/em>-coordinate of the point at which the two graphs intersect is close to 20. In other words [latex]{e}^{3}\\approx 20.[\/latex] A calculator gives a better approximation: [latex]{e}^{3}\\approx 20.0855.[\/latex]<\/p>\r\n\r\n<div id=\"CNX_Precalc_Figure_04_06_003\" class=\"small\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/06220640\/CNX_Precalc_Figure_04_06_003.jpg\" alt=\"Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).\" width=\"487\" height=\"288\" \/> <strong>Figure 3.\u00a0<\/strong>The graphs of [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] and [latex]y=3[\/latex] cross at the point [latex]\\left(e^{3},3\\right),[\/latex] which is approximately (20.0855, 3).[\/caption]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137401539\" class=\"precalculus tryit\">\r\n<h3>Try it #10<\/h3>\r\n<div id=\"ti_04_06_11\">\r\n<div id=\"fs-id1165134375704\">\r\n<p id=\"fs-id1165134375706\">Use a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[\/latex] to 2 decimal places.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137633980\">[reveal-answer q=\"fs-id1165137633980\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165137633980\"]\r\n<p id=\"fs-id1165137633982\">[latex]x\\approx 9.97[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137755280\" class=\"bc-section section\"><span style=\"color: #6c64ad;font-size: 1em;font-weight: 600\">Solving Applied Problems Using Exponential Equations<\/span><\/div>\r\n<div id=\"fs-id1165137828382\" class=\"bc-section section\">\r\n<p id=\"fs-id1165137828387\">In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. 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\r\n<div class=\"textbox examples\">\r\n<h3>Example 12:\u00a0 Real World application; Depreciation<\/h3>\r\nIn 2018, a car was purchased for $32,000 and depreciates 20% per year.\u00a0 When will the car be worth $8,000?\r\n\r\n[reveal-answer q=\"310526\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"310526\"]\r\n\r\nWe begin by writing a formula for the situation.\u00a0Since a noncontinuous rate is given, we use the formula [latex]f\\left(t\\right)=ab^t.[\/latex] Let [latex]t=0[\/latex] represent 2019.\u00a0 The initial condition is [latex]a = 32000,[\/latex] the rate is [latex]r=-0.2,[\/latex] and the growth factor is [latex]b=1+r=1-0.2=0.8.[\/latex] Therefore, the formula is\r\n<p style=\"text-align: center\">[latex]f\\left(t\\right)=32000\\left(0.8\\right)^t.[\/latex]<\/p>\r\nNext, we need to know when the function's output will be 8,000.\u00a0 Set the formula equal to 8000 and solve.\r\n<p style=\"text-align: center\">[latex]\\begin{align*}8000&amp;=32000\\left(0.8\\right)^t&amp;&amp;\\text{ }\\\\0.25&amp;=0.8^t&amp;&amp;\\text{Divide both sides by 32000.}\\\\\\mathrm{ln}\\left(0.25\\right)&amp;=\\mathrm{ln}\\left(0.8^t\\right)&amp;&amp;\\text{Take the natural logarithm of both sides.}\\\\\\mathrm{ln}\\left(0.25\\right)&amp;=t\\mathrm{ln}\\left(0.8\\right)&amp;&amp;\\text{Use the power rule for logarithms.}\\\\t&amp;=\\frac{\\mathrm{ln}\\left(0.25\\right)}{\\mathrm{ln}\\left(0.8\\right)}\\approx6.21&amp;&amp;\\text{Divide both sides by }\\mathrm{ln}\\left(0.8\\right).\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\r\nIt will take approximately 6.2 years for the car to depreciate to $8,000.\u00a0 In 2024, the car will be worth $8,000.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 13:\u00a0 Real World Application; BLood Alcohol Content<\/h3>\r\nA person's blood alcohol content (BAC) is a measure of how much alcohol is in the bloodstream.\u00a0 When a person stops drinking, over time the BAC will decay exponentially.\u00a0 For a particular individual, the formula [latex]f\\left(t\\right)=0.1e^{-0.0067t}[\/latex] models their BAC over time, t, measured in minutes.\u00a0 When will their BAC be 0.04?\r\n\r\n[reveal-answer q=\"267570\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"267570\"]\r\n\r\nWe are asked to find the input when the output is 0.04 so solve the equation [latex]0.04=0.1e^{-0.0067t}.[\/latex]\r\n<p style=\"text-align: center\">[latex]\\begin{align*}0.4&amp;=e^{-0.0067 t}&amp;&amp;\\text{Divide both sides by 0.1.}\\\\\\mathrm{ln}\\left(0.4\\right)&amp;=-0.0067t&amp;&amp;\\text{Take the natural logarithm and simplify.}\\\\t&amp;=\\frac{\\mathrm{ln}\\left(0.4\\right)}{-0.0067}&amp;&amp;\\text{Divide both sides by }-0.0067.\\\\\\text{ }&amp;\\approx136.8&amp;&amp;\\text{ }\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\r\nIt will take approximately 137 minutes for the BAC to drop to 0.04 for this individual.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 14:\u00a0 Real World Application; Tuition<\/h3>\r\nCollege A is charging $40,000 tuition in 2019 and it is increasing at a continuous rate of 7% per year.\u00a0 College B charges $45,000 in 2019, but it is increasing at 4% per year.\u00a0 When will college A cost more than college B?\r\n\r\n[reveal-answer q=\"136880\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"136880\"]\r\n\r\nFirst, we determine the model representing each of the tuition amounts.\u00a0 College A has a continuous rate so use [latex]A(t)=ae^{kt}[\/latex] with [latex]a=40000[\/latex] and [latex]k=0.07.[\/latex]\u00a0 Therefore, the model for college A is [latex]A(t)=40000e^{0.07t}.[\/latex]\u00a0 College B has a noncontinuous rate so use [latex]B(t)=ab^t[\/latex] where [latex]a=45000[\/latex] and [latex]b=1+r=1+0.04=1.04.[\/latex]\u00a0 The model for college B is [latex]B(t)=45000\\left(1.04\\right)^t.[\/latex]\r\n\r\nWe need to know when these two models are equal so we set the equations equal to each other and solve.\r\n\r\n[latex]\\begin{align*}40000e^{0.07t}&amp;=45000\\left(1.04\\right)^t&amp;&amp;\\text{ }\\\\\\frac{8}{9}e^{0.07t}&amp;=1.04^t&amp;&amp;\\text{Divide both sides by 45000 and simplify.}\\\\\\mathrm{ln}\\left(\\frac{8}{9}e^{0.07t}\\right)&amp;=\\mathrm{ln}\\left(1.04^t\\right)&amp;&amp;\\text{Take the natural logarithm of both sides.}\\\\\\mathrm{ln}\\left(\\frac{8}{9}\\right)+0.07t&amp;=\\mathrm{ln}\\left(1.04^t\\right)&amp;&amp;\\text{Use the product rule and simplify.}\\\\\\mathrm{ln}\\left(\\frac{8}{9}\\right)+0.07t&amp;=t\\mathrm{ln}\\left(1.04\\right)&amp;&amp;\\text{Use the power rule on the right side.}\\\\t\\left(0.07-\\mathrm{ln}\\left(1.04\\right)\\right)&amp;=-\\mathrm{ln}\\left(\\frac{8}{9}\\right)&amp;&amp;\\text{Collect like terms and factor }t.\\\\t&amp;=\\frac{-\\mathrm{ln}\\left(\\frac{8}{9}\\right)}{0.07-\\mathrm{ln}\\left(1.04\\right)}&amp;&amp;\\text{Divide by the coefficient of }t.\\\\\\text{ }&amp;\\approx3.83.&amp;&amp;\\text{ }\\\\\\end{align*}[\/latex][latex]\\\\[\/latex]\r\n\r\nCollege B will have a higher tuition in approximately 4 years.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It #11<\/h3>\r\nA town's population is 14,000 people in 2017 and is increasing at a rate of 2.1% each year.\u00a0 When will the town's population reach 18,000 people?\r\n\r\n[reveal-answer q=\"649450\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"649450\"]In approximately 12.1 years or in 2029, the population will be 18,000 people.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Conversions Between Continuous and Noncontinuous Growth Rates<\/h3>\r\nRecall that there are two possible formulas that can be used to represent an exponential function: [latex]f(x)=ab^x=a{\\left(b\\right)}^x[\/latex] for noncontinuous growth and [latex]f(x)=ae^{kx}=a{\\left(e^k\\right)}^x[\/latex] for continuous growth.\u00a0 When comparing the two forms, we see that [latex]b=e^k.[\/latex]\u00a0 Further, since [latex]b=1+r[\/latex] where [latex]r[\/latex] is the noncontinuous growth rate, we have that [latex]1+r=e^k.[\/latex]\r\n<div class=\"textbox examples\">\r\n<h3>Example 15:\u00a0 Continuous Growth to Noncontinuous Growth<\/h3>\r\nGiven a continuous growth rate of 12%, find the noncontinuous rate.\r\n\r\n[reveal-answer q=\"522523\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"522523\"]\r\n\r\nThe continuous growth rate [latex]k[\/latex] is given as [latex]k=0.12.[\/latex]\u00a0 We solve the equation\u00a0[latex]1+r=e^k[\/latex] with this value of [latex]k[\/latex] plugged in.\r\n<p style=\"text-align: center\">[latex]\\begin{align*}1+r&amp;=e^{0.12}\\\\r&amp;=e^{0.12}-1\\approx0.1275.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\r\nThe noncontinuous rate is approximately 12.75%. Notice that since both rates are modeling the same growth, the noncontinuous rate must be slightly higher than the continuous rate, because the continuous rate allows for growth on the growth immediately.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 16:\u00a0 Noncontinuous Growth to Continuous Growth<\/h3>\r\nGiven a noncontinuous growth rate of 15%, find the continuous rate.\r\n\r\n[reveal-answer q=\"671517\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"671517\"]\r\n\r\nThe noncontinuous growth rate [latex]r[\/latex] is given as [latex]r=0.15.[\/latex] Solve the equation\u00a0[latex]1+r=e^k[\/latex] with this value of [latex]r[\/latex] plugged in.\r\n<p style=\"text-align: center\">[latex]\\begin{align*}1+0.15&amp;=e^{k}\\\\k&amp;=\\mathrm{ln}\\left(1.15\\right)\\approx0.1398.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\r\nWe took the natural logarithm of both sides to solve the equation.\r\n\r\nThe continuous rate is approximately 13.98%.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It #12<\/h3>\r\na.\u00a0 Given a continuous decreasing rate of 5%, find the noncontinuous rate.\r\n\r\nb.\u00a0 Given a noncontinuous increasing rate of 7%, find the continuous rate.\r\n\r\n[reveal-answer q=\"78291\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"78291\"]\r\n\r\na.\u00a0 The noncontinuous rate of decrease is approximately 4.877% when the continuous rate of decrease is 5%.\r\n\r\nb.\u00a0 The continuous rate of increase is approximately 6.766%, when the noncontinuous rate is 7%.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Solving Logarithmic Applications<\/h3>\r\nExponential growth and decay often involve very large or very small numbers. It is common to use a logarithmic scale when measurements result in extremely large values or extremely small values.\u00a0 The Richter Scale, pH and decibels are examples of such scales.\r\n\r\nTo describe these numbers, we often use <strong>orders of magnitude<\/strong>. The\u00a0order of magnitude\u00a0is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star,\u00a0<span class=\"no-emphasis\">Proxima Centauri<\/span>, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is 4.01134972 x 10<sup>13<\/sup>\u00a0.\u00a0So, we could describe this number as having order of magnitude of 13<span id=\"MathJax-Element-1564-Frame\" class=\"MathJax\" style=\"font-style: normal;font-weight: 400;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px;color: #555555;font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;background-color: #ffffff\" role=\"presentation\"><span id=\"MathJax-Span-26891\" class=\"math\"><span id=\"MathJax-Span-26892\" class=\"mrow\"><span id=\"MathJax-Span-26893\" class=\"semantics\"><span id=\"MathJax-Span-26894\" class=\"mrow\"><span id=\"MathJax-Span-26895\" class=\"mrow\"><span id=\"MathJax-Span-26901\" class=\"mo\">.<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\r\n<p id=\"fs-id1165137557013\">The magnitude (size) of an earthquake is measured on a scale known as the Richter Scale.\u00a0<sup id=\"footnote-ref4\"><\/sup>The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 6 is not twice as great as an earthquake of magnitude 3.\u00a0 It is 10<sup>6\u22123\u00a0<\/sup>= 10<sup>3\u00a0<\/sup>= 1,000\u00a0times as great!<\/p>\r\nThe Richter scale strength of an earthquake, M, is given by [latex]M=\\mathrm{log}\\left(\\frac{W}{W_0}\\right),[\/latex] where [latex]W[\/latex] is the strength of the seismic waves of an earthquake and [latex]W_0[\/latex] is the strength of normally occurring earthquakes.\u00a0 Minor earthquakes occur regularly allowing [latex]W_0[\/latex] to be determined.\r\n<div class=\"textbox examples\">\r\n<h3>Example 17:\u00a0 The Richter Scale<\/h3>\r\nAn earthquake Richter Scale strength of 5 is considered a moderate strength earthquake.\u00a0 How much stronger was the 2010 magnitude 5.5 earthquake that occurred between Ontario and Quebec compared to a standard earthquake?\r\n\r\n[reveal-answer q=\"397328\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"397328\"]\r\n\r\nFirst we note that [latex]M=5.5[\/latex] and substitute this into the Richter Scale equation.\u00a0 [latex]5.5=\\mathrm{log}\\left(\\frac{W}{W_0}\\right)[\/latex] written as an exponential equation is [latex]10^{5.5}=\\frac{W}{W_0}.[\/latex] This can be written as [latex]W=10^{5.5}W_0[\/latex] or\u00a0 [latex]W\\approx316228W_0.[\/latex]\r\n\r\nThe earthquake was 316,228 times stronger than a standard earthquake.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1165137759741\">In chemistry,\u00a0<span class=\"no-emphasis\">pH<\/span>\u00a0is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered <em>acidic<\/em>, and substances with a pH greater than 7 are said to be <em>alkaline<\/em>. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:<\/p>\r\n\r\n<ul id=\"fs-id1165135253210\">\r\n \t<li>Battery acid: 0.8<\/li>\r\n \t<li>Stomach acid: 2.7<\/li>\r\n \t<li>Orange juice: 3.3<\/li>\r\n \t<li>Pure water: 7 (at 25\u00b0 C)<\/li>\r\n \t<li>Human blood: 7.35<\/li>\r\n \t<li>Fresh coconut: 7.8<\/li>\r\n \t<li>Sodium hydroxide (lye): 14<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137540406\">To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where [latex]\\left[H^+\\right][\/latex]<span id=\"MathJax-Element-982-Frame\" class=\"MathJax\" style=\"font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px\" role=\"presentation\"><span id=\"MathJax-Span-13524\" class=\"math\"><span id=\"MathJax-Span-13525\" class=\"mrow\"><span id=\"MathJax-Span-13526\" class=\"semantics\"><span id=\"MathJax-Span-13527\" class=\"mrow\"><span id=\"MathJax-Span-13528\" class=\"mrow\"><span id=\"MathJax-Span-13529\" class=\"mtext\"><\/span><span id=\"MathJax-Span-13530\" class=\"mi\"><\/span><span id=\"MathJax-Span-13531\" class=\"mtext\"><\/span><\/span><\/span><\/span><\/span><\/span> <\/span>is the <em>concentration of hydrogen ions<\/em> in the solution measured in moles per liter.<\/p>\r\n<p style=\"text-align: center\">[latex]pH=-\\mathrm{log}\\left(\\left[H^+\\right]\\right)=\\mathrm{log}\\left(\\frac{1}{\\left[H^+\\right]}\\right)[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 18:\u00a0 pH<\/h3>\r\nIf the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?\r\n\r\n[reveal-answer q=\"63528\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"63528\"]\r\n\r\nSuppose [latex]c[\/latex] is the original concentration of hydrogen ions, and [latex]p[\/latex] is the original pH of the liquid.\u00a0 Then [latex]p=-\\mathrm{log}\\left(c\\right).[\/latex]\u00a0 If the concentration is doubled, the new concentration is [latex]2c[\/latex].\u00a0 Then the pH of the new liquid is\r\n<p style=\"text-align: center\">[latex]pH=-\\mathrm{log}\\left(2c\\right).[\/latex]<\/p>\r\nUsing the product rule of logarithms,\r\n<p style=\"text-align: center\">[latex]pH=-\\left(\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(c\\right)\\right)=-\\mathrm{log}\\left(2\\right)-\\mathrm{log}\\left(c\\right).[\/latex]<\/p>\r\nSince [latex]p=-log(c),[\/latex] the new pH is\r\n<p style=\"text-align: center\">[latex]pH=p-\\mathrm{log}\\left(2\\right)\\approx p-0.301.[\/latex]<\/p>\r\nWhen the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 19:\u00a0 pH<\/h3>\r\nOrange juice has a pH of approximately 3.3.\u00a0 Determine the hydrogen ion concentration.\r\n\r\n[reveal-answer q=\"574782\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"574782\"]\r\n\r\nSince the pH of orange juice is 3.3, we substitute 3.3 into the pH formula and get [latex]3.3=-\\mathrm{log}\\left(\\left[H^+\\right]\\right).[\/latex] Rewriting this as an exponential function, [latex]\\left[H^+\\right]=10^{-3.3}\\approx5.012\\times10^{-4}.[\/latex]\u00a0 Therefore the concentration is [latex]\\left[H^+\\right]\\approx5\\times10^{-4}=0.0005,[\/latex] or 0.0005 moles per liter.[\/hidden-answer]\r\n\r\n<\/div>\r\nA <em>decibel<\/em> (dB) is a measure of how loud a sound is when compared to a reference value.\u00a0 A commonly used reference value is the sound intensity of the softest sound a human can typically hear; usually that of a child.\u00a0 We will call this value [latex]I_0.[\/latex]\u00a0 Since there is a very wide range of sounds that humans can hear, the logarithmic scale is used. The formula for decibels is\r\n<p style=\"text-align: center\">Sound level in decibels [latex]=10\\mathrm{log}\\left(\\frac{I}{I_0}\\right),[\/latex]<\/p>\r\nwhere [latex]I[\/latex] is the sound intensity of the sound being measured.\r\n<div class=\"textbox examples\">\r\n<h3>Example 20:\u00a0 Decibels<\/h3>\r\nA vacuum cleaner sound level measures at 75dB and a balloon popping measures 125dB.\u00a0 \u00a0A balloon popping is how many times more intense than the sound intensity of the vacuum cleaner?\r\n\r\n[reveal-answer q=\"485947\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"485947\"]\r\n\r\nLet [latex]I_B[\/latex] be the sound intensity of the balloon popping and [latex]I_V[\/latex] be the sound intensity of the vacuum cleaner.\u00a0 We then look at the difference.\r\n<p style=\"text-align: center\">[latex]\\begin{align*}125\\text{ dB }-75\\text{ dB }&amp;=10\\mathrm{log}\\left(\\frac{I_B}{I_0}\\right)-10\\mathrm{log}\\left(\\frac{I_V}{I_0}\\right)&amp;&amp;\\text{ }\\\\50\\text{ dB }&amp;=10\\mathrm{log}\\left(\\frac{\\frac{I_B}{I_0}}{\\frac{I_V}{I_0}}\\right)&amp;&amp;\\text{Use the quotient rule for logarithms.}\\\\5\\text{ dB }&amp;=\\mathrm{log}\\left(\\frac{I_B}{I_V}\\right)&amp;&amp;\\text{Simplify.}\\\\10^5&amp;=\\frac{I_B}{I_V}&amp;&amp;\\text{Write as an exponential equation.}\\\\100000I_V&amp;=I_B&amp;&amp;\\text{Multiply both sides by }I_V.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\r\nThe balloon's sound intensity is 100,000 times more than the vacuum cleaner's sound intensity.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1165137410373\" class=\"precalculus media\">\r\n<p id=\"fs-id1165137722668\">Access these online resources for additional instruction and practice with exponential and logarithmic equations.<\/p>\r\n\r\n<ul id=\"fs-id1165137722673\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/solvelogeq\">Solving Logarithmic Equations<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/solveexplog\">Solving Exponential Equations with Logarithms<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137431646\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\r\n<ul id=\"fs-id1165137450888\">\r\n \t<li>An exponential equation can be solved by taking the logarithm of each side.<\/li>\r\n \t<li>We can solve exponential equations with base [latex]e,[\/latex] by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other.<\/li>\r\n \t<li>After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions.<\/li>\r\n \t<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c,[\/latex] where [latex]S[\/latex] is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation [latex]{b}^{c}=S,[\/latex] and solve for the unknown.<\/li>\r\n \t<li>We can also use graphing to solve equations with the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c.[\/latex] We graph both equations [latex]y={\\mathrm{log}}_{b}\\left(S\\right)[\/latex] and [latex]y=c[\/latex] on the same coordinate plane and identify the solution as the <em>x-<\/em>value of the intersecting point.<\/li>\r\n \t<li>Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Glossary<\/h3>\r\n<dl id=\"fs-id1165135149311\">\r\n \t<dt>extraneous solution<\/dt>\r\n \t<dd id=\"fs-id1165134059776\">a solution introduced while solving an equation that does not satisfy the conditions of the original equation<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Use logarithms to solve exponential equations.<\/li>\n<li>Use the definition of a logarithm to solve logarithmic equations.<\/li>\n<li>Solve applied problems involving exponential and logarithmic equations.<\/li>\n<\/ul>\n<\/div>\n<div id=\"CNX_Precalc_Figure_04_06_001\" class=\"small\">\n<div style=\"width: 498px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/06220634\/CNX_Precalc_Figure_04_06_001.jpg\" alt=\"Seven rabbits in front of a brick building.\" width=\"488\" height=\"324\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 1.\u00a0<\/strong>Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the \u201crabbit plague.\u201d (credit: Richard Taylor, Flickr)<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137871518\">In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.<\/p>\n<p id=\"fs-id1165135695212\">Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential equations.<\/p>\n<h3>Solving Exponential Equations Using Logarithms<\/h3>\n<p>Many times we need to solve equations of the form [latex]n=ab^x[\/latex] where [latex]n[\/latex] is a real number.\u00a0 To do this we will divide both sides by [latex]a[\/latex] to get\u00a0[latex]\\frac{n}{a}=b^x[\/latex] and then take the logarithm of both sides giving the equation\u00a0[latex]\\mathrm{log}\\left(\\frac{n}{a}\\right)=\\mathrm{log}\\left(b^x\\right).[\/latex] Next, we use the power rule for logarithms to get\u00a0[latex]\\mathrm{log}\\left(\\frac{n}{a}\\right)=x\\mathrm{log}\\left(b\\right)[\/latex].\u00a0 Finally, divide both sides by\u00a0[latex]\\mathrm{log}\\left(b\\right)[\/latex]\u00a0 to get\u00a0[latex]x=\\frac{\\mathrm{log}\\left(\\frac{n}{a}\\right)}{\\mathrm{log}\\left(b\\right)}[\/latex].\u00a0 Note that any base for the logarithm can be used but base 10 and base [latex]e[\/latex] are most commonly used.<\/p>\n<div class=\"textbox examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137673524\"><strong>Given an exponential equation in the form [latex]n=ab^x[\/latex], solve for the unknown.<\/strong><\/p>\n<ol id=\"fs-id1165137784632\" type=\"1\">\n<li>Divide both sides by [latex]a[\/latex] or the initial condition.<\/li>\n<li>Apply the logarithm of both sides of the equation.\n<ul id=\"fs-id1165137824134\">\n<li>If one of the terms in the equation has base 10, use the common logarithm.<\/li>\n<li>If none of the terms in the equation has base 10, use the natural logarithm.<\/li>\n<\/ul>\n<\/li>\n<li>Use the rules of logarithms to solve for the unknown.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 1:\u00a0 Solving a Basic Exponential Equation<\/h3>\n<p>Solve [latex]5=3\\left(2\\right)^x.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q280679\">Show Solution<\/span><\/p>\n<div id=\"q280679\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by dividing both sides by 3 to get<\/p>\n<p style=\"text-align: center\">[latex]5\/3=\\left(2\\right)^x.[\/latex]<\/p>\n<p>Then, take the natural logarithm of both sides to get the equation<\/p>\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left(5\/3\\right)=\\mathrm{ln}\\left(2^x\\right).[\/latex]<\/p>\n<p>Use the power rule of logarithms [latex]\\mathrm{ln}\\left(b^x\\right)=x\\mathrm{ln}\\left(b\\right)[\/latex] to get<\/p>\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left(5\/3\\right)=x\\mathrm{ln}\\left(2\\right).[\/latex]<\/p>\n<p>Finally, divide both sides by [latex]\\mathrm{ln}\\left(2\\right)[\/latex] to get<\/p>\n<p style=\"text-align: center\">[latex]x=\\frac{\\mathrm{ln}\\left(5\/3\\right)}{\\mathrm{ln}\\left(2\\right)}\\approx0.7370.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It #1<\/h3>\n<p>Solve [latex]7=15\\left(4\\right)^x.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q261872\">Show Solution<\/span><\/p>\n<div id=\"q261872\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\frac{\\mathrm{ln}\\left(7\/15\\right)}{\\mathrm{ln}\\left(4\\right)}\\approx-0.5498.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 2:\u00a0 Solving an Exponential Equation with an Algebraic Expression in the Exponent<\/h3>\n<p>Solve [latex]15=3\\left(0.5\\right)^{x+1}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q346617\">Show Solution<\/span><\/p>\n<div id=\"q346617\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the algebraic expression is in the exponent, this equation will be solved using the same steps as above but the x+1 should be kept in parenthesis until we are ready to simplify at the end.\u00a0 Begin by dividing both sides by 3 to get<\/p>\n<p style=\"text-align: center\">[latex]5=\\left(0.5\\right)^{\\left(x+1\\right)}.[\/latex]<\/p>\n<p>Then, take the natural logarithm of both sides to get the equation<\/p>\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left(5\\right)=\\mathrm{ln}\\left(0.5^{\\left(x+1\\right)}\\right).[\/latex]<\/p>\n<p>Use the power rule of logarithms [latex]\\mathrm{ln}\\left(b^x\\right)=x\\mathrm{ln}\\left(b\\right)[\/latex] so<\/p>\n<p style=\"text-align: center\">[latex]\\mathrm{ln}\\left(5\\right)={\\left(x+1\\right)}\\mathrm{ln}\\left(0.5\\right).[\/latex]<\/p>\n<p>Divide both sides of the equation by [latex]\\mathrm{ln}\\left(0.5\\right)[\/latex] to get the equation<\/p>\n<p style=\"text-align: center\">[latex]\\left(x+1\\right)=\\frac{\\mathrm{ln}\\left(5\\right)}{\\mathrm{ln}\\left(0.5\\right)}.[\/latex]<\/p>\n<p>Note that this is just a linear equation so we subtract 1 from both sides and the solution is<\/p>\n<p style=\"text-align: center\">[latex]x=\\frac{\\mathrm{ln}\\left(5\\right)}{\\mathrm{ln}\\left(0.5\\right)}-1\\approx-3.3219.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It #2<\/h3>\n<p>Solve [latex]15=2\\left(7\\right)^{x^2+1}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q653920\">Show Solution<\/span><\/p>\n<div id=\"q653920\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\pm\\sqrt{\\frac{\\mathrm{ln}\\left(7.5\\right)}{\\mathrm{ln}\\left(7\\right)}-1}\\approx\\pm0.1883[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h4>Equations Containing <em><span class=\"e2\">e<\/span><\/em><\/h4>\n<p id=\"fs-id1165137606150\">One common type of exponential equations uses base [latex]e.[\/latex] This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base [latex]e[\/latex] on either side, we can use the <span class=\"no-emphasis\">natural logarithm<\/span> to solve it.<\/p>\n<div id=\"fs-id1165137727175\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137656412\"><strong>Given an equation of the form [latex]y=a{e}^{kt}\\text{,}[\/latex] solve for [latex]t.[\/latex]<\/strong><\/p>\n<ol id=\"fs-id1165135189859\" type=\"1\">\n<li>Divide both sides of the equation by [latex]a.[\/latex]<\/li>\n<li>Apply the natural logarithm of both sides of the equation.<\/li>\n<li>Divide both sides of the equation by [latex]k.[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_06_06\" class=\"textbox examples\">\n<div id=\"fs-id1165137846466\">\n<div id=\"fs-id1165137846468\">\n<h3>Example 3:\u00a0 Solve an Equation with Continuous Growth<\/h3>\n<p id=\"fs-id1165137581969\">Solve [latex]100=20{e}^{2t}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q538063\">Show Solution<\/span><\/p>\n<div id=\"q538063\" class=\"hidden-answer\" style=\"display: none\">\n<p>For this problem, use the natural logarithm since the equation contains base [latex]e.[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}100&=20{e}^{2t}&& \\hfill \\\\5&={e}^{2t}&&\\textrm{Divide both sides by 20}\\text{.}\\hfill \\\\ \\mathrm{ln}\\left(5\\right)&=2t&&\\textrm{Take the natural logarithm of both sides.}\\\\ \\text{ }&\\text{ } &&\\text{Use the fact that }\\mathrm{ln}\\left(e^{x}\\right)=x\\text{.}\\hfill \\\\t&=\\frac{\\mathrm{ln}\\left(5\\right)}{2}&&\\textrm{Divide by the coefficient of }t\\text{.}\\hfill \\end{align*}[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<p>Using laws of logs, we can also write this answer in the form [latex]t=\\mathrm{ln}\\sqrt[\\leftroot{1}\\uproot{2} ]{5}.[\/latex] If we want a decimal approximation of the answer, we use a calculator.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135186732\" class=\"precalculus tryit\">\n<h3>Try it #3<\/h3>\n<div id=\"ti_04_06_06\">\n<div id=\"fs-id1165137737732\">\n<p id=\"fs-id1165137431561\">Solve [latex]3{e}^{0.5t}=11.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137726884\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137726884\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137698501\">[latex]t=2\\mathrm{ln}\\left(\\frac{11}{3}\\right)[\/latex] or [latex]\\mathrm{ln}\\left({\\left(\\frac{11}{3}\\right)}^{2}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137748966\" class=\"bc-section section\">\n<div id=\"fs-id1165137667260\" class=\"bc-section section\">\n<div id=\"fs-id1165135369638\" class=\"precalculus qa key-takeaways\">\n<h3>Q&amp;A<\/h3>\n<p id=\"fs-id1165137547216\"><strong>Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?<\/strong><\/p>\n<p id=\"fs-id1165137435646\"><em>No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined.\u00a0<\/em><\/p>\n<\/div>\n<div id=\"Example_04_06_04\" class=\"textbox examples\">\n<div id=\"fs-id1165137405247\">\n<div id=\"fs-id1165137635295\">\n<h3>Example 4:\u00a0 Solving an Equation with Positive and Negative Powers<\/h3>\n<p id=\"fs-id1165137805073\">Solve [latex]{3}^{x+1}=-2.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137723720\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137723720\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137723720\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137874850\">This equation has no solution. There is no real value of [latex]x[\/latex] that will make the equation a true statement because any power of a positive number is positive.<\/p>\n<p id=\"fs-id1165137578263\"><a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_06_002\">Figure 2<\/a> shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.<\/p>\n<div id=\"CNX_Precalc_Figure_04_06_002\" class=\"small\">\n<div style=\"width: 340px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/06220637\/CNX_Precalc_Figure_04_06_002.jpg\" alt=\"Graph of 3^(x+1)=-2 and y=-2. The graph notes that they do not cross.\" width=\"330\" height=\"297\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137849213\">\n<div id=\"CNX_Precalc_Figure_04_06_002\" class=\"small\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137920668\" class=\"precalculus tryit\">\n<h3>Try it #4<\/h3>\n<div id=\"ti_04_06_04\">\n<div id=\"fs-id1165137761184\">\n<p id=\"fs-id1165137847264\">Solve [latex]{2}^{x}=-100.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137844325\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137844325\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137844325\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135208714\">The equation has no solution.\u00a0 We know that\u00a0[latex]{2}^{x}[\/latex] is always greater than 0.\u00a0 Therefore, there is no value for the input [latex]x[\/latex] which produces an output of\u00a0 -100.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137641700\" class=\"bc-section section\">\n<h3><\/h3>\n<p id=\"fs-id1165137939689\">Sometimes there are exponential functions on both sides of the equation such as [latex]ab^x=cd^x[\/latex].\u00a0 The process is similar to solving [latex]n=ab^x[\/latex] but we will need to be careful to use the product rule of logarithms before applying the power rule of logarithms.<\/p>\n<div id=\"fs-id1165137939925\" class=\"precalculus howto examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1165137673524\"><strong>Given an equation with exponential expression on each side, solve for the unknown.<\/strong><\/p>\n<ol id=\"fs-id1165137784632\" type=\"1\">\n<li>Divide both sides by one of the initial conditions.<\/li>\n<li>Apply the logarithm of both sides of the equation.\n<ul id=\"fs-id1165137824134\">\n<li>If one of the terms in the equation has base 10, use the common logarithm.<\/li>\n<li>If none of the terms in the equation has base 10, use the natural logarithm.<\/li>\n<\/ul>\n<\/li>\n<li>Use the product and power rules of logarithms and then solve for the unknown.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_04_06_05\" class=\"textbox examples\">\n<div id=\"fs-id1165137653913\">\n<div id=\"fs-id1165137715230\">\n<h3>Example 5:\u00a0 Solving an Equation Containing Powers of Different Bases<\/h3>\n<p id=\"fs-id1165137549801\">Solve [latex]6\\left(5\\right)^{x+2}=2\\left(4\\right)^{x}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q336145\">Show Solution<\/span><\/p>\n<div id=\"q336145\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align*}6\\left(5\\right)^{x+2}&=2\\left(4\\right)^x&&\\text{ }\\\\3\\left(5\\right)^{x+2}&=\\left(4\\right)^x&&\\text{Divide both sides by 2.}\\\\\\mathrm{ln}\\left(3\\left(5\\right)^{x+2}\\right)&=\\mathrm{ln}\\left(4^x\\right)&&\\text{Take the natural logarithm of both sides.}\\\\\\mathrm{ln}\\left(3\\right)+\\mathrm{ln}\\left(5^{x+2}\\right)&=\\mathrm{ln}\\left(4^x\\right)&&\\text{Use the product rule of logarithms.}\\\\\\mathrm{ln}\\left(3\\right)+\\left(x+2\\right)\\mathrm{ln}\\left(5\\right)&=x\\mathrm{ln}\\left(4\\right)&&\\text{Use the power law of logarithms.}\\\\\\mathrm{ln}\\left(3\\right)+x\\mathrm{ln}\\left(5\\right)+2\\mathrm{ln}\\left(5\\right)&=x\\mathrm{ln}\\left(4\\right)&&\\text{Use the distributive law.}\\\\x\\mathrm{ln}\\left(5\\right)-x\\mathrm{ln}\\left(4\\right)&=-2\\mathrm{ln}\\left(5\\right)-\\mathrm{ln}\\left(3\\right)&&\\text{Move terms containing x on one side, }\\\\&\\text{}&&\\text{and terms without x on the other.}\\\\x\\left(\\mathrm{ln}\\left(5\\right)-\\mathrm{ln}\\left(4\\right)\\right)&=-2\\mathrm{ln}\\left(5\\right)-\\mathrm{ln}\\left(3\\right)&& \\text{On the left hand side, factor out an }x.\\\\x\\mathrm{ln}\\left(\\frac{5}{4}\\right)&=\\mathrm{ln}\\left(\\frac{1}{75}\\right)&&\\text{Use the laws of logarithms.}\\\\x&=\\frac{\\mathrm{ln}\\left(\\frac{1}{75}\\right)}{\\mathrm{ln}\\left(\\frac{5}{4}\\right)}&& \\text{Divide by the coefficient of }x.\\\\\\end{align*}[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<p>Notice that the product rule of logarithms was used before the power rule because the power rule cannot be applied to [latex]ab^x[\/latex]; [latex]a[\/latex] is not raised to the [latex]x[\/latex] power.\u00a0 Also, note that when the power rule is used on [latex]\\mathrm{ln}\\left(5^{x+2}\\right)[\/latex] the [latex]\\left(x+2\\right)[\/latex] has parenthesis so that the [latex]\\mathrm{ln}\\left(5\\right)[\/latex] gets properly distributed.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137473583\" class=\"precalculus tryit\">\n<h3>Try it #5<\/h3>\n<div id=\"ti_04_06_05\">\n<div id=\"fs-id1165137402173\">\n<p id=\"fs-id1165137402175\">Solve [latex]{2}^{x}={3}^{x+1}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137605315\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137605315\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137605315\" class=\"hidden-answer\" style=\"display: none\">\n[latex]x=\\frac{\\mathrm{ln}\\left(3\\right)}{\\mathrm{ln}\\left(\\frac{2}{3}\\right)}[\/latex]\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137469838\" class=\"bc-section section\">\n<div id=\"fs-id1165137532335\" class=\"precalculus qa key-takeaways\">\n<h3>Q&amp;A<\/h3>\n<p id=\"eip-id1165135593037\"><strong>Does every equation of the form\u00a0<\/strong>[latex]y=A{e}^{kt}[\/latex]\u00a0<strong>have a solution?<\/strong><\/p>\n<p id=\"fs-id1165137423829\"><em>No. There is a solution when [latex]k\\ne 0,[\/latex] and when [latex]y[\/latex] and [latex]A[\/latex] are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is [latex]2=-3{e}^{t}.[\/latex]<\/em><\/p>\n<\/div>\n<div id=\"Example_04_06_07\" class=\"textbox examples\">\n<div id=\"fs-id1165137828228\">\n<div id=\"fs-id1165137704438\">\n<h3>Example 6:\u00a0 Solving an Equation That Requires Algebra First<\/h3>\n<p id=\"fs-id1165134223341\">Solve<\/p>\n<ol>\n<li>[latex]4{e}^{2x}+5=12.[\/latex]<\/li>\n<li>[latex]{e}^{2t}-3=-4{e}^{2t}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1165137675352\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137675352\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137675352\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li style=\"text-align: center\"><span style=\"font-size: 1rem;text-align: initial\">[latex]\\\\[\/latex][latex]\\begin{align*}4{e}^{2x}+5&=12&&\\text{ }\\\\4{e}^{2x}&=7&&\\text{Combine like terms.}\\\\e^{2x}&=\\frac{7}{4}&&\\text{Divide by the coefficient.}\\\\2x&=\\mathrm{ln}\\left(\\frac{7}{4}\\right)&&\\text{Take the natural logarithm of both sides.}\\\\x&=\\frac{1}{2}\\mathrm{ln}\\left(\\frac{7}{4}\\right)&&\\text{Solve for }x. \\end{align*}[\/latex]<\/span><\/li>\n<li style=\"text-align: center\">[latex]\\\\[\/latex][latex]\\begin{align*}{e}^{2t}-3&=-4{e}^{2t}&&\\text{ }\\\\5{e}^{2t}-3&=0&&\\text{Add }4{e}^{2t}\\text{ to both sides.}\\\\5{e}^{2t}&=3&&\\text{Add 3 to both sides.}\\\\{e}^{2t}&=\\frac{3}{5}&&\\text{Divide both sides by 5.}\\\\2t&=\\mathrm{ln}\\left(\\frac{3}{5}\\right)&&\\text{Take the natural logarithm of both sides.}\\\\t&=\\frac{1}{2}\\mathrm{ln}\\left(\\frac{3}{5}\\right)&&\\text{Solve for }x.\\\\\\end{align*}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1165137828228\">\n<div id=\"fs-id1165137675352\">\n<div class=\"unnumbered\" style=\"text-align: left\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137827163\" class=\"precalculus tryit\">\n<h3>Try it #6<\/h3>\n<div id=\"ti_04_06_07\">\n<div id=\"fs-id1165137936638\">\n<p id=\"fs-id1165137936640\">Solve [latex]3+{e}^{2t}=7{e}^{2t}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137819982\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137819982\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137819982\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137819984\">[latex]t=\\mathrm{ln}\\left(\\frac{1}{\\sqrt[\\leftroot{1}\\uproot{2} ]{2}}\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137665482\" class=\"bc-section section\">\n<h4>Extraneous Solutions<\/h4>\n<p id=\"fs-id1165137742403\">Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.<\/p>\n<p>These extraneous solutions frequently occur when the exponential function is a quadratic form. Recall that quadratic equations can be solved by factoring and setting each factor equal to zero, or by the quadratic equation.\u00a0 W will look for the pattern of the quadratic and then choose which technique can most easily be used.<\/p>\n<div id=\"Example_04_06_08\" class=\"textbox examples\">\n<div id=\"fs-id1165137828517\">\n<div id=\"fs-id1165137828519\">\n<h3>Example 7:\u00a0 Solving Exponential Functions in Quadratic Form<\/h3>\n<p id=\"fs-id1165137443096\">Solve [latex]{e}^{2x}-{e}^{x}=56.[\/latex]<\/p>\n<\/div>\n<div>\n<div id=\"eip-id1165133221795\" class=\"unnumbered\" style=\"text-align: left\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q884164\">Show Solution<\/span><\/p>\n<div id=\"q884164\" class=\"hidden-answer\" style=\"display: none\">We first re-write the equation using the fact that [latex]e^{2x}={\\left({e^x}\\right)}^2.[\/latex] Notice that this equation has the form [latex]e^x[\/latex] squared minus\u00a0[latex]e^x[\/latex] equals 56. This is a quadratic equation which can be solved by factoring.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}{e}^{2x}-{e}^{x}&=56&& \\hfill \\\\ {e}^{2x}-{e}^{x}-56&=0&&\\text{Get one side of the equation equal to zero}.\\hfill \\\\ \\left({e}^{x}+7\\right)\\left({e}^{x}-8\\right)&=0&&\\text{Factor by the FOIL method}.\\hfill \\\\{e}^{x}+7&=0\\text{ or }{e}^{x}-8=0&& \\text{If a product is zero, then one factor must be zero}.\\hfill \\\\{e}^{x}&=-7{\\text{ or e}}^{x}=8&& \\text{Isolate the exponentials}.\\hfill \\\\ {e}^{x}&=8&&\\text{Reject the equation that has no solution}.\\hfill \\\\x&=\\mathrm{ln}\\left(8\\right)&&\\text{Write as a logarmithm}.\\hfill \\end{align*}[\/latex]<\/p>\n<h3>Analysis<\/h3>\n<div>When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation [latex]{e}^{x}=-7[\/latex] because a positive number never equals a negative number. The solution [latex]\\mathrm{ln}\\left(-7\\right)[\/latex] is not a real number, and in the real number system this solution is rejected as an extraneous solution.<\/div>\n<div class=\"unnumbered\" style=\"text-align: left\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135560773\" class=\"precalculus tryit\">\n<h3>Try it #7<\/h3>\n<div id=\"ti_04_06_08\">\n<div id=\"fs-id1165134378706\">\n<p id=\"fs-id1165134378708\">Solve [latex]{e}^{2x}={e}^{x}+2.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135638513\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135638513\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135638513\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135638516\">[latex]x=\\mathrm{ln}2.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137862548\" class=\"bc-section section\">\n<h3>Using the Definition of a Logarithm to Solve Logarithmic Equations<\/h3>\n<p id=\"fs-id1165137862553\">We have already seen that the\u00a0<span class=\"no-emphasis\">logarithmic equation\u00a0<\/span>[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equivalent to the exponential equation [latex]{b}^{y}=x,[\/latex] for [latex]x>0.[\/latex] We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 8:\u00a0 Solving a Logarithmic Equation<\/h3>\n<p id=\"fs-id1165134148350\">Solve the equation [latex]{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x-5\\right)=3.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q809039\">Show Solution<\/span><\/p>\n<div id=\"q809039\" class=\"hidden-answer\" style=\"display: none\">\n<p>To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for [latex]x:[\/latex]<\/p>\n<div id=\"eip-id2205910\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{log}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x-5\\right)&=3&&\\text{ }\\\\\\mathrm{log}_{2}\\left(2\\left(3x-5\\right)\\right)&=3&&\\text{Apply the product rule of logarithms}.\\\\{\\mathrm{log}}_{2}\\left(6x-10\\right)&=3&&\\text{Distribute}.\\\\{2}^{3}&=6x-10&&\\text{Apply the definition of a logarithm}.\\\\8&=6x-10&&\\text{Calculate }{2}^{3}.\\\\18&=6x&&\\text{Add 10 to both sides}.\\\\x&=3&& \\text{Divide by 6}.\\end{align*}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_06_09\" class=\"textbox examples\">\n<div id=\"fs-id1165137841585\">\n<div id=\"fs-id1165137725474\">\n<h3>Example 9:\u00a0 Using Algebra to Solve a Logarithmic Equation<\/h3>\n<p id=\"fs-id1165135152135\">Solve [latex]2\\mathrm{ln}\\left(x\\right)+3=7.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q414654\">Show Solution<\/span><\/p>\n<div id=\"q414654\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center\">[latex]\\begin{align*}2\\mathrm{ln}\\left(x\\right)+3&=7&&\\text{ }\\\\2\\mathrm{ln}\\left(x\\right)&=4&&\\text{Subtract 3}.\\\\\\mathrm{ln}\\left(x\\right)&=2&&\\text{Divide by 2}.\\\\x&={e}^{2}&&\\text{Rewrite in exponential form}.\\end{align*}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div id=\"eip-id1165135466384\" class=\"unnumbered\" style=\"text-align: center\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137571063\" class=\"precalculus tryit\">\n<h3>Try it #8<\/h3>\n<div id=\"ti_04_06_09\">\n<div id=\"fs-id1165137437378\">\n<p id=\"fs-id1165137437380\">Solve [latex]6+\\mathrm{ln}\\left(x\\right)=10.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165135209084\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165135209084\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165135209084\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137673564\">[latex]x={e}^{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_06_10\" class=\"textbox examples\">\n<div id=\"fs-id1165137482839\">\n<div id=\"fs-id1165137557104\">\n<h3>Example 10:\u00a0 Using Algebra Before and After Using the Definition of the Natural Logarithm<\/h3>\n<p id=\"fs-id1165137557109\">Solve [latex]2\\mathrm{ln}\\left(6x\\right)=7.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137501502\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137501502\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137501502\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"eip-id1165135388424\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\mathrm{ln}\\left(6x\\right)&=7&&\\text{ }\\\\\\mathrm{ln}\\left(6x\\right)&=\\frac{7}{2}&& \\text{Divide by 2}.\\\\6x&={e}^{\\left(\\frac{7}{2}\\right)}&&\\text{Use the definition of the natural logarithm}.\\\\x&=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}&&\\text{Divide by 6}.\\end{align*}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"precalculus tryit\">\n<h3>Try it #9<\/h3>\n<div id=\"ti_04_06_10\">\n<div id=\"fs-id1165137862628\">\n<p id=\"fs-id1165137862630\">Solve [latex]2\\mathrm{ln}\\left(x+1\\right)=10.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1165137530181\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137530181\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137530181\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137530183\">[latex]x={e}^{5}-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_06_11\" class=\"textbox examples\">\n<div id=\"fs-id1165137531831\">\n<div id=\"fs-id1165137805003\">\n<h3>Example 11:\u00a0 Using a Graph to Understand the Solution to a Logarithmic Equation<\/h3>\n<p id=\"fs-id1165137805008\">Solve [latex]\\mathrm{ln}\\left(x\\right)=3.[\/latex]<\/p>\n<\/div>\n<div>\n<div class=\"wp-caption-text\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q409852\">Show Solution<\/span><\/p>\n<div id=\"q409852\" class=\"hidden-answer\" style=\"display: none\">\n<div class=\"wp-caption-text\">\n<p>To solve [latex]\\mathrm{ln}\\left(x\\right)=3,[\/latex] write the equation in exponential form using the definition of the natural logarithm to get [latex]x={e}^{3}.[\/latex][latex]\\\\[\/latex]<\/p>\n<p id=\"fs-id1165137443165\"><a class=\"autogenerated-content\" href=\"#CNX_Precalc_Figure_04_06_003\">Figure 3<\/a>\u00a0represents the graph of the equation. On the graph, the <em>x<\/em>-coordinate of the point at which the two graphs intersect is close to 20. In other words [latex]{e}^{3}\\approx 20.[\/latex] A calculator gives a better approximation: [latex]{e}^{3}\\approx 20.0855.[\/latex]<\/p>\n<div id=\"CNX_Precalc_Figure_04_06_003\" class=\"small\">\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/06220640\/CNX_Precalc_Figure_04_06_003.jpg\" alt=\"Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).\" width=\"487\" height=\"288\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 3.\u00a0<\/strong>The graphs of [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] and [latex]y=3[\/latex] cross at the point [latex]\\left(e^{3},3\\right),[\/latex] which is approximately (20.0855, 3).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137401539\" class=\"precalculus tryit\">\n<h3>Try it #10<\/h3>\n<div id=\"ti_04_06_11\">\n<div id=\"fs-id1165134375704\">\n<p id=\"fs-id1165134375706\">Use a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[\/latex] to 2 decimal places.<\/p>\n<\/div>\n<div id=\"fs-id1165137633980\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165137633980\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165137633980\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137633982\">[latex]x\\approx 9.97[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137755280\" class=\"bc-section section\"><span style=\"color: #6c64ad;font-size: 1em;font-weight: 600\">Solving Applied Problems Using Exponential Equations<\/span><\/div>\n<div id=\"fs-id1165137828382\" class=\"bc-section section\">\n<p id=\"fs-id1165137828387\">In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. 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<div class=\"textbox examples\">\n<h3>Example 12:\u00a0 Real World application; Depreciation<\/h3>\n<p>In 2018, a car was purchased for $32,000 and depreciates 20% per year.\u00a0 When will the car be worth $8,000?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q310526\">Show Solution<\/span><\/p>\n<div id=\"q310526\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by writing a formula for the situation.\u00a0Since a noncontinuous rate is given, we use the formula [latex]f\\left(t\\right)=ab^t.[\/latex] Let [latex]t=0[\/latex] represent 2019.\u00a0 The initial condition is [latex]a = 32000,[\/latex] the rate is [latex]r=-0.2,[\/latex] and the growth factor is [latex]b=1+r=1-0.2=0.8.[\/latex] Therefore, the formula is<\/p>\n<p style=\"text-align: center\">[latex]f\\left(t\\right)=32000\\left(0.8\\right)^t.[\/latex]<\/p>\n<p>Next, we need to know when the function&#8217;s output will be 8,000.\u00a0 Set the formula equal to 8000 and solve.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}8000&=32000\\left(0.8\\right)^t&&\\text{ }\\\\0.25&=0.8^t&&\\text{Divide both sides by 32000.}\\\\\\mathrm{ln}\\left(0.25\\right)&=\\mathrm{ln}\\left(0.8^t\\right)&&\\text{Take the natural logarithm of both sides.}\\\\\\mathrm{ln}\\left(0.25\\right)&=t\\mathrm{ln}\\left(0.8\\right)&&\\text{Use the power rule for logarithms.}\\\\t&=\\frac{\\mathrm{ln}\\left(0.25\\right)}{\\mathrm{ln}\\left(0.8\\right)}\\approx6.21&&\\text{Divide both sides by }\\mathrm{ln}\\left(0.8\\right).\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\n<p>It will take approximately 6.2 years for the car to depreciate to $8,000.\u00a0 In 2024, the car will be worth $8,000.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 13:\u00a0 Real World Application; BLood Alcohol Content<\/h3>\n<p>A person&#8217;s blood alcohol content (BAC) is a measure of how much alcohol is in the bloodstream.\u00a0 When a person stops drinking, over time the BAC will decay exponentially.\u00a0 For a particular individual, the formula [latex]f\\left(t\\right)=0.1e^{-0.0067t}[\/latex] models their BAC over time, t, measured in minutes.\u00a0 When will their BAC be 0.04?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q267570\">Show Solution<\/span><\/p>\n<div id=\"q267570\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are asked to find the input when the output is 0.04 so solve the equation [latex]0.04=0.1e^{-0.0067t}.[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}0.4&=e^{-0.0067 t}&&\\text{Divide both sides by 0.1.}\\\\\\mathrm{ln}\\left(0.4\\right)&=-0.0067t&&\\text{Take the natural logarithm and simplify.}\\\\t&=\\frac{\\mathrm{ln}\\left(0.4\\right)}{-0.0067}&&\\text{Divide both sides by }-0.0067.\\\\\\text{ }&\\approx136.8&&\\text{ }\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\n<p>It will take approximately 137 minutes for the BAC to drop to 0.04 for this individual.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 14:\u00a0 Real World Application; Tuition<\/h3>\n<p>College A is charging $40,000 tuition in 2019 and it is increasing at a continuous rate of 7% per year.\u00a0 College B charges $45,000 in 2019, but it is increasing at 4% per year.\u00a0 When will college A cost more than college B?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q136880\">Show Solution<\/span><\/p>\n<div id=\"q136880\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we determine the model representing each of the tuition amounts.\u00a0 College A has a continuous rate so use [latex]A(t)=ae^{kt}[\/latex] with [latex]a=40000[\/latex] and [latex]k=0.07.[\/latex]\u00a0 Therefore, the model for college A is [latex]A(t)=40000e^{0.07t}.[\/latex]\u00a0 College B has a noncontinuous rate so use [latex]B(t)=ab^t[\/latex] where [latex]a=45000[\/latex] and [latex]b=1+r=1+0.04=1.04.[\/latex]\u00a0 The model for college B is [latex]B(t)=45000\\left(1.04\\right)^t.[\/latex]<\/p>\n<p>We need to know when these two models are equal so we set the equations equal to each other and solve.<\/p>\n<p>[latex]\\begin{align*}40000e^{0.07t}&=45000\\left(1.04\\right)^t&&\\text{ }\\\\\\frac{8}{9}e^{0.07t}&=1.04^t&&\\text{Divide both sides by 45000 and simplify.}\\\\\\mathrm{ln}\\left(\\frac{8}{9}e^{0.07t}\\right)&=\\mathrm{ln}\\left(1.04^t\\right)&&\\text{Take the natural logarithm of both sides.}\\\\\\mathrm{ln}\\left(\\frac{8}{9}\\right)+0.07t&=\\mathrm{ln}\\left(1.04^t\\right)&&\\text{Use the product rule and simplify.}\\\\\\mathrm{ln}\\left(\\frac{8}{9}\\right)+0.07t&=t\\mathrm{ln}\\left(1.04\\right)&&\\text{Use the power rule on the right side.}\\\\t\\left(0.07-\\mathrm{ln}\\left(1.04\\right)\\right)&=-\\mathrm{ln}\\left(\\frac{8}{9}\\right)&&\\text{Collect like terms and factor }t.\\\\t&=\\frac{-\\mathrm{ln}\\left(\\frac{8}{9}\\right)}{0.07-\\mathrm{ln}\\left(1.04\\right)}&&\\text{Divide by the coefficient of }t.\\\\\\text{ }&\\approx3.83.&&\\text{ }\\\\\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\n<p>College B will have a higher tuition in approximately 4 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It #11<\/h3>\n<p>A town&#8217;s population is 14,000 people in 2017 and is increasing at a rate of 2.1% each year.\u00a0 When will the town&#8217;s population reach 18,000 people?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q649450\">Show Solution<\/span><\/p>\n<div id=\"q649450\" class=\"hidden-answer\" style=\"display: none\">In approximately 12.1 years or in 2029, the population will be 18,000 people.<\/div>\n<\/div>\n<\/div>\n<h3>Conversions Between Continuous and Noncontinuous Growth Rates<\/h3>\n<p>Recall that there are two possible formulas that can be used to represent an exponential function: [latex]f(x)=ab^x=a{\\left(b\\right)}^x[\/latex] for noncontinuous growth and [latex]f(x)=ae^{kx}=a{\\left(e^k\\right)}^x[\/latex] for continuous growth.\u00a0 When comparing the two forms, we see that [latex]b=e^k.[\/latex]\u00a0 Further, since [latex]b=1+r[\/latex] where [latex]r[\/latex] is the noncontinuous growth rate, we have that [latex]1+r=e^k.[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Example 15:\u00a0 Continuous Growth to Noncontinuous Growth<\/h3>\n<p>Given a continuous growth rate of 12%, find the noncontinuous rate.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q522523\">Show Solution<\/span><\/p>\n<div id=\"q522523\" class=\"hidden-answer\" style=\"display: none\">\n<p>The continuous growth rate [latex]k[\/latex] is given as [latex]k=0.12.[\/latex]\u00a0 We solve the equation\u00a0[latex]1+r=e^k[\/latex] with this value of [latex]k[\/latex] plugged in.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}1+r&=e^{0.12}\\\\r&=e^{0.12}-1\\approx0.1275.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\n<p>The noncontinuous rate is approximately 12.75%. Notice that since both rates are modeling the same growth, the noncontinuous rate must be slightly higher than the continuous rate, because the continuous rate allows for growth on the growth immediately.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 16:\u00a0 Noncontinuous Growth to Continuous Growth<\/h3>\n<p>Given a noncontinuous growth rate of 15%, find the continuous rate.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q671517\">Show Solution<\/span><\/p>\n<div id=\"q671517\" class=\"hidden-answer\" style=\"display: none\">\n<p>The noncontinuous growth rate [latex]r[\/latex] is given as [latex]r=0.15.[\/latex] Solve the equation\u00a0[latex]1+r=e^k[\/latex] with this value of [latex]r[\/latex] plugged in.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}1+0.15&=e^{k}\\\\k&=\\mathrm{ln}\\left(1.15\\right)\\approx0.1398.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\n<p>We took the natural logarithm of both sides to solve the equation.<\/p>\n<p>The continuous rate is approximately 13.98%.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It #12<\/h3>\n<p>a.\u00a0 Given a continuous decreasing rate of 5%, find the noncontinuous rate.<\/p>\n<p>b.\u00a0 Given a noncontinuous increasing rate of 7%, find the continuous rate.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q78291\">Show Solution<\/span><\/p>\n<div id=\"q78291\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a0 The noncontinuous rate of decrease is approximately 4.877% when the continuous rate of decrease is 5%.<\/p>\n<p>b.\u00a0 The continuous rate of increase is approximately 6.766%, when the noncontinuous rate is 7%.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Solving Logarithmic Applications<\/h3>\n<p>Exponential growth and decay often involve very large or very small numbers. It is common to use a logarithmic scale when measurements result in extremely large values or extremely small values.\u00a0 The Richter Scale, pH and decibels are examples of such scales.<\/p>\n<p>To describe these numbers, we often use <strong>orders of magnitude<\/strong>. The\u00a0order of magnitude\u00a0is the power of ten, when the number is expressed in scientific notation, with one digit to the left of the decimal. For example, the distance to the nearest star,\u00a0<span class=\"no-emphasis\">Proxima Centauri<\/span>, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific notation, this is 4.01134972 x 10<sup>13<\/sup>\u00a0.\u00a0So, we could describe this number as having order of magnitude of 13<span id=\"MathJax-Element-1564-Frame\" class=\"MathJax\" style=\"font-style: normal;font-weight: 400;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px;color: #555555;font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;background-color: #ffffff\" role=\"presentation\"><span id=\"MathJax-Span-26891\" class=\"math\"><span id=\"MathJax-Span-26892\" class=\"mrow\"><span id=\"MathJax-Span-26893\" class=\"semantics\"><span id=\"MathJax-Span-26894\" class=\"mrow\"><span id=\"MathJax-Span-26895\" class=\"mrow\"><span id=\"MathJax-Span-26901\" class=\"mo\">.<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n<p id=\"fs-id1165137557013\">The magnitude (size) of an earthquake is measured on a scale known as the Richter Scale.\u00a0<sup id=\"footnote-ref4\"><\/sup>The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 6 is not twice as great as an earthquake of magnitude 3.\u00a0 It is 10<sup>6\u22123\u00a0<\/sup>= 10<sup>3\u00a0<\/sup>= 1,000\u00a0times as great!<\/p>\n<p>The Richter scale strength of an earthquake, M, is given by [latex]M=\\mathrm{log}\\left(\\frac{W}{W_0}\\right),[\/latex] where [latex]W[\/latex] is the strength of the seismic waves of an earthquake and [latex]W_0[\/latex] is the strength of normally occurring earthquakes.\u00a0 Minor earthquakes occur regularly allowing [latex]W_0[\/latex] to be determined.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 17:\u00a0 The Richter Scale<\/h3>\n<p>An earthquake Richter Scale strength of 5 is considered a moderate strength earthquake.\u00a0 How much stronger was the 2010 magnitude 5.5 earthquake that occurred between Ontario and Quebec compared to a standard earthquake?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q397328\">Show Solution<\/span><\/p>\n<div id=\"q397328\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we note that [latex]M=5.5[\/latex] and substitute this into the Richter Scale equation.\u00a0 [latex]5.5=\\mathrm{log}\\left(\\frac{W}{W_0}\\right)[\/latex] written as an exponential equation is [latex]10^{5.5}=\\frac{W}{W_0}.[\/latex] This can be written as [latex]W=10^{5.5}W_0[\/latex] or\u00a0 [latex]W\\approx316228W_0.[\/latex]<\/p>\n<p>The earthquake was 316,228 times stronger than a standard earthquake.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1165137759741\">In chemistry,\u00a0<span class=\"no-emphasis\">pH<\/span>\u00a0is used as a measure of the acidity or alkalinity of a substance. The pH scale runs from 0 to 14. Substances with a pH less than 7 are considered <em>acidic<\/em>, and substances with a pH greater than 7 are said to be <em>alkaline<\/em>. Our bodies, for instance, must maintain a pH close to 7.35 in order for enzymes to work properly. To get a feel for what is acidic and what is alkaline, consider the following pH levels of some common substances:<\/p>\n<ul id=\"fs-id1165135253210\">\n<li>Battery acid: 0.8<\/li>\n<li>Stomach acid: 2.7<\/li>\n<li>Orange juice: 3.3<\/li>\n<li>Pure water: 7 (at 25\u00b0 C)<\/li>\n<li>Human blood: 7.35<\/li>\n<li>Fresh coconut: 7.8<\/li>\n<li>Sodium hydroxide (lye): 14<\/li>\n<\/ul>\n<p id=\"fs-id1165137540406\">To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where [latex]\\left[H^+\\right][\/latex]<span id=\"MathJax-Element-982-Frame\" class=\"MathJax\" style=\"font-style: normal;font-weight: normal;line-height: normal;font-size: 14px;text-indent: 0px;text-align: left;letter-spacing: normal;float: none;direction: ltr;max-width: none;max-height: none;min-width: 0px;min-height: 0px;border: 0px;padding: 0px;margin: 0px\" role=\"presentation\"><span id=\"MathJax-Span-13524\" class=\"math\"><span id=\"MathJax-Span-13525\" class=\"mrow\"><span id=\"MathJax-Span-13526\" class=\"semantics\"><span id=\"MathJax-Span-13527\" class=\"mrow\"><span id=\"MathJax-Span-13528\" class=\"mrow\"><span id=\"MathJax-Span-13529\" class=\"mtext\"><\/span><span id=\"MathJax-Span-13530\" class=\"mi\"><\/span><span id=\"MathJax-Span-13531\" class=\"mtext\"><\/span><\/span><\/span><\/span><\/span><\/span> <\/span>is the <em>concentration of hydrogen ions<\/em> in the solution measured in moles per liter.<\/p>\n<p style=\"text-align: center\">[latex]pH=-\\mathrm{log}\\left(\\left[H^+\\right]\\right)=\\mathrm{log}\\left(\\frac{1}{\\left[H^+\\right]}\\right)[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Example 18:\u00a0 pH<\/h3>\n<p>If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q63528\">Show Solution<\/span><\/p>\n<div id=\"q63528\" class=\"hidden-answer\" style=\"display: none\">\n<p>Suppose [latex]c[\/latex] is the original concentration of hydrogen ions, and [latex]p[\/latex] is the original pH of the liquid.\u00a0 Then [latex]p=-\\mathrm{log}\\left(c\\right).[\/latex]\u00a0 If the concentration is doubled, the new concentration is [latex]2c[\/latex].\u00a0 Then the pH of the new liquid is<\/p>\n<p style=\"text-align: center\">[latex]pH=-\\mathrm{log}\\left(2c\\right).[\/latex]<\/p>\n<p>Using the product rule of logarithms,<\/p>\n<p style=\"text-align: center\">[latex]pH=-\\left(\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(c\\right)\\right)=-\\mathrm{log}\\left(2\\right)-\\mathrm{log}\\left(c\\right).[\/latex]<\/p>\n<p>Since [latex]p=-log(c),[\/latex] the new pH is<\/p>\n<p style=\"text-align: center\">[latex]pH=p-\\mathrm{log}\\left(2\\right)\\approx p-0.301.[\/latex]<\/p>\n<p>When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 19:\u00a0 pH<\/h3>\n<p>Orange juice has a pH of approximately 3.3.\u00a0 Determine the hydrogen ion concentration.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q574782\">Show Solution<\/span><\/p>\n<div id=\"q574782\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the pH of orange juice is 3.3, we substitute 3.3 into the pH formula and get [latex]3.3=-\\mathrm{log}\\left(\\left[H^+\\right]\\right).[\/latex] Rewriting this as an exponential function, [latex]\\left[H^+\\right]=10^{-3.3}\\approx5.012\\times10^{-4}.[\/latex]\u00a0 Therefore the concentration is [latex]\\left[H^+\\right]\\approx5\\times10^{-4}=0.0005,[\/latex] or 0.0005 moles per liter.<\/p><\/div>\n<\/div>\n<\/div>\n<p>A <em>decibel<\/em> (dB) is a measure of how loud a sound is when compared to a reference value.\u00a0 A commonly used reference value is the sound intensity of the softest sound a human can typically hear; usually that of a child.\u00a0 We will call this value [latex]I_0.[\/latex]\u00a0 Since there is a very wide range of sounds that humans can hear, the logarithmic scale is used. The formula for decibels is<\/p>\n<p style=\"text-align: center\">Sound level in decibels [latex]=10\\mathrm{log}\\left(\\frac{I}{I_0}\\right),[\/latex]<\/p>\n<p>where [latex]I[\/latex] is the sound intensity of the sound being measured.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 20:\u00a0 Decibels<\/h3>\n<p>A vacuum cleaner sound level measures at 75dB and a balloon popping measures 125dB.\u00a0 \u00a0A balloon popping is how many times more intense than the sound intensity of the vacuum cleaner?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q485947\">Show Solution<\/span><\/p>\n<div id=\"q485947\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]I_B[\/latex] be the sound intensity of the balloon popping and [latex]I_V[\/latex] be the sound intensity of the vacuum cleaner.\u00a0 We then look at the difference.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}125\\text{ dB }-75\\text{ dB }&=10\\mathrm{log}\\left(\\frac{I_B}{I_0}\\right)-10\\mathrm{log}\\left(\\frac{I_V}{I_0}\\right)&&\\text{ }\\\\50\\text{ dB }&=10\\mathrm{log}\\left(\\frac{\\frac{I_B}{I_0}}{\\frac{I_V}{I_0}}\\right)&&\\text{Use the quotient rule for logarithms.}\\\\5\\text{ dB }&=\\mathrm{log}\\left(\\frac{I_B}{I_V}\\right)&&\\text{Simplify.}\\\\10^5&=\\frac{I_B}{I_V}&&\\text{Write as an exponential equation.}\\\\100000I_V&=I_B&&\\text{Multiply both sides by }I_V.\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\n<p>The balloon&#8217;s sound intensity is 100,000 times more than the vacuum cleaner&#8217;s sound intensity.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1165137410373\" class=\"precalculus media\">\n<p id=\"fs-id1165137722668\">Access these online resources for additional instruction and practice with exponential and logarithmic equations.<\/p>\n<ul id=\"fs-id1165137722673\">\n<li><a href=\"http:\/\/openstax.org\/l\/solvelogeq\">Solving Logarithmic Equations<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/solveexplog\">Solving Exponential Equations with Logarithms<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137431646\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id1165137450888\">\n<li>An exponential equation can be solved by taking the logarithm of each side.<\/li>\n<li>We can solve exponential equations with base [latex]e,[\/latex] by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other.<\/li>\n<li>After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions.<\/li>\n<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c,[\/latex] where [latex]S[\/latex] is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation [latex]{b}^{c}=S,[\/latex] and solve for the unknown.<\/li>\n<li>We can also use graphing to solve equations with the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c.[\/latex] We graph both equations [latex]y={\\mathrm{log}}_{b}\\left(S\\right)[\/latex] and [latex]y=c[\/latex] on the same coordinate plane and identify the solution as the <em>x-<\/em>value of the intersecting point.<\/li>\n<li>Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Glossary<\/h3>\n<dl id=\"fs-id1165135149311\">\n<dt>extraneous solution<\/dt>\n<dd id=\"fs-id1165134059776\">a solution introduced while solving an equation that does not satisfy the conditions of the original equation<\/dd>\n<\/dl>\n<\/div>\n","protected":false},"author":311,"menu_order":9,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-456","chapter","type-chapter","status-publish","hentry"],"part":223,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/456","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":35,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/456\/revisions"}],"predecessor-version":[{"id":3186,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/456\/revisions\/3186"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/parts\/223"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/456\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/media?parent=456"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=456"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/contributor?post=456"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/license?post=456"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}