{"id":3521,"date":"2016-08-05T05:30:12","date_gmt":"2016-08-05T05:30:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=3521"},"modified":"2016-09-28T21:37:44","modified_gmt":"2016-09-28T21:37:44","slug":"use-compound-interest-formulas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tallahassee-intermediatealgebra\/chapter\/use-compound-interest-formulas\/","title":{"raw":"Exponential and Logarithmic Equations","rendered":"Exponential and Logarithmic Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Exponential Equations with Unlike Bases\r\n<ul>\r\n \t<li>Identify an exponential equation whose terms all have the same base<\/li>\r\n \t<li>Idenitfy cases where equations can be rewritten so all terms have the same base<\/li>\r\n \t<li>Apply the one-to-one property of exponents to solve an exponential equation<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Exponential Equations with Unlike Bases\r\n<ul>\r\n \t<li>Use logarithms to solve exponential equations whose terms\u00a0cannot be rewritten with the same base<\/li>\r\n \t<li>Solve exponential equations of the form \u00a0[latex]y=A{e}^{kt}[\/latex] for t<\/li>\r\n \t<li>Recognize when there may be extraneous solutions, or no solutions for exponential equations<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Logarithmic Equations\r\n<ul>\r\n \t<li>Use the definition of a logarithm to solve logarithmic equations<\/li>\r\n \t<li>Use a graph to verify or analyze the solution to a logarithmic equation<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Applied Exponential and Logarithmic Equations\r\n<ul>\r\n \t<li>Solve half-life problems<\/li>\r\n \t<li>Solve pH problems<\/li>\r\n \t<li>Solve problems involving Richter scale readings<\/li>\r\n \t<li>Solve problems involving decibels<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n\r\n[caption id=\"\" align=\"alignleft\" width=\"450\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201933\/CNX_Precalc_Figure_04_06_0012.jpg\" alt=\"Seven rabbits in front of a brick building.\" width=\"450\" height=\"298\" \/> Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the \"rabbit plague.\" (credit: Richard Taylor, Flickr)[\/caption]\r\n\r\nIn 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.\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 functions.<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1165134354674\">When an <strong>exponential equation<\/strong> has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we can\u00a0set the exponents equal to one another, and solve for the unknown.<\/p>\r\n<p id=\"fs-id1165135192889\">For example, consider the equation [latex]{3}^{4x - 7}=\\frac{{3}^{2x}}{3}[\/latex]. To solve for <i>x<\/i>, we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for <em>x<\/em>:<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{3}^{4x - 7}\\hfill &amp; =\\frac{{3}^{2x}}{3}\\hfill &amp; \\hfill \\\\ {3}^{4x - 7}\\hfill &amp; =\\frac{{3}^{2x}}{{3}^{1}}\\hfill &amp; {\\text{Rewrite 3 as 3}}^{1}.\\hfill \\\\ {3}^{4x - 7}\\hfill &amp; ={3}^{2x - 1}\\hfill &amp; \\text{Use the division property of exponents}\\text{.}\\hfill \\\\ 4x - 7\\hfill &amp; =2x - 1\\text{ }\\hfill &amp; \\text{Apply the one-to-one property of exponents}\\text{.}\\hfill \\\\ 2x\\hfill &amp; =6\\hfill &amp; \\text{Subtract 2}x\\text{ and add 7 to both sides}\\text{.}\\hfill \\\\ x\\hfill &amp; =3\\hfill &amp; \\text{Divide by 3}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\r\nIn our first example, we solve an exponential equation whose terms all have a common base.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]{2}^{x - 1}={2}^{2x - 4}[\/latex].\r\n[reveal-answer q=\"579160\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"579160\"][latex]\\begin{array}{c} {2}^{x - 1}={2}^{2x - 4}\\hfill &amp; \\text{The common base is }2.\\hfill \\\\ \\text{ }x - 1=2x - 4\\hfill &amp; \\text{By the one-to-one property the exponents must be equal}.\\hfill \\\\ \\text{ }x=3\\hfill &amp; \\text{Solve for }x.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn general, we can summarize solving exponential equations whose terms all have the same base in this way:\r\nFor any algebraic expressions <em>S<\/em>\u00a0and <em>T<\/em>, and any positive real number [latex]b\\ne 1[\/latex]\r\n<div id=\"fs-id1165137702126\" class=\"equation\" data-type=\"equation\">[latex]{b}^{S}={b}^{T}\\text{ if and only if }S=T[\/latex]<\/div>\r\n<div class=\"equation\" data-type=\"equation\">\r\n<ul>\r\n \t<li>Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[\/latex].<\/li>\r\n \t<li>Use the one-to-one property to set the exponents equal.<\/li>\r\n \t<li>Solve the resulting equation, <em>S\u00a0<\/em>= <em>T<\/em>, for the unknown.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1165137730366\" class=\"solution\" data-type=\"solution\"><section data-depth=\"1\"><section id=\"fs-id1165137667260\" data-depth=\"2\">\r\n<h2 data-type=\"title\">Rewriting Equations So All Powers Have the Same Base<\/h2>\r\n<p id=\"fs-id1165137725147\">Sometimes\u00a0we can rewrite the terms in an equation as powers with a common base, and solve using the one-to-one property. This takes a keen eye for recognizing common powers. \u00a0For example, you can rewrite 8 as [latex]2^3[\/latex] or 36 as [latex]6^2[\/latex] or [latex]\\frac{1}{4}[\/latex] as [latex]\\left(\\frac{1}{2}\\right)^{2}[\/latex]<\/p>\r\n<p id=\"fs-id1165137784867\">Consider the equation [latex]256={4}^{x - 5}[\/latex]. We can rewrite both sides of this equation as a power of 2. Then we apply the rules of exponents, along with the one-to-one property, to solve for <em>x<\/em>:<\/p>\r\n\r\n<div id=\"eip-687\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}256={4}^{x - 5}\\hfill &amp; \\hfill \\\\ {2}^{8}={\\left({2}^{2}\\right)}^{x - 5}\\hfill &amp; \\text{Rewrite each side as a power with base 2}.\\hfill \\\\ {2}^{8}={2}^{2x - 10}\\hfill &amp; \\text{Use the one-to-one property of exponents}.\\hfill \\\\ 8=2x - 10\\hfill &amp; \\text{Apply the one-to-one property of exponents}.\\hfill \\\\ 18=2x\\hfill &amp; \\text{Add 10 to both sides}.\\hfill \\\\ x=9\\hfill &amp; \\text{Divide by 2}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">In the next example, we show how to find a common base for two expressions whose bases are 8, and 16. \u00a0We can then solve the resulting equation using the one-to-one property of exponents.<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]{8}^{x+2}={16}^{x+1}[\/latex].\r\n[reveal-answer q=\"731579\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"731579\"][latex]\\begin{array}{c}\\text{ }{8}^{x+2}={16}^{x+1}\\hfill &amp; \\hfill \\\\ {\\left({2}^{3}\\right)}^{x+2}={\\left({2}^{4}\\right)}^{x+1}\\hfill &amp; \\text{Write }8\\text{ and }16\\text{ as powers of }2.\\hfill \\\\ \\text{ }{2}^{3x+6}={2}^{4x+4}\\hfill &amp; \\text{To take a power of a power, multiply exponents}.\\hfill \\\\ \\text{ }3x+6=4x+4\\hfill &amp; \\text{Use the one-to-one property to set the exponents equal}.\\hfill \\\\ \\text{ }x=2\\hfill &amp; \\text{Solve for }x.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example, we are given an exponential equation that contains a square root. \u00a0Remember that you can write roots as rational exponents, so you may be able to find like bases when it is not completely obvious at first.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]{2}^{5x}=\\sqrt{2}[\/latex].\r\n[reveal-answer q=\"507738\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"507738\"][latex]\\begin{array}{c}{2}^{5x}={2}^{\\frac{1}{2}}\\hfill &amp; \\text{Write the square root of 2 as a power of }2.\\hfill \\\\ 5x=\\frac{1}{2}\\hfill &amp; \\text{Use the one-to-one property}.\\hfill \\\\ x=\\frac{1}{10}\\hfill &amp; \\text{Solve for }x.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nBy changing [latex]\\sqrt{2}[\/latex] to [latex]{2}^{\\frac{1}{2}}[\/latex] we were able to solve the equation in the previous example. In general, here are some steps to consider when you are solving exponential equations. \u00a0A good first step is always to determine whether you can rewrite the terms with a common base.\r\n<ol id=\"fs-id1165137663646\" data-number-style=\"arabic\">\r\n \t<li>Rewrite each side in the equation as a power with a common base.<\/li>\r\n \t<li>Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[\/latex].<\/li>\r\n \t<li>Use the one-to-one property to set the exponents equal.<\/li>\r\n \t<li>Solve the resulting equation, <em>S\u00a0<\/em>= <em>T<\/em>, for the unknown.<\/li>\r\n<\/ol>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nDo all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process? Write your thoughts in the textbox below before you check our proposed answer.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"711116\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"711116\"]No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined<em data-effect=\"italics\">.<\/em>[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example we show you a case where there is no solution to an exponential equation. \u00a0Remember how exponential functions are defined and ask yourself - \"does this make sense\" before diving into solving exponential equations.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]{3}^{x+1}=-2[\/latex].\r\n[reveal-answer q=\"152201\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"152201\"]This equation has no solution. There is no real value of <em>x<\/em>\u00a0that will make the equation a true statement because any power of a positive number is positive.\r\n\r\nFor example [latex]3^2=9[\/latex], and [latex]2^4=16[\/latex]. \u00a0Remember that we have defined exponential functions as having a base that is greater than 0.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h3>\u00a0Analysis of the Solution<\/h3>\r\n<\/div>\r\n<\/section><\/section><\/div>\r\n<div id=\"Example_04_06_04\" class=\"example\" data-type=\"example\">\r\n<div id=\"fs-id1165137405247\" class=\"exercise\" data-type=\"exercise\">\r\n<div id=\"fs-id1165137849213\" class=\"commentary\" data-type=\"commentary\">\r\n<p id=\"fs-id1165137578263\">The figure below\u00a0shows the graphs of the two separate expressions in the equation [latex]{3}^{x+1}=-2[\/latex] as [latex]y={3}^{x+1}[\/latex] and [latex]y=-2[\/latex]. The two graphs do not cross showing us that\u00a0the left side is never equal to the right side. Thus the equation has no solution.<\/p>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201934\/CNX_Precalc_Figure_04_06_0022.jpg\" alt=\"Graph of 3^(x+1)=-2 and y=-2. The graph notes that they do not cross.\" width=\"487\" height=\"438\" data-media-type=\"image\/jpg\" \/>\r\n\r\n<\/div>\r\n<h2>Exponential Equations with unlike Bases<\/h2>\r\nSometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since [latex]\\mathrm{log}\\left(a\\right)=\\mathrm{log}\\left(b\\right)[\/latex] is equivalent to <em>a\u00a0<\/em>= <em>b<\/em>, we may apply logarithms with the same base on both sides of an exponential equation.\r\n\r\nIn our first example we will use the law of logs combined with factoring to solve an exponential equation whose terms do not have the same base. Note how first, we rewrite the exponential terms as logarithms.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]{5}^{x+2}={4}^{x}[\/latex].\r\n[reveal-answer q=\"281663\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"281663\"]\r\n\r\nThere is no easy way to get the powers to have the same base for this equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\text{ }{5}^{x+2}={4}^{x}. \\\\ \\text{ }\\mathrm{ln}{5}^{x+2}=\\mathrm{ln}{4}^{x}\\hfill &amp; \\text{Take ln of both sides}.\\hfill \\\\ \\text{ }\\left(x+2\\right)\\mathrm{ln}5=x\\mathrm{ln}4\\hfill &amp; \\text{Use laws of logs}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5+2\\mathrm{ln}5=x\\mathrm{ln}4\\hfill &amp; \\text{Use the distributive law}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5-x\\mathrm{ln}4=-2\\mathrm{ln}5\\hfill &amp; \\text{Get terms containing }x\\text{ on one side, terms without }x\\text{ on the other}.\\hfill \\\\ x\\left(\\mathrm{ln}5-\\mathrm{ln}4\\right)=-2\\mathrm{ln}5\\hfill &amp; \\text{On the left hand side, factor out an }x.\\hfill \\\\ \\text{ }x\\mathrm{ln}\\left(\\frac{5}{4}\\right)=\\mathrm{ln}\\left(\\frac{1}{25}\\right)\\hfill &amp; \\text{Use the laws of logs}.\\hfill \\\\ \\text{ }x=\\frac{\\mathrm{ln}\\left(\\frac{1}{25}\\right)}{\\mathrm{ln}\\left(\\frac{5}{4}\\right)}\\hfill &amp; \\text{Divide by the coefficient of }x.\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn general we can solve exponential equations whose terms do not have like bases in the following way:\r\n<ol id=\"fs-id1165137784632\" data-number-style=\"arabic\">\r\n \t<li>Apply the logarithm of both sides of the equation.\r\n<ul id=\"fs-id1165137824134\" data-bullet-style=\"bullet\">\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 class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nIs there any way to solve [latex]{2}^{x}={3}^{x}[\/latex]?\r\n\r\nUse the textbox below to formulate an answer or example before you look at the solution.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"608971\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"608971\"]Yes. The solution is x = 0.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Equations Containing [latex]e[\/latex]<\/h2>\r\n<section id=\"fs-id1165137469838\" data-depth=\"2\">Base <em>e\u00a0<\/em>is a very common base found in science, finance and engineering applications.\u00a0When we have an equation with a base <em>e<\/em>\u00a0on either side, we can use the <strong>natural logarithm<\/strong> to solve it. Earlier, we introduced a formula that models continuous growth,\u00a0[latex]y=A{e}^{kt}[\/latex]. \u00a0This formula is found in business, finance, and many biological and physical science applications. In our next example we will show how to solve this equation for t, the elapsed time for the behavior in question.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]100=20{e}^{2t}[\/latex].\r\n[reveal-answer q=\"714083\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"714083\"][latex]\\begin{array}{c}100\\hfill &amp; =20{e}^{2t}\\hfill &amp; \\hfill \\\\ 5\\hfill &amp; ={e}^{2t}\\hfill &amp; \\text{Divide by the coefficient of the power}\\text{.}\\hfill \\\\ \\mathrm{ln}5\\hfill &amp; =2t\\hfill &amp; \\text{Take ln of both sides}\\text{. Use the fact that }\\mathrm{ln}\\left(x\\right)\\text{ and }{e}^{x}\\text{ are inverse functions}\\text{.}\\hfill \\\\ t\\hfill &amp; =\\frac{\\mathrm{ln}5}{2}\\hfill &amp; \\text{Divide by the coefficient of }t\\text{.}\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_06_06\" class=\"example\" data-type=\"example\">\r\n<div id=\"fs-id1165137846466\" class=\"exercise\" data-type=\"exercise\">\r\n<div id=\"fs-id1165137705083\" class=\"commentary\" data-type=\"commentary\">\r\n<h3 data-type=\"title\">Analysis of the Solution<\/h3>\r\nUsing laws of logs, we can also write this answer in the form [latex]t=\\mathrm{ln}\\sqrt{5}[\/latex]. If we want a decimal approximation of the answer, we use a calculator.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nDoes every equation of the form [latex]y=A{e}^{kt}[\/latex] have a solution? Write your thoughts or an example in the textbox below before you check the answer.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"483016\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"483016\"]No. There is a solution when [latex]k\\ne 0[\/latex], and when y and A 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].[\/hidden-answer]\r\n\r\n<\/div>\r\nWe will provide one more example using the continuous growth formula, but this time we have to do a couple steps of algebra to get in it a form that can be solved.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]4{e}^{2x}+5=12[\/latex].\r\n[reveal-answer q=\"593052\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"593052\"][latex]\\begin{array}{c}4{e}^{2x}+5=12\\hfill &amp; \\hfill \\\\ 4{e}^{2x}=7\\hfill &amp; \\text{Combine like terms}.\\hfill \\\\ {e}^{2x}=\\frac{7}{4}\\hfill &amp; \\text{Divide by the coefficient of the power}.\\hfill \\\\ 2x=\\mathrm{ln}\\left(\\frac{7}{4}\\right)\\hfill &amp; \\text{Take ln of both sides}.\\hfill \\\\ x=\\frac{1}{2}\\mathrm{ln}\\left(\\frac{7}{4}\\right)\\hfill &amp; \\text{Solve for }x.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\"><\/div>\r\n<div class=\"bcc-box bcc-success\"><\/div>\r\n<\/section><section id=\"fs-id1165137665482\" data-depth=\"2\">\r\n<h2 data-type=\"title\">Extraneous Solutions<\/h2>\r\n<p id=\"fs-id1165137742403\">Sometimes the methods used to solve an equation introduce an <strong>extraneous solution<\/strong>, 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\nIn the next example we will solve an exponential equation that is quadratic in form. We will factor first and then use the zero product principle. Note how we find two solutions, but reject one that does not satisfy the original equaiton.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]{e}^{2x}-{e}^{x}=56[\/latex].\r\n[reveal-answer q=\"152760\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"152760\"][latex]\\begin{array}{c}{e}^{2x}-{e}^{x}\\hfill &amp; =56\\hfill &amp; \\hfill \\\\ {e}^{2x}-{e}^{x}-56\\hfill &amp; =0\\hfill &amp; \\text{Get one side of the equation equal to zero}.\\hfill \\\\ \\left({e}^{x}+7\\right)\\left({e}^{x}-8\\right)\\hfill &amp; =0\\hfill &amp; \\text{Factor by the FOIL method}.\\hfill \\\\ {e}^{x}+7\\hfill &amp; =0\\text{ or }{e}^{x}-8=0 &amp; \\text{If a product is zero, then one factor must be zero}.\\hfill \\\\ {e}^{x}\\hfill &amp; =-7{\\text{ or e}}^{x}=8\\hfill &amp; \\text{Isolate the exponentials}.\\hfill \\\\ {e}^{x}\\hfill &amp; =8\\hfill &amp; \\text{Reject the equation in which the power equals a negative number}.\\hfill \\\\ x\\hfill &amp; =\\mathrm{ln}8\\hfill &amp; \\text{Solve the equation in which the power equals a positive number}.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>\u00a0Analysis of the Solution<\/h3>\r\n<div id=\"Example_04_06_08\" class=\"example\" data-type=\"example\">\r\n<div id=\"fs-id1165137828517\" class=\"exercise\" data-type=\"exercise\">\r\n<div id=\"fs-id1165137806430\" class=\"commentary\" data-type=\"commentary\">\r\n<p id=\"fs-id1165137806436\">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]x=\\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.<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nDoes every logarithmic equation have a solution? Write your ideas, or a counter example in the box below.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"810736\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"810736\"]No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Logarithmic Equations<\/h2>\r\nWe have already seen that every <strong>logarithmic equation<\/strong> [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equivalent to the exponential equation [latex]{b}^{y}=x[\/latex]. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.\r\n<p id=\"fs-id1165134148350\">For example, consider the equation [latex]{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3[\/latex]. 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 <em>x<\/em>:<\/p>\r\n\r\n<div id=\"eip-id2205910\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3\\hfill &amp; \\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(2\\left(3x - 5\\right)\\right)=3\\hfill &amp; \\text{Apply the product rule of logarithms}.\\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(6x - 10\\right)=3\\hfill &amp; \\text{Distribute}.\\hfill \\\\ \\text{ }{2}^{3}=6x - 10\\hfill &amp; \\text{Apply the definition of a logarithm}.\\hfill \\\\ \\text{ }8=6x - 10\\hfill &amp; \\text{Calculate }{2}^{3}.\\hfill \\\\ \\text{ }18=6x\\hfill &amp; \\text{Add 10 to both sides}.\\hfill \\\\ \\text{ }x=3\\hfill &amp; \\text{Divide by 6}.\\hfill \\end{array}[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\"><\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">In our first example we will show how to use techniques from solving linear equations to solve a logarithmic equation.<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]2\\mathrm{ln}x+3=7[\/latex].\r\n[reveal-answer q=\"1854\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"1854\"][latex]\\begin{array}{c}2\\mathrm{ln}x+3=7\\hfill &amp; \\hfill \\\\ \\text{ }2\\mathrm{ln}x=4\\hfill &amp; \\text{Subtract 3}.\\hfill \\\\ \\text{ }\\mathrm{ln}x=2\\hfill &amp; \\text{Divide by 2}.\\hfill \\\\ \\text{ }x={e}^{2}\\hfill &amp; \\text{Rewrite in exponential form}.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Analysis of the solution<\/h3>\r\nThe solution for the previous equation was [latex]x={e}^{2}[\/latex], this is often referred to as the exact solution. \u00a0Sometimes, you may be asked to give an approximation, which would be more useful to someone using the result for a financial, scientific or engineering application. The approximation for\u00a0[latex]x={e}^{2}[\/latex] can be found with a calculator.\u00a0[latex]x={e}^{2}\\approx{7.4}[\/latex]\r\n\r\nIn general, we can describe using the definition of a logarithm to solve logarithmic equations as follows:\r\n<p class=\"title\" data-type=\"title\">For any algebraic expression <em>S<\/em> and real numbers <em>b<\/em> and <em>c<\/em>, where [latex]b&gt;0,\\text{ }b\\ne 1[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165137732219\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{ if and only if }{b}^{c}=S[\/latex]<\/div>\r\n<div class=\"equation\" data-type=\"equation\">Here is another example of what you may encounter when solving logarithmic equations.<\/div>\r\n<\/div>\r\n<div class=\"equation\" style=\"text-align: left;\" data-type=\"equation\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]2\\mathrm{ln}\\left(6x\\right)=7[\/latex].\r\n[reveal-answer q=\"663498\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"663498\"][latex]\\begin{array}{c}2\\mathrm{ln}\\left(6x\\right)=7\\hfill &amp; \\hfill \\\\ \\text{ }\\mathrm{ln}\\left(6x\\right)=\\frac{7}{2}\\hfill &amp; \\text{Divide by 2}.\\hfill \\\\ \\text{ }6x={e}^{\\left(\\frac{7}{2}\\right)}\\hfill &amp; \\text{Use the definition of }\\mathrm{ln}.\\hfill \\\\ \\text{ }x=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}\\hfill &amp; \\text{Divide by 6}.\\hfill \\end{array}[\/latex]\r\n<h4>Answer<\/h4>\r\nExact answer: [latex]x=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}[\/latex]\r\n\r\nApproximate answer: [latex]x\\approx{5.5}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example we will show you the power of using graphs to analyze solutions to logarithmic equations. By turning each side of the equation into a function and plotting them on the same set of axes, we can see how they interact with each other. \u00a0In this case, we get a logarithmic function and a horizontal line, and find that the solution is the point where the two intersect.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]\\mathrm{ln}x=3[\/latex].\r\n[reveal-answer q=\"691419\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"691419\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\mathrm{ln}x=3\\hfill &amp; \\hfill \\\\ x={e}^{3}\\hfill &amp; \\text{Use the definition of the natural logarithm}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\r\n<p id=\"fs-id1165137443165\">The graph below\u00a0represents the graph of the equation. On the graph, the <em data-effect=\"italics\">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<figure class=\"small\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201935\/CNX_Precalc_Figure_04_06_0032.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\" data-media-type=\"image\/jpg\" \/><\/figure><figure id=\"CNX_Precalc_Figure_04_06_003\" class=\"small\"><figcaption>\u00a0The graphs of [latex]y=\\mathrm{ln}x[\/latex] and <em>y\u00a0<\/em>= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex], which is approximately (20.0855, 3).<\/figcaption><\/figure>[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the same way that a story can help you understand a complex concept, graphs can help us understand a wide variety of concepts in mathematics. Visual representations of the parts of an equation can be created by turning each side into a function and plotting the functions on the same set of axes.\r\n\r\n<\/div>\r\n<h2>Use the one-to-one property of logarithms to solve logarithmic equations<\/h2>\r\n<section id=\"fs-id1165137755280\" data-depth=\"1\">\r\n<p id=\"fs-id1165135237092\">As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers <em>x\u00a0<\/em>&gt; 0, <em>S\u00a0<\/em>&gt; 0, <em>T\u00a0<\/em>&gt; 0 and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],<\/p>\r\n\r\n<div id=\"eip-674\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex].<\/div>\r\n<p id=\"eip-625\">For example,<\/p>\r\n\r\n<div id=\"eip-453\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{If }{\\mathrm{log}}_{2}\\left(x - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right),\\text{then }x - 1=8[\/latex].<\/div>\r\n<p id=\"fs-id1165137874962\">So, if [latex]x - 1=8[\/latex], then we can solve for <em>x<\/em>, and we get <em>x\u00a0<\/em>= 9. To check, we can substitute <em>x\u00a0<\/em>= 9 into the original equation: [latex]{\\mathrm{log}}_{2}\\left(9 - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex]. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)[\/latex]\r\n\r\nCheck your results.\r\n[reveal-answer q=\"874129\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"874129\"][latex]\\begin{array}{c}\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)\\hfill &amp; \\hfill \\\\ \\text{ }\\mathrm{log}\\left(\\frac{3x - 2}{2}\\right)=\\mathrm{log}\\left(x+4\\right)\\hfill &amp; \\text{Apply the quotient rule of logarithms}.\\hfill \\\\ \\text{ }\\frac{3x - 2}{2}=x+4\\hfill &amp; \\text{Apply the one to one property of a logarithm}.\\hfill \\\\ \\text{ }3x - 2=2x+8\\hfill &amp; \\text{Multiply both sides of the equation by }2.\\hfill \\\\ \\text{ }x=10\\hfill &amp; \\text{Subtract 2}x\\text{ and add 2}.\\hfill \\end{array}[\/latex]\r\n<p id=\"fs-id1165135172191\">To check the result, substitute <em>x\u00a0<\/em>= 10 into [latex]\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)[\/latex].<\/p>\r\n\r\n<div id=\"eip-316\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\mathrm{log}\\left(3\\left(10\\right)-2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(\\left(10\\right)+4\\right)\\hfill &amp; \\hfill \\\\ \\text{ }\\mathrm{log}\\left(28\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(14\\right)\\hfill &amp; \\hfill \\\\ \\text{ }\\mathrm{log}\\left(\\frac{28}{2}\\right)=\\mathrm{log}\\left(14\\right)\\hfill &amp; \\text{The solution checks}.\\hfill \\end{array}[\/latex]<\/div>\r\n<h4 class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\">Answer<\/h4>\r\n<em>x\u00a0<\/em>= 10\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNote, when solving an equation involving logarithms, always check to see if the answer is correct or if there is an extraneous solution.\r\n\r\nIn general, we can summarize solving logarithmic equations as follows:\r\n\r\nFor\u00a0[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]\r\n<ol data-number-style=\"arabic\">\r\n \t<li>Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form [latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex].<\/li>\r\n \t<li>Use the one-to-one property to set the arguments equal.<\/li>\r\n \t<li>Solve the resulting equation, <em>S<\/em> =\u00a0<em>T<\/em>, for the unknown.<\/li>\r\n<\/ol>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]\r\n[reveal-answer q=\"306816\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"306816\"][latex]\\begin{array}{c}\\text{ }\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}\\left(2x+3\\right)\\hfill &amp; \\hfill \\\\ \\text{ }{x}^{2}=2x+3\\hfill &amp; \\text{Use the one-to-one property of the logarithm}.\\hfill \\\\ \\text{ }{x}^{2}-2x - 3=0\\hfill &amp; \\text{Get zero on one side before factoring}.\\hfill \\\\ \\left(x - 3\\right)\\left(x+1\\right)=0\\hfill &amp; \\text{Factor using FOIL}.\\hfill \\\\ \\text{ }x - 3=0\\text{ or }x+1=0\\hfill &amp; \\text{If a product is zero, one of the factors must be zero}.\\hfill \\\\ \\text{ }x=3\\text{ or }x=-1\\hfill &amp; \\text{Solve for }x.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3 data-type=\"title\">Analysis of the Solution<\/h3>\r\n<p id=\"fs-id1165134505611\">There are two solutions: <em>x\u00a0<\/em>= 3 or <em>x\u00a0<\/em>= \u20131. The solution <em>x\u00a0<\/em>= \u20131 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.<\/p>\r\nApplications of Logarithmic Equations\r\n\r\n<section id=\"fs-id1165137828382\" data-depth=\"1\">\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<img class=\" wp-image-3765 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/11191452\/HEUraniumC-300x240.jpg\" alt=\"Gloved hands holding a dish of highly enrich uranium metal.\" width=\"393\" height=\"314\" \/>\r\n<p id=\"fs-id1165134192326\">One such application is called <strong>half-life, <\/strong>which refers to the amount of time it takes for half a given quantity of radioactive material to decay. The table below\u00a0lists the half-life for several of the more common radioactive substances.<\/p>\r\n\r\n<table style=\"width: 50%;\" summary=\"Seven rows and three columns. The first column is labeled,\">\r\n<thead>\r\n<tr>\r\n<th style=\"text-align: center;\" data-align=\"center\">Substance<\/th>\r\n<th style=\"text-align: center;\" data-align=\"center\">Use<\/th>\r\n<th style=\"text-align: center;\" data-align=\"center\">Half-life<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>gallium-67<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>80 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>cobalt-60<\/td>\r\n<td>manufacturing<\/td>\r\n<td>5.3 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>technetium-99m<\/td>\r\n<td>nuclear medicine<\/td>\r\n<td>6 hours<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>americium-241<\/td>\r\n<td>construction<\/td>\r\n<td>432 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>carbon-14<\/td>\r\n<td>archeological dating<\/td>\r\n<td>5,715 years<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>uranium-235<\/td>\r\n<td>atomic power<\/td>\r\n<td>703,800,000 years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165135209669\">We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:<\/p>\r\n\r\n<div id=\"eip-247\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}A\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{T}t}\\hfill \\\\ A\\left(t\\right)={A}_{0}{e}^{\\mathrm{ln}\\left(0.5\\right)\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left({e}^{\\mathrm{ln}\\left(0.5\\right)}\\right)}^{\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left(\\frac{1}{2}\\right)}^{\\frac{t}{T}}\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137408740\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137408743\">\r\n \t<li>[latex]{A}_{0}[\/latex] is the amount initially present<\/li>\r\n \t<li><em>T<\/em>\u00a0is the half-life of the substance<\/li>\r\n \t<li><em>t<\/em>\u00a0is the time period over which the substance is studied<\/li>\r\n \t<li>[latex]A(t)[\/latex]\u00a0is the amount of the substance present after time\u00a0<em>t<\/em><\/li>\r\n<\/ul>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nHow long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?\r\n[reveal-answer q=\"774904\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"774904\"][latex]\\begin{array}{c}\\text{ }y=\\text{1000}e\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill &amp; \\hfill \\\\ \\text{ }900=1000{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill &amp; \\text{After 10% decays, 900 grams are left}.\\hfill \\\\ \\text{ }0.9={e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill &amp; \\text{Divide by 1000}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\mathrm{ln}\\left({e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\right)\\hfill &amp; \\text{Take ln of both sides}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill &amp; \\text{ln}\\left({e}^{M}\\right)=M\\hfill \\\\ \\text{ }\\text{ }t=\\text{703,800,000}\\times \\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{years}\\begin{array}{c}{cccc}&amp; &amp; &amp; \\end{array}\\hfill &amp; \\text{Solve for }t.\\hfill \\\\ \\text{ }\\text{ }t\\approx \\text{106,979,777 years}\\hfill &amp; \\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_04_06_13\" class=\"example\" data-type=\"example\">\r\n<div id=\"fs-id1165137628651\" class=\"exercise\" data-type=\"exercise\">\r\n<div id=\"fs-id1165137453455\" class=\"commentary\" data-type=\"commentary\">\r\n<h3 data-type=\"title\">Analysis of the Solution<\/h3>\r\n<p id=\"fs-id1165137453460\">Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>pH<\/h2>\r\n<img class=\" wp-image-3767 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/11193831\/Lemon-300x212.jpg\" alt=\"Lemons\" width=\"259\" height=\"183\" \/>\r\n\r\nIn chemistry, pH is 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 acidic, and substances with a pH greater than 7 are said to be alkaline. In our next example we will find how\u00a0doubling the concentration of hydrogen ions in a liquid\u00a0affects pH.\r\n\r\n<\/section><section id=\"fs-id1165137828382\" data-depth=\"1\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn chemistry, [latex]\\text{pH}=-\\mathrm{log}\\left[{H}^{+}\\right][\/latex]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?\r\n[reveal-answer q=\"994106\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"994106\"]\r\n<p id=\"fs-id1165135415820\">Suppose <em>C<\/em>\u00a0is the original concentration of hydrogen ions, and <em>P<\/em>\u00a0is the original pH of the liquid. Then [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex]. If the concentration is doubled, the new concentration is 2<em>C<\/em>. Then the pH of the new liquid is<\/p>\r\n\r\n<div id=\"eip-id1165134547456\" class=\"equation unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)[\/latex]<\/div>\r\n<p id=\"fs-id1165135571875\">Using the product rule of logs<\/p>\r\n\r\n<div id=\"eip-id1165134547498\" class=\"equation unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)=-\\left(\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(C\\right)\\right)=-\\mathrm{log}\\left(2\\right)-\\mathrm{log}\\left(C\\right)[\/latex]<\/div>\r\n<p id=\"fs-id1165135443976\">Since [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex], the new pH is<\/p>\r\n\r\n<div id=\"eip-id1165135440065\" class=\"equation unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\text{pH}=P-\\mathrm{log}\\left(2\\right)\\approx P - 0.301[\/latex]<\/div>\r\n<p id=\"fs-id1165135251361\">When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Richter Scale<\/h2>\r\n[caption id=\"attachment_3770\" align=\"alignleft\" width=\"412\"]<img class=\" wp-image-3770\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/11194210\/Earthquake_Richter_Scale-300x223.jpg\" alt=\"Richter Scale of Earthquake Energy\" width=\"412\" height=\"306\" \/> Richter Scale of Earthquake Energy[\/caption]\r\n\r\nThe Richter scale is a logarithmic function that is used to measure the magnitude of earthquakes. The magnitude of an earthquake is related to how much energy is released by the quake. Instruments called seismographs detect movement in the earth; the smallest movement that can be detected shows on a seismograph as a wave with amplitude [latex]A_{0}[\/latex].\r\n\r\nA \u2013 the measure of the amplitude of the earthquake wave\r\n[latex]A_{0}[\/latex]\u00a0\u2013 the amplitude of the smallest detectable wave (or standard wave)\r\n\r\nFrom this you can find R, the Richter scale measure of the magnitude of the earthquake using the formula:\r\n<p style=\"text-align: center;\">[latex]R=\\mathrm{log}\\left(\\frac{A}{A_{0}}\\right)[\/latex]<\/p>\r\nThe intensity of an earthquake will typically measure between 2 and 10 on the Richter scale. Any earthquakes registering below a 5 are fairly minor; they may shake the ground a bit, but are seldom strong enough to cause much damage. Earthquakes with a Richter rating of between 5 and 7.9 are much more severe, and any quake above an 8 is likely to cause massive damage. (The highest rating ever recorded for an earthquake is 9.5 during the 1960 Valdivia earthquake in Chile.)\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nAn earthquake is measured with a wave amplitude 392 times as great as [latex]A_{0}[\/latex]. What is the magnitude of this earthquake using the Richter scale, to the nearest tenth?\r\n[reveal-answer q=\"677160\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"677160\"]\r\n\r\nUse the Richter scale equation.\r\n\r\n[latex]R=\\mathrm{log}\\left(\\frac{A}{A_{0}}\\right)[\/latex]\r\n\r\nSince A is 392 times as large as[latex]A_{0}[\/latex], [latex]A=392A_{0}[\/latex]. Substitute this expression in for A.\r\n\r\n[latex]R=\\mathrm{log}\\left(\\frac{392A_{0}}{A_{0}}\\right)[\/latex]\r\n\r\nSimplify the argument of the logarithm, then use a calculator to evaluate.\r\n\r\n[latex]\\begin{array}{c}R=\\mathrm{log}\\left(\\frac{392A_{0}}{A_{0}}\\right)\\\\=\\mathrm{log}\\left(392\\right)\\\\=2.5932\\\\\\approx{2.6}\\end{array}[\/latex]\r\n<h4>Answer<\/h4>\r\nThe earthquake registered 2.6 on the Richter scale.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nA difference of 1 point on the Richter scale equates to a 10-fold difference in the amplitude of the earthquake (which is related to the wave strength). This means that an earthquake that measures 3.6 on the Richter scale has 10 times the amplitude of one that measures 2.6.\r\n\r\nIn the Richter scale example, the wave amplitude of the earthquake was 392 times normal. What if it were 10 times that, or 3,920 times normal? To find the measurement of that size earthquake on the Richter scale, you find log 3920. A calculator gives a value of 3.5932...or 3.6, when rounded to the nearest tenth. One extra point on the Richter scale can mean a lot more shaking!\r\n<h2>Decibels<\/h2>\r\n[caption id=\"attachment_3772\" align=\"alignleft\" width=\"300\"]<img class=\"size-medium wp-image-3772\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/11195710\/Rui%CC%81do_Noise_041113GFDL-300x225.jpeg\" alt=\"Child covering his ears with his hands\" width=\"300\" height=\"225\" \/> Turn it Down[\/caption]\r\n\r\nSound is measured in a logarithmic scale using a unit called a decibel. The formula looks similar to the Richter scale:\r\n<p style=\"text-align: center;\">[latex]d=10\\mathrm{log}\\left(\\frac{P}{P_{0}}\\right)[\/latex]<\/p>\r\nwhere P is the power or intensity of the sound and P0 is the weakest sound that the human ear can hear. In the next example we will find how much more intense the noise from a dishwasher is than the noise from a hot water pump.\r\n\r\n<\/section><section id=\"fs-id1165137828382\" data-depth=\"1\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nOne hot water pump has a noise rating of 50 decibels. One dishwasher, however, has a noise rating of 62 decibels. The dishwasher noise is how many times more intense than the hot water pump noise?\r\n[reveal-answer q=\"789189\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"789189\"]\r\nYou can\u2019t easily compare the two noises using the formula, but you can compare them to [latex]P_{0}[\/latex]. Start by finding the intensity of noise for the hot water pump. Use h for the intensity of the hot water pump\u2019s noise.\r\n<p style=\"text-align: center;\">[latex]50=10\\mathrm{log}\\left(\\frac{h}{P_{0}}\\right)[\/latex]<\/p>\r\nDivide the equations by 10 to get the log by itself.\r\n<p style=\"text-align: center;\">[latex]5=\\mathrm{log}\\left(\\frac{h}{P_{0}}\\right)[\/latex]<\/p>\r\nRewrite the equation as an exponential equation.\r\n<p style=\"text-align: center;\">[latex]10^5=\\frac{h}{P_{0}}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Solve for h, the intensity of the water pump.<\/p>\r\n<p style=\"text-align: center;\">[latex]h=P_{0}10^5[\/latex]<\/p>\r\nRepeat the same process to find the intensity of the noise for the dishwasher, use d to represent the intensity of the sound of the dishwasher. Remember that [latex]{P_{0}}[\/latex] is a baseline for the most faint sound the human ear can hear.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}62=10\\mathrm{log}\\left(\\frac{d}{P_{0}}\\right)\\\\6.2=\\mathrm{log}\\left(\\frac{d}{P_{0}}\\right)\\\\10^{6.2}=\\frac{d}{P_{0}}\\\\d=P_{0}10^{6.2}\\end{array}[\/latex]<\/p>\r\nTo compare d to h, you can divide. (Think: if the dishwasher\u2019s noise is twice as intense as the pump\u2019s,\u00a0then d should be 2h\u2014that is, [latex]\\frac{d}{h}[\/latex]\u00a0should be 2.)\r\nUse the laws of exponents to simplify the quotient.\r\n<p style=\"text-align: center;\">[latex]\\Large\\frac{d}{h}=\\frac{P_{0}10^{6.2}}{P_{0}10^5}=\\frac{10^{6.2}}{10^{5}}=\\normalsize10^{6.2-5}=10^{1.2}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe dishwasher\u2019s noise is [latex]10^{1.2}[\/latex]\u00a0(or about 15.85) times as intense as the hot water pump.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nApplications of logarithms and exponentials are everywhere in science. \u00a0We hope the examples here have given you an idea of how useful they can be.\r\n\r\n<\/section>\r\n<h2>Summary<\/h2>\r\n<\/section><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>We can use the one-to-one property of exponents to solve exponential equations whose bases are the same. \u00a0The terms in some exponential equations can be rewritten with the same base, allowing us to use the same principle. There are exponential equations that do not have solutions because we define exponential functions as having a positive base. When restrictions are placed on the inputs of a function, it is natural that there will be restrictions on the output as well.\r\n\r\nThe inverse operation of exponentiation is the logarithm, so we can use logarithms to solve exponential equations whose terms do not have the same bases. \u00a0This is similar to using multiplication to \"undo\" division or addition to \"undo\" subtraction. It is important to check exponential equations for extraneous solutions or no solutions.\r\n\r\n<section id=\"fs-id1165137755280\" data-depth=\"1\">We can combine the definitions of logarithms and algebraic tools used for solving linear equations to solve logarithmic equations. \u00a0Graphing each side of a logarithmic equation helps you analyze the solution.<\/section><\/div>\r\n<section data-depth=\"1\"><\/section>\r\n<h2 data-depth=\"1\">Put it Together<\/h2>\r\n[caption id=\"attachment_4055\" align=\"aligncenter\" width=\"599\"]<img class=\" wp-image-4055\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/18192910\/Screen-Shot-2016-08-18-at-12.28.05-PM-300x291.png\" alt=\"Ground Motion recorded on seismic station at Penn State, University Park Campus (Deike Bldg)\" width=\"599\" height=\"581\" \/> Ground Motion recorded on seismic station at Penn State, University Park Campus (Deike Bldg)[\/caption]\r\n\r\nIn the beginning of this module we left Joan with a plan to try and stump her grandfather, the geologist, with the following math problem.\r\n<div class=\"textbox\">\r\n<p style=\"text-align: left;\">The Alaska quake of 1964 had a Richter scale value of 8.5, how many times greater than baseline was the measured wave amplitude recorded by a seismograph?<\/p>\r\n<p style=\"text-align: center;\">[latex]8.5=\\mathrm{log}\\left(\\frac{A}{A_{0}}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<p style=\"text-align: left;\">He was so excited to have found something he and Joan could talk about, he beamed when he saw that she had a problem for him to solve.<\/p>\r\n<p style=\"text-align: left;\">\"I probably knew how to solve that once, but it has been a long time,\" said Grandpa. \"Will you show me?\"<\/p>\r\n<p style=\"text-align: left;\">First, Joan explained to her grandpa that the goal of the problem was to get the argument out of the logarithm. She remembered that a logarithm is an exponent, so her first step was to rewrite the logarithm as an exponential:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}8.5=\\mathrm{log}\\left(\\frac{A}{A_{0}}\\right)\\\\{10}^{8.5}=\\frac{A}{A_{0}}\\\\{10}^{8.5}\\cdot{A_{0}}={A}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Joan entered [latex]{10}^{8.5}[\/latex] into a calculator and got the following number:<\/p>\r\n<p style=\"text-align: center;\">[latex]316,227,766.017[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Wow, the amplitude of the waves that caused the Alaska earthquake of 1964 was 3 million times greater than the baseline reading!<\/p>\r\n<p style=\"text-align: left;\">When working with values that are very large or very small, it is very helpful to work in logarithmic or exponential scales. It helps scientists avoid having to do calculations with massive numbers. \u00a0Additionally, it makes comparisons between different measurements easier to understand.<\/p>\r\n\r\n<div>\r\n<p style=\"text-align: left;\">Joan had such a good time explaining to Grandpa how the math problem worked, she forgot all about having stumped him. Later that night, Hobbes didn't wake her up as he let himself out of the window and down the ladder Anne and Joan made for him. And Joan was happily saving away for the graduation trip that she was planning to take with Hazel (and maybe even her next dreamy boyfriend). Algebra kept Joan from driving while intoxicated, taught her how food poisoning worked, and even helped her choose a good cell phone plan. But this isn't what Joan will remember now that she's finished her math class. What she'll remember is that she now has the power and knowledge to figure out solutions to so many of the challenges in her real life. And we hope you remember that, too! (Don't worry; if you forget some of the math rules along the way, you can always check back and refresh your memory. We'll be waiting.)<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137405247\" class=\"exercise\" data-type=\"exercise\"><\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Exponential Equations with Unlike Bases\n<ul>\n<li>Identify an exponential equation whose terms all have the same base<\/li>\n<li>Idenitfy cases where equations can be rewritten so all terms have the same base<\/li>\n<li>Apply the one-to-one property of exponents to solve an exponential equation<\/li>\n<\/ul>\n<\/li>\n<li>Exponential Equations with Unlike Bases\n<ul>\n<li>Use logarithms to solve exponential equations whose terms\u00a0cannot be rewritten with the same base<\/li>\n<li>Solve exponential equations of the form \u00a0[latex]y=A{e}^{kt}[\/latex] for t<\/li>\n<li>Recognize when there may be extraneous solutions, or no solutions for exponential equations<\/li>\n<\/ul>\n<\/li>\n<li>Logarithmic Equations\n<ul>\n<li>Use the definition of a logarithm to solve logarithmic equations<\/li>\n<li>Use a graph to verify or analyze the solution to a logarithmic equation<\/li>\n<\/ul>\n<\/li>\n<li>Applied Exponential and Logarithmic Equations\n<ul>\n<li>Solve half-life problems<\/li>\n<li>Solve pH problems<\/li>\n<li>Solve problems involving Richter scale readings<\/li>\n<li>Solve problems involving decibels<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<div style=\"width: 460px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201933\/CNX_Precalc_Figure_04_06_0012.jpg\" alt=\"Seven rabbits in front of a brick building.\" width=\"450\" height=\"298\" \/><\/p>\n<p class=\"wp-caption-text\">Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the &#8220;rabbit plague.&#8221; (credit: Richard Taylor, Flickr)<\/p>\n<\/div>\n<p>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 functions.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165134354674\">When an <strong>exponential equation<\/strong> has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we can\u00a0set the exponents equal to one another, and solve for the unknown.<\/p>\n<p id=\"fs-id1165135192889\">For example, consider the equation [latex]{3}^{4x - 7}=\\frac{{3}^{2x}}{3}[\/latex]. To solve for <i>x<\/i>, we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for <em>x<\/em>:<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{3}^{4x - 7}\\hfill & =\\frac{{3}^{2x}}{3}\\hfill & \\hfill \\\\ {3}^{4x - 7}\\hfill & =\\frac{{3}^{2x}}{{3}^{1}}\\hfill & {\\text{Rewrite 3 as 3}}^{1}.\\hfill \\\\ {3}^{4x - 7}\\hfill & ={3}^{2x - 1}\\hfill & \\text{Use the division property of exponents}\\text{.}\\hfill \\\\ 4x - 7\\hfill & =2x - 1\\text{ }\\hfill & \\text{Apply the one-to-one property of exponents}\\text{.}\\hfill \\\\ 2x\\hfill & =6\\hfill & \\text{Subtract 2}x\\text{ and add 7 to both sides}\\text{.}\\hfill \\\\ x\\hfill & =3\\hfill & \\text{Divide by 3}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<p>In our first example, we solve an exponential equation whose terms all have a common base.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]{2}^{x - 1}={2}^{2x - 4}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q579160\">Show Answer<\/span><\/p>\n<div id=\"q579160\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c} {2}^{x - 1}={2}^{2x - 4}\\hfill & \\text{The common base is }2.\\hfill \\\\ \\text{ }x - 1=2x - 4\\hfill & \\text{By the one-to-one property the exponents must be equal}.\\hfill \\\\ \\text{ }x=3\\hfill & \\text{Solve for }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In general, we can summarize solving exponential equations whose terms all have the same base in this way:<br \/>\nFor any algebraic expressions <em>S<\/em>\u00a0and <em>T<\/em>, and any positive real number [latex]b\\ne 1[\/latex]<\/p>\n<div id=\"fs-id1165137702126\" class=\"equation\" data-type=\"equation\">[latex]{b}^{S}={b}^{T}\\text{ if and only if }S=T[\/latex]<\/div>\n<div class=\"equation\" data-type=\"equation\">\n<ul>\n<li>Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[\/latex].<\/li>\n<li>Use the one-to-one property to set the exponents equal.<\/li>\n<li>Solve the resulting equation, <em>S\u00a0<\/em>= <em>T<\/em>, for the unknown.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1165137730366\" class=\"solution\" data-type=\"solution\">\n<section data-depth=\"1\">\n<section id=\"fs-id1165137667260\" data-depth=\"2\">\n<h2 data-type=\"title\">Rewriting Equations So All Powers Have the Same Base<\/h2>\n<p id=\"fs-id1165137725147\">Sometimes\u00a0we can rewrite the terms in an equation as powers with a common base, and solve using the one-to-one property. This takes a keen eye for recognizing common powers. \u00a0For example, you can rewrite 8 as [latex]2^3[\/latex] or 36 as [latex]6^2[\/latex] or [latex]\\frac{1}{4}[\/latex] as [latex]\\left(\\frac{1}{2}\\right)^{2}[\/latex]<\/p>\n<p id=\"fs-id1165137784867\">Consider the equation [latex]256={4}^{x - 5}[\/latex]. We can rewrite both sides of this equation as a power of 2. Then we apply the rules of exponents, along with the one-to-one property, to solve for <em>x<\/em>:<\/p>\n<div id=\"eip-687\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}256={4}^{x - 5}\\hfill & \\hfill \\\\ {2}^{8}={\\left({2}^{2}\\right)}^{x - 5}\\hfill & \\text{Rewrite each side as a power with base 2}.\\hfill \\\\ {2}^{8}={2}^{2x - 10}\\hfill & \\text{Use the one-to-one property of exponents}.\\hfill \\\\ 8=2x - 10\\hfill & \\text{Apply the one-to-one property of exponents}.\\hfill \\\\ 18=2x\\hfill & \\text{Add 10 to both sides}.\\hfill \\\\ x=9\\hfill & \\text{Divide by 2}.\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">In the next example, we show how to find a common base for two expressions whose bases are 8, and 16. \u00a0We can then solve the resulting equation using the one-to-one property of exponents.<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]{8}^{x+2}={16}^{x+1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q731579\">Show Answer<\/span><\/p>\n<div id=\"q731579\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c}\\text{ }{8}^{x+2}={16}^{x+1}\\hfill & \\hfill \\\\ {\\left({2}^{3}\\right)}^{x+2}={\\left({2}^{4}\\right)}^{x+1}\\hfill & \\text{Write }8\\text{ and }16\\text{ as powers of }2.\\hfill \\\\ \\text{ }{2}^{3x+6}={2}^{4x+4}\\hfill & \\text{To take a power of a power, multiply exponents}.\\hfill \\\\ \\text{ }3x+6=4x+4\\hfill & \\text{Use the one-to-one property to set the exponents equal}.\\hfill \\\\ \\text{ }x=2\\hfill & \\text{Solve for }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next example, we are given an exponential equation that contains a square root. \u00a0Remember that you can write roots as rational exponents, so you may be able to find like bases when it is not completely obvious at first.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]{2}^{5x}=\\sqrt{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q507738\">Show Answer<\/span><\/p>\n<div id=\"q507738\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c}{2}^{5x}={2}^{\\frac{1}{2}}\\hfill & \\text{Write the square root of 2 as a power of }2.\\hfill \\\\ 5x=\\frac{1}{2}\\hfill & \\text{Use the one-to-one property}.\\hfill \\\\ x=\\frac{1}{10}\\hfill & \\text{Solve for }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>By changing [latex]\\sqrt{2}[\/latex] to [latex]{2}^{\\frac{1}{2}}[\/latex] we were able to solve the equation in the previous example. In general, here are some steps to consider when you are solving exponential equations. \u00a0A good first step is always to determine whether you can rewrite the terms with a common base.<\/p>\n<ol id=\"fs-id1165137663646\" data-number-style=\"arabic\">\n<li>Rewrite each side in the equation as a power with a common base.<\/li>\n<li>Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[\/latex].<\/li>\n<li>Use the one-to-one property to set the exponents equal.<\/li>\n<li>Solve the resulting equation, <em>S\u00a0<\/em>= <em>T<\/em>, for the unknown.<\/li>\n<\/ol>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process? Write your thoughts in the textbox below before you check our proposed answer.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q711116\">Show Answer<\/span><\/p>\n<div id=\"q711116\" class=\"hidden-answer\" style=\"display: none\">No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined<em data-effect=\"italics\">.<\/em><\/div>\n<\/div>\n<\/div>\n<p>In the next example we show you a case where there is no solution to an exponential equation. \u00a0Remember how exponential functions are defined and ask yourself &#8211; &#8220;does this make sense&#8221; before diving into solving exponential equations.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]{3}^{x+1}=-2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q152201\">Show Answer<\/span><\/p>\n<div id=\"q152201\" class=\"hidden-answer\" style=\"display: none\">This equation has no solution. There is no real value of <em>x<\/em>\u00a0that will make the equation a true statement because any power of a positive number is positive.<\/p>\n<p>For example [latex]3^2=9[\/latex], and [latex]2^4=16[\/latex]. \u00a0Remember that we have defined exponential functions as having a base that is greater than 0.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h3>\u00a0Analysis of the Solution<\/h3>\n<\/div>\n<\/section>\n<\/section>\n<\/div>\n<div id=\"Example_04_06_04\" class=\"example\" data-type=\"example\">\n<div id=\"fs-id1165137405247\" class=\"exercise\" data-type=\"exercise\">\n<div id=\"fs-id1165137849213\" class=\"commentary\" data-type=\"commentary\">\n<p id=\"fs-id1165137578263\">The figure below\u00a0shows the graphs of the two separate expressions in the equation [latex]{3}^{x+1}=-2[\/latex] as [latex]y={3}^{x+1}[\/latex] and [latex]y=-2[\/latex]. The two graphs do not cross showing us that\u00a0the left side is never equal to the right side. Thus the equation has no solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201934\/CNX_Precalc_Figure_04_06_0022.jpg\" alt=\"Graph of 3^(x+1)=-2 and y=-2. The graph notes that they do not cross.\" width=\"487\" height=\"438\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<h2>Exponential Equations with unlike Bases<\/h2>\n<p>Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since [latex]\\mathrm{log}\\left(a\\right)=\\mathrm{log}\\left(b\\right)[\/latex] is equivalent to <em>a\u00a0<\/em>= <em>b<\/em>, we may apply logarithms with the same base on both sides of an exponential equation.<\/p>\n<p>In our first example we will use the law of logs combined with factoring to solve an exponential equation whose terms do not have the same base. Note how first, we rewrite the exponential terms as logarithms.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]{5}^{x+2}={4}^{x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q281663\">Show Answer<\/span><\/p>\n<div id=\"q281663\" class=\"hidden-answer\" style=\"display: none\">\n<p>There is no easy way to get the powers to have the same base for this equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\text{ }{5}^{x+2}={4}^{x}. \\\\ \\text{ }\\mathrm{ln}{5}^{x+2}=\\mathrm{ln}{4}^{x}\\hfill & \\text{Take ln of both sides}.\\hfill \\\\ \\text{ }\\left(x+2\\right)\\mathrm{ln}5=x\\mathrm{ln}4\\hfill & \\text{Use laws of logs}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5+2\\mathrm{ln}5=x\\mathrm{ln}4\\hfill & \\text{Use the distributive law}.\\hfill \\\\ \\text{ }x\\mathrm{ln}5-x\\mathrm{ln}4=-2\\mathrm{ln}5\\hfill & \\text{Get terms containing }x\\text{ on one side, terms without }x\\text{ on the other}.\\hfill \\\\ x\\left(\\mathrm{ln}5-\\mathrm{ln}4\\right)=-2\\mathrm{ln}5\\hfill & \\text{On the left hand side, factor out an }x.\\hfill \\\\ \\text{ }x\\mathrm{ln}\\left(\\frac{5}{4}\\right)=\\mathrm{ln}\\left(\\frac{1}{25}\\right)\\hfill & \\text{Use the laws of logs}.\\hfill \\\\ \\text{ }x=\\frac{\\mathrm{ln}\\left(\\frac{1}{25}\\right)}{\\mathrm{ln}\\left(\\frac{5}{4}\\right)}\\hfill & \\text{Divide by the coefficient of }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In general we can solve exponential equations whose terms do not have like bases in the following way:<\/p>\n<ol id=\"fs-id1165137784632\" data-number-style=\"arabic\">\n<li>Apply the logarithm of both sides of the equation.\n<ul id=\"fs-id1165137824134\" data-bullet-style=\"bullet\">\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 class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Is there any way to solve [latex]{2}^{x}={3}^{x}[\/latex]?<\/p>\n<p>Use the textbox below to formulate an answer or example before you look at the solution.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q608971\">Show Answer<\/span><\/p>\n<div id=\"q608971\" class=\"hidden-answer\" style=\"display: none\">Yes. The solution is x = 0.<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Equations Containing [latex]e[\/latex]<\/h2>\n<section id=\"fs-id1165137469838\" data-depth=\"2\">Base <em>e\u00a0<\/em>is a very common base found in science, finance and engineering applications.\u00a0When we have an equation with a base <em>e<\/em>\u00a0on either side, we can use the <strong>natural logarithm<\/strong> to solve it. Earlier, we introduced a formula that models continuous growth,\u00a0[latex]y=A{e}^{kt}[\/latex]. \u00a0This formula is found in business, finance, and many biological and physical science applications. In our next example we will show how to solve this equation for t, the elapsed time for the behavior in question.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>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=\"q714083\">Show Answer<\/span><\/p>\n<div id=\"q714083\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c}100\\hfill & =20{e}^{2t}\\hfill & \\hfill \\\\ 5\\hfill & ={e}^{2t}\\hfill & \\text{Divide by the coefficient of the power}\\text{.}\\hfill \\\\ \\mathrm{ln}5\\hfill & =2t\\hfill & \\text{Take ln of both sides}\\text{. Use the fact that }\\mathrm{ln}\\left(x\\right)\\text{ and }{e}^{x}\\text{ are inverse functions}\\text{.}\\hfill \\\\ t\\hfill & =\\frac{\\mathrm{ln}5}{2}\\hfill & \\text{Divide by the coefficient of }t\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_06_06\" class=\"example\" data-type=\"example\">\n<div id=\"fs-id1165137846466\" class=\"exercise\" data-type=\"exercise\">\n<div id=\"fs-id1165137705083\" class=\"commentary\" data-type=\"commentary\">\n<h3 data-type=\"title\">Analysis of the Solution<\/h3>\n<p>Using laws of logs, we can also write this answer in the form [latex]t=\\mathrm{ln}\\sqrt{5}[\/latex]. If we want a decimal approximation of the answer, we use a calculator.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Does every equation of the form [latex]y=A{e}^{kt}[\/latex] have a solution? Write your thoughts or an example in the textbox below before you check the answer.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q483016\">Show Answer<\/span><\/p>\n<div id=\"q483016\" class=\"hidden-answer\" style=\"display: none\">No. There is a solution when [latex]k\\ne 0[\/latex], and when y and A 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].<\/div>\n<\/div>\n<\/div>\n<p>We will provide one more example using the continuous growth formula, but this time we have to do a couple steps of algebra to get in it a form that can be solved.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]4{e}^{2x}+5=12[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q593052\">Show Answer<\/span><\/p>\n<div id=\"q593052\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c}4{e}^{2x}+5=12\\hfill & \\hfill \\\\ 4{e}^{2x}=7\\hfill & \\text{Combine like terms}.\\hfill \\\\ {e}^{2x}=\\frac{7}{4}\\hfill & \\text{Divide by the coefficient of the power}.\\hfill \\\\ 2x=\\mathrm{ln}\\left(\\frac{7}{4}\\right)\\hfill & \\text{Take ln of both sides}.\\hfill \\\\ x=\\frac{1}{2}\\mathrm{ln}\\left(\\frac{7}{4}\\right)\\hfill & \\text{Solve for }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\"><\/div>\n<div class=\"bcc-box bcc-success\"><\/div>\n<\/section>\n<section id=\"fs-id1165137665482\" data-depth=\"2\">\n<h2 data-type=\"title\">Extraneous Solutions<\/h2>\n<p id=\"fs-id1165137742403\">Sometimes the methods used to solve an equation introduce an <strong>extraneous solution<\/strong>, 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>In the next example we will solve an exponential equation that is quadratic in form. We will factor first and then use the zero product principle. Note how we find two solutions, but reject one that does not satisfy the original equaiton.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]{e}^{2x}-{e}^{x}=56[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q152760\">Show Answer<\/span><\/p>\n<div id=\"q152760\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c}{e}^{2x}-{e}^{x}\\hfill & =56\\hfill & \\hfill \\\\ {e}^{2x}-{e}^{x}-56\\hfill & =0\\hfill & \\text{Get one side of the equation equal to zero}.\\hfill \\\\ \\left({e}^{x}+7\\right)\\left({e}^{x}-8\\right)\\hfill & =0\\hfill & \\text{Factor by the FOIL method}.\\hfill \\\\ {e}^{x}+7\\hfill & =0\\text{ or }{e}^{x}-8=0 & \\text{If a product is zero, then one factor must be zero}.\\hfill \\\\ {e}^{x}\\hfill & =-7{\\text{ or e}}^{x}=8\\hfill & \\text{Isolate the exponentials}.\\hfill \\\\ {e}^{x}\\hfill & =8\\hfill & \\text{Reject the equation in which the power equals a negative number}.\\hfill \\\\ x\\hfill & =\\mathrm{ln}8\\hfill & \\text{Solve the equation in which the power equals a positive number}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>\u00a0Analysis of the Solution<\/h3>\n<div id=\"Example_04_06_08\" class=\"example\" data-type=\"example\">\n<div id=\"fs-id1165137828517\" class=\"exercise\" data-type=\"exercise\">\n<div id=\"fs-id1165137806430\" class=\"commentary\" data-type=\"commentary\">\n<p id=\"fs-id1165137806436\">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]x=\\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.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Does every logarithmic equation have a solution? Write your ideas, or a counter example in the box below.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q810736\">Show Answer<\/span><\/p>\n<div id=\"q810736\" class=\"hidden-answer\" style=\"display: none\">No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.<\/div>\n<\/div>\n<\/div>\n<h2>Logarithmic Equations<\/h2>\n<p>We have already seen that every <strong>logarithmic equation<\/strong> [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is equivalent to the exponential equation [latex]{b}^{y}=x[\/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<p id=\"fs-id1165134148350\">For example, consider the equation [latex]{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3[\/latex]. 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 <em>x<\/em>:<\/p>\n<div id=\"eip-id2205910\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}{\\mathrm{log}}_{2}\\left(2\\right)+{\\mathrm{log}}_{2}\\left(3x - 5\\right)=3\\hfill & \\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(2\\left(3x - 5\\right)\\right)=3\\hfill & \\text{Apply the product rule of logarithms}.\\hfill \\\\ \\text{ }{\\mathrm{log}}_{2}\\left(6x - 10\\right)=3\\hfill & \\text{Distribute}.\\hfill \\\\ \\text{ }{2}^{3}=6x - 10\\hfill & \\text{Apply the definition of a logarithm}.\\hfill \\\\ \\text{ }8=6x - 10\\hfill & \\text{Calculate }{2}^{3}.\\hfill \\\\ \\text{ }18=6x\\hfill & \\text{Add 10 to both sides}.\\hfill \\\\ \\text{ }x=3\\hfill & \\text{Divide by 6}.\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\"><\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">In our first example we will show how to use techniques from solving linear equations to solve a logarithmic equation.<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]2\\mathrm{ln}x+3=7[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q1854\">Show Answer<\/span><\/p>\n<div id=\"q1854\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c}2\\mathrm{ln}x+3=7\\hfill & \\hfill \\\\ \\text{ }2\\mathrm{ln}x=4\\hfill & \\text{Subtract 3}.\\hfill \\\\ \\text{ }\\mathrm{ln}x=2\\hfill & \\text{Divide by 2}.\\hfill \\\\ \\text{ }x={e}^{2}\\hfill & \\text{Rewrite in exponential form}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Analysis of the solution<\/h3>\n<p>The solution for the previous equation was [latex]x={e}^{2}[\/latex], this is often referred to as the exact solution. \u00a0Sometimes, you may be asked to give an approximation, which would be more useful to someone using the result for a financial, scientific or engineering application. The approximation for\u00a0[latex]x={e}^{2}[\/latex] can be found with a calculator.\u00a0[latex]x={e}^{2}\\approx{7.4}[\/latex]<\/p>\n<p>In general, we can describe using the definition of a logarithm to solve logarithmic equations as follows:<\/p>\n<p class=\"title\" data-type=\"title\">For any algebraic expression <em>S<\/em> and real numbers <em>b<\/em> and <em>c<\/em>, where [latex]b>0,\\text{ }b\\ne 1[\/latex],<\/p>\n<div id=\"fs-id1165137732219\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{ if and only if }{b}^{c}=S[\/latex]<\/div>\n<div class=\"equation\" data-type=\"equation\">Here is another example of what you may encounter when solving logarithmic equations.<\/div>\n<\/div>\n<div class=\"equation\" style=\"text-align: left;\" data-type=\"equation\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]2\\mathrm{ln}\\left(6x\\right)=7[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q663498\">Show Answer<\/span><\/p>\n<div id=\"q663498\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c}2\\mathrm{ln}\\left(6x\\right)=7\\hfill & \\hfill \\\\ \\text{ }\\mathrm{ln}\\left(6x\\right)=\\frac{7}{2}\\hfill & \\text{Divide by 2}.\\hfill \\\\ \\text{ }6x={e}^{\\left(\\frac{7}{2}\\right)}\\hfill & \\text{Use the definition of }\\mathrm{ln}.\\hfill \\\\ \\text{ }x=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}\\hfill & \\text{Divide by 6}.\\hfill \\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>Exact answer: [latex]x=\\frac{1}{6}{e}^{\\left(\\frac{7}{2}\\right)}[\/latex]<\/p>\n<p>Approximate answer: [latex]x\\approx{5.5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next example we will show you the power of using graphs to analyze solutions to logarithmic equations. By turning each side of the equation into a function and plotting them on the same set of axes, we can see how they interact with each other. \u00a0In this case, we get a logarithmic function and a horizontal line, and find that the solution is the point where the two intersect.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\mathrm{ln}x=3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q691419\">Show Answer<\/span><\/p>\n<div id=\"q691419\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\mathrm{ln}x=3\\hfill & \\hfill \\\\ x={e}^{3}\\hfill & \\text{Use the definition of the natural logarithm}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<p id=\"fs-id1165137443165\">The graph below\u00a0represents the graph of the equation. On the graph, the <em data-effect=\"italics\">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<figure class=\"small\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201935\/CNX_Precalc_Figure_04_06_0032.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\" data-media-type=\"image\/jpg\" \/><\/figure>\n<figure id=\"CNX_Precalc_Figure_04_06_003\" class=\"small\"><figcaption>\u00a0The graphs of [latex]y=\\mathrm{ln}x[\/latex] and <em>y\u00a0<\/em>= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex], which is approximately (20.0855, 3).<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/div>\n<p>In the same way that a story can help you understand a complex concept, graphs can help us understand a wide variety of concepts in mathematics. Visual representations of the parts of an equation can be created by turning each side into a function and plotting the functions on the same set of axes.<\/p>\n<\/div>\n<h2>Use the one-to-one property of logarithms to solve logarithmic equations<\/h2>\n<section id=\"fs-id1165137755280\" data-depth=\"1\">\n<p id=\"fs-id1165135237092\">As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers <em>x\u00a0<\/em>&gt; 0, <em>S\u00a0<\/em>&gt; 0, <em>T\u00a0<\/em>&gt; 0 and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],<\/p>\n<div id=\"eip-674\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex].<\/div>\n<p id=\"eip-625\">For example,<\/p>\n<div id=\"eip-453\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{If }{\\mathrm{log}}_{2}\\left(x - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right),\\text{then }x - 1=8[\/latex].<\/div>\n<p id=\"fs-id1165137874962\">So, if [latex]x - 1=8[\/latex], then we can solve for <em>x<\/em>, and we get <em>x\u00a0<\/em>= 9. To check, we can substitute <em>x\u00a0<\/em>= 9 into the original equation: [latex]{\\mathrm{log}}_{2}\\left(9 - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex]. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)[\/latex]<\/p>\n<p>Check your results.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q874129\">Show Answer<\/span><\/p>\n<div id=\"q874129\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c}\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)\\hfill & \\hfill \\\\ \\text{ }\\mathrm{log}\\left(\\frac{3x - 2}{2}\\right)=\\mathrm{log}\\left(x+4\\right)\\hfill & \\text{Apply the quotient rule of logarithms}.\\hfill \\\\ \\text{ }\\frac{3x - 2}{2}=x+4\\hfill & \\text{Apply the one to one property of a logarithm}.\\hfill \\\\ \\text{ }3x - 2=2x+8\\hfill & \\text{Multiply both sides of the equation by }2.\\hfill \\\\ \\text{ }x=10\\hfill & \\text{Subtract 2}x\\text{ and add 2}.\\hfill \\end{array}[\/latex]<\/p>\n<p id=\"fs-id1165135172191\">To check the result, substitute <em>x\u00a0<\/em>= 10 into [latex]\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)[\/latex].<\/p>\n<div id=\"eip-316\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}\\mathrm{log}\\left(3\\left(10\\right)-2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(\\left(10\\right)+4\\right)\\hfill & \\hfill \\\\ \\text{ }\\mathrm{log}\\left(28\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(14\\right)\\hfill & \\hfill \\\\ \\text{ }\\mathrm{log}\\left(\\frac{28}{2}\\right)=\\mathrm{log}\\left(14\\right)\\hfill & \\text{The solution checks}.\\hfill \\end{array}[\/latex]<\/div>\n<h4 class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\">Answer<\/h4>\n<p><em>x\u00a0<\/em>= 10<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Note, when solving an equation involving logarithms, always check to see if the answer is correct or if there is an extraneous solution.<\/p>\n<p>In general, we can summarize solving logarithmic equations as follows:<\/p>\n<p>For\u00a0[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/p>\n<ol data-number-style=\"arabic\">\n<li>Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form [latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T[\/latex].<\/li>\n<li>Use the one-to-one property to set the arguments equal.<\/li>\n<li>Solve the resulting equation, <em>S<\/em> =\u00a0<em>T<\/em>, for the unknown.<\/li>\n<\/ol>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q306816\">Show Answer<\/span><\/p>\n<div id=\"q306816\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c}\\text{ }\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}\\left(2x+3\\right)\\hfill & \\hfill \\\\ \\text{ }{x}^{2}=2x+3\\hfill & \\text{Use the one-to-one property of the logarithm}.\\hfill \\\\ \\text{ }{x}^{2}-2x - 3=0\\hfill & \\text{Get zero on one side before factoring}.\\hfill \\\\ \\left(x - 3\\right)\\left(x+1\\right)=0\\hfill & \\text{Factor using FOIL}.\\hfill \\\\ \\text{ }x - 3=0\\text{ or }x+1=0\\hfill & \\text{If a product is zero, one of the factors must be zero}.\\hfill \\\\ \\text{ }x=3\\text{ or }x=-1\\hfill & \\text{Solve for }x.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3 data-type=\"title\">Analysis of the Solution<\/h3>\n<p id=\"fs-id1165134505611\">There are two solutions: <em>x\u00a0<\/em>= 3 or <em>x\u00a0<\/em>= \u20131. The solution <em>x\u00a0<\/em>= \u20131 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.<\/p>\n<p>Applications of Logarithmic Equations<\/p>\n<section id=\"fs-id1165137828382\" data-depth=\"1\">\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<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3765 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/11191452\/HEUraniumC-300x240.jpg\" alt=\"Gloved hands holding a dish of highly enrich uranium metal.\" width=\"393\" height=\"314\" \/><\/p>\n<p id=\"fs-id1165134192326\">One such application is called <strong>half-life, <\/strong>which refers to the amount of time it takes for half a given quantity of radioactive material to decay. The table below\u00a0lists the half-life for several of the more common radioactive substances.<\/p>\n<table style=\"width: 50%;\" summary=\"Seven rows and three columns. The first column is labeled,\">\n<thead>\n<tr>\n<th style=\"text-align: center;\" data-align=\"center\">Substance<\/th>\n<th style=\"text-align: center;\" data-align=\"center\">Use<\/th>\n<th style=\"text-align: center;\" data-align=\"center\">Half-life<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>gallium-67<\/td>\n<td>nuclear medicine<\/td>\n<td>80 hours<\/td>\n<\/tr>\n<tr>\n<td>cobalt-60<\/td>\n<td>manufacturing<\/td>\n<td>5.3 years<\/td>\n<\/tr>\n<tr>\n<td>technetium-99m<\/td>\n<td>nuclear medicine<\/td>\n<td>6 hours<\/td>\n<\/tr>\n<tr>\n<td>americium-241<\/td>\n<td>construction<\/td>\n<td>432 years<\/td>\n<\/tr>\n<tr>\n<td>carbon-14<\/td>\n<td>archeological dating<\/td>\n<td>5,715 years<\/td>\n<\/tr>\n<tr>\n<td>uranium-235<\/td>\n<td>atomic power<\/td>\n<td>703,800,000 years<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165135209669\">We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:<\/p>\n<div id=\"eip-247\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}A\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{T}t}\\hfill \\\\ A\\left(t\\right)={A}_{0}{e}^{\\mathrm{ln}\\left(0.5\\right)\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left({e}^{\\mathrm{ln}\\left(0.5\\right)}\\right)}^{\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left(\\frac{1}{2}\\right)}^{\\frac{t}{T}}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137408740\">where<\/p>\n<ul id=\"fs-id1165137408743\">\n<li>[latex]{A}_{0}[\/latex] is the amount initially present<\/li>\n<li><em>T<\/em>\u00a0is the half-life of the substance<\/li>\n<li><em>t<\/em>\u00a0is the time period over which the substance is studied<\/li>\n<li>[latex]A(t)[\/latex]\u00a0is the amount of the substance present after time\u00a0<em>t<\/em><\/li>\n<\/ul>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q774904\">Show Answer<\/span><\/p>\n<div id=\"q774904\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{c}\\text{ }y=\\text{1000}e\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill & \\hfill \\\\ \\text{ }900=1000{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill & \\text{After 10% decays, 900 grams are left}.\\hfill \\\\ \\text{ }0.9={e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill & \\text{Divide by 1000}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\mathrm{ln}\\left({e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\right)\\hfill & \\text{Take ln of both sides}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill & \\text{ln}\\left({e}^{M}\\right)=M\\hfill \\\\ \\text{ }\\text{ }t=\\text{703,800,000}\\times \\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{years}\\begin{array}{c}{cccc}& & & \\end{array}\\hfill & \\text{Solve for }t.\\hfill \\\\ \\text{ }\\text{ }t\\approx \\text{106,979,777 years}\\hfill & \\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_04_06_13\" class=\"example\" data-type=\"example\">\n<div id=\"fs-id1165137628651\" class=\"exercise\" data-type=\"exercise\">\n<div id=\"fs-id1165137453455\" class=\"commentary\" data-type=\"commentary\">\n<h3 data-type=\"title\">Analysis of the Solution<\/h3>\n<p id=\"fs-id1165137453460\">Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>pH<\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3767 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/11193831\/Lemon-300x212.jpg\" alt=\"Lemons\" width=\"259\" height=\"183\" \/><\/p>\n<p>In chemistry, pH is 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 acidic, and substances with a pH greater than 7 are said to be alkaline. In our next example we will find how\u00a0doubling the concentration of hydrogen ions in a liquid\u00a0affects pH.<\/p>\n<\/section>\n<section id=\"fs-id1165137828382\" data-depth=\"1\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In chemistry, [latex]\\text{pH}=-\\mathrm{log}\\left[{H}^{+}\\right][\/latex]. 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=\"q994106\">Show Answer<\/span><\/p>\n<div id=\"q994106\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135415820\">Suppose <em>C<\/em>\u00a0is the original concentration of hydrogen ions, and <em>P<\/em>\u00a0is the original pH of the liquid. Then [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex]. If the concentration is doubled, the new concentration is 2<em>C<\/em>. Then the pH of the new liquid is<\/p>\n<div id=\"eip-id1165134547456\" class=\"equation unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)[\/latex]<\/div>\n<p id=\"fs-id1165135571875\">Using the product rule of logs<\/p>\n<div id=\"eip-id1165134547498\" class=\"equation unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)=-\\left(\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(C\\right)\\right)=-\\mathrm{log}\\left(2\\right)-\\mathrm{log}\\left(C\\right)[\/latex]<\/div>\n<p id=\"fs-id1165135443976\">Since [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex], the new pH is<\/p>\n<div id=\"eip-id1165135440065\" class=\"equation unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\text{pH}=P-\\mathrm{log}\\left(2\\right)\\approx P - 0.301[\/latex]<\/div>\n<p id=\"fs-id1165135251361\">When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Richter Scale<\/h2>\n<div id=\"attachment_3770\" style=\"width: 422px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3770\" class=\"wp-image-3770\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/11194210\/Earthquake_Richter_Scale-300x223.jpg\" alt=\"Richter Scale of Earthquake Energy\" width=\"412\" height=\"306\" \/><\/p>\n<p id=\"caption-attachment-3770\" class=\"wp-caption-text\">Richter Scale of Earthquake Energy<\/p>\n<\/div>\n<p>The Richter scale is a logarithmic function that is used to measure the magnitude of earthquakes. The magnitude of an earthquake is related to how much energy is released by the quake. Instruments called seismographs detect movement in the earth; the smallest movement that can be detected shows on a seismograph as a wave with amplitude [latex]A_{0}[\/latex].<\/p>\n<p>A \u2013 the measure of the amplitude of the earthquake wave<br \/>\n[latex]A_{0}[\/latex]\u00a0\u2013 the amplitude of the smallest detectable wave (or standard wave)<\/p>\n<p>From this you can find R, the Richter scale measure of the magnitude of the earthquake using the formula:<\/p>\n<p style=\"text-align: center;\">[latex]R=\\mathrm{log}\\left(\\frac{A}{A_{0}}\\right)[\/latex]<\/p>\n<p>The intensity of an earthquake will typically measure between 2 and 10 on the Richter scale. Any earthquakes registering below a 5 are fairly minor; they may shake the ground a bit, but are seldom strong enough to cause much damage. Earthquakes with a Richter rating of between 5 and 7.9 are much more severe, and any quake above an 8 is likely to cause massive damage. (The highest rating ever recorded for an earthquake is 9.5 during the 1960 Valdivia earthquake in Chile.)<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>An earthquake is measured with a wave amplitude 392 times as great as [latex]A_{0}[\/latex]. What is the magnitude of this earthquake using the Richter scale, to the nearest tenth?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q677160\">Show Answer<\/span><\/p>\n<div id=\"q677160\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the Richter scale equation.<\/p>\n<p>[latex]R=\\mathrm{log}\\left(\\frac{A}{A_{0}}\\right)[\/latex]<\/p>\n<p>Since A is 392 times as large as[latex]A_{0}[\/latex], [latex]A=392A_{0}[\/latex]. Substitute this expression in for A.<\/p>\n<p>[latex]R=\\mathrm{log}\\left(\\frac{392A_{0}}{A_{0}}\\right)[\/latex]<\/p>\n<p>Simplify the argument of the logarithm, then use a calculator to evaluate.<\/p>\n<p>[latex]\\begin{array}{c}R=\\mathrm{log}\\left(\\frac{392A_{0}}{A_{0}}\\right)\\\\=\\mathrm{log}\\left(392\\right)\\\\=2.5932\\\\\\approx{2.6}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The earthquake registered 2.6 on the Richter scale.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>A difference of 1 point on the Richter scale equates to a 10-fold difference in the amplitude of the earthquake (which is related to the wave strength). This means that an earthquake that measures 3.6 on the Richter scale has 10 times the amplitude of one that measures 2.6.<\/p>\n<p>In the Richter scale example, the wave amplitude of the earthquake was 392 times normal. What if it were 10 times that, or 3,920 times normal? To find the measurement of that size earthquake on the Richter scale, you find log 3920. A calculator gives a value of 3.5932&#8230;or 3.6, when rounded to the nearest tenth. One extra point on the Richter scale can mean a lot more shaking!<\/p>\n<h2>Decibels<\/h2>\n<div id=\"attachment_3772\" style=\"width: 310px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3772\" class=\"size-medium wp-image-3772\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/11195710\/Rui%CC%81do_Noise_041113GFDL-300x225.jpeg\" alt=\"Child covering his ears with his hands\" width=\"300\" height=\"225\" \/><\/p>\n<p id=\"caption-attachment-3772\" class=\"wp-caption-text\">Turn it Down<\/p>\n<\/div>\n<p>Sound is measured in a logarithmic scale using a unit called a decibel. The formula looks similar to the Richter scale:<\/p>\n<p style=\"text-align: center;\">[latex]d=10\\mathrm{log}\\left(\\frac{P}{P_{0}}\\right)[\/latex]<\/p>\n<p>where P is the power or intensity of the sound and P0 is the weakest sound that the human ear can hear. In the next example we will find how much more intense the noise from a dishwasher is than the noise from a hot water pump.<\/p>\n<\/section>\n<section id=\"fs-id1165137828382\" data-depth=\"1\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>One hot water pump has a noise rating of 50 decibels. One dishwasher, however, has a noise rating of 62 decibels. The dishwasher noise is how many times more intense than the hot water pump noise?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q789189\">Show Answer<\/span><\/p>\n<div id=\"q789189\" class=\"hidden-answer\" style=\"display: none\">\nYou can\u2019t easily compare the two noises using the formula, but you can compare them to [latex]P_{0}[\/latex]. Start by finding the intensity of noise for the hot water pump. Use h for the intensity of the hot water pump\u2019s noise.<\/p>\n<p style=\"text-align: center;\">[latex]50=10\\mathrm{log}\\left(\\frac{h}{P_{0}}\\right)[\/latex]<\/p>\n<p>Divide the equations by 10 to get the log by itself.<\/p>\n<p style=\"text-align: center;\">[latex]5=\\mathrm{log}\\left(\\frac{h}{P_{0}}\\right)[\/latex]<\/p>\n<p>Rewrite the equation as an exponential equation.<\/p>\n<p style=\"text-align: center;\">[latex]10^5=\\frac{h}{P_{0}}[\/latex]<\/p>\n<p style=\"text-align: left;\">Solve for h, the intensity of the water pump.<\/p>\n<p style=\"text-align: center;\">[latex]h=P_{0}10^5[\/latex]<\/p>\n<p>Repeat the same process to find the intensity of the noise for the dishwasher, use d to represent the intensity of the sound of the dishwasher. Remember that [latex]{P_{0}}[\/latex] is a baseline for the most faint sound the human ear can hear.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}62=10\\mathrm{log}\\left(\\frac{d}{P_{0}}\\right)\\\\6.2=\\mathrm{log}\\left(\\frac{d}{P_{0}}\\right)\\\\10^{6.2}=\\frac{d}{P_{0}}\\\\d=P_{0}10^{6.2}\\end{array}[\/latex]<\/p>\n<p>To compare d to h, you can divide. (Think: if the dishwasher\u2019s noise is twice as intense as the pump\u2019s,\u00a0then d should be 2h\u2014that is, [latex]\\frac{d}{h}[\/latex]\u00a0should be 2.)<br \/>\nUse the laws of exponents to simplify the quotient.<\/p>\n<p style=\"text-align: center;\">[latex]\\Large\\frac{d}{h}=\\frac{P_{0}10^{6.2}}{P_{0}10^5}=\\frac{10^{6.2}}{10^{5}}=\\normalsize10^{6.2-5}=10^{1.2}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The dishwasher\u2019s noise is [latex]10^{1.2}[\/latex]\u00a0(or about 15.85) times as intense as the hot water pump.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Applications of logarithms and exponentials are everywhere in science. \u00a0We hope the examples here have given you an idea of how useful they can be.<\/p>\n<\/section>\n<h2>Summary<\/h2>\n<\/section>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p>We can use the one-to-one property of exponents to solve exponential equations whose bases are the same. \u00a0The terms in some exponential equations can be rewritten with the same base, allowing us to use the same principle. There are exponential equations that do not have solutions because we define exponential functions as having a positive base. When restrictions are placed on the inputs of a function, it is natural that there will be restrictions on the output as well.<\/p>\n<p>The inverse operation of exponentiation is the logarithm, so we can use logarithms to solve exponential equations whose terms do not have the same bases. \u00a0This is similar to using multiplication to &#8220;undo&#8221; division or addition to &#8220;undo&#8221; subtraction. It is important to check exponential equations for extraneous solutions or no solutions.<\/p>\n<section id=\"fs-id1165137755280\" data-depth=\"1\">We can combine the definitions of logarithms and algebraic tools used for solving linear equations to solve logarithmic equations. \u00a0Graphing each side of a logarithmic equation helps you analyze the solution.<\/section>\n<\/div>\n<section data-depth=\"1\"><\/section>\n<h2 data-depth=\"1\">Put it Together<\/h2>\n<div id=\"attachment_4055\" style=\"width: 609px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4055\" class=\"wp-image-4055\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/18192910\/Screen-Shot-2016-08-18-at-12.28.05-PM-300x291.png\" alt=\"Ground Motion recorded on seismic station at Penn State, University Park Campus (Deike Bldg)\" width=\"599\" height=\"581\" \/><\/p>\n<p id=\"caption-attachment-4055\" class=\"wp-caption-text\">Ground Motion recorded on seismic station at Penn State, University Park Campus (Deike Bldg)<\/p>\n<\/div>\n<p>In the beginning of this module we left Joan with a plan to try and stump her grandfather, the geologist, with the following math problem.<\/p>\n<div class=\"textbox\">\n<p style=\"text-align: left;\">The Alaska quake of 1964 had a Richter scale value of 8.5, how many times greater than baseline was the measured wave amplitude recorded by a seismograph?<\/p>\n<p style=\"text-align: center;\">[latex]8.5=\\mathrm{log}\\left(\\frac{A}{A_{0}}\\right)[\/latex]<\/p>\n<\/div>\n<p style=\"text-align: left;\">He was so excited to have found something he and Joan could talk about, he beamed when he saw that she had a problem for him to solve.<\/p>\n<p style=\"text-align: left;\">&#8220;I probably knew how to solve that once, but it has been a long time,&#8221; said Grandpa. &#8220;Will you show me?&#8221;<\/p>\n<p style=\"text-align: left;\">First, Joan explained to her grandpa that the goal of the problem was to get the argument out of the logarithm. She remembered that a logarithm is an exponent, so her first step was to rewrite the logarithm as an exponential:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}8.5=\\mathrm{log}\\left(\\frac{A}{A_{0}}\\right)\\\\{10}^{8.5}=\\frac{A}{A_{0}}\\\\{10}^{8.5}\\cdot{A_{0}}={A}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Joan entered [latex]{10}^{8.5}[\/latex] into a calculator and got the following number:<\/p>\n<p style=\"text-align: center;\">[latex]316,227,766.017[\/latex]<\/p>\n<p style=\"text-align: left;\">Wow, the amplitude of the waves that caused the Alaska earthquake of 1964 was 3 million times greater than the baseline reading!<\/p>\n<p style=\"text-align: left;\">When working with values that are very large or very small, it is very helpful to work in logarithmic or exponential scales. It helps scientists avoid having to do calculations with massive numbers. \u00a0Additionally, it makes comparisons between different measurements easier to understand.<\/p>\n<div>\n<p style=\"text-align: left;\">Joan had such a good time explaining to Grandpa how the math problem worked, she forgot all about having stumped him. Later that night, Hobbes didn&#8217;t wake her up as he let himself out of the window and down the ladder Anne and Joan made for him. And Joan was happily saving away for the graduation trip that she was planning to take with Hazel (and maybe even her next dreamy boyfriend). Algebra kept Joan from driving while intoxicated, taught her how food poisoning worked, and even helped her choose a good cell phone plan. But this isn&#8217;t what Joan will remember now that she&#8217;s finished her math class. What she&#8217;ll remember is that she now has the power and knowledge to figure out solutions to so many of the challenges in her real life. And we hope you remember that, too! (Don&#8217;t worry; if you forget some of the math rules along the way, you can always check back and refresh your memory. We&#8217;ll be waiting.)<\/p>\n<\/div>\n<div id=\"fs-id1165137405247\" class=\"exercise\" data-type=\"exercise\"><\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3521\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Screenshot: WRPS Records Virginia Earthquake. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><li>Lemon. <strong>Authored by<\/strong>: Andru00e9 Karwath. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/commons.wikimedia.org\/w\/index.php?curid=59992\">https:\/\/commons.wikimedia.org\/w\/index.php?curid=59992<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>1964 Alaska earthquake. <strong>Provided by<\/strong>: Wikipedia. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/en.wikipedia.org\/wiki\/1964_Alaska_earthquake\">https:\/\/en.wikipedia.org\/wiki\/1964_Alaska_earthquake<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">Public domain content<\/div><ul class=\"citation-list\"><li>Earthquake_Richter_Scale. <strong>Authored by<\/strong>: Webber. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/commons.wikimedia.org\/w\/index.php?curid=1567023\">https:\/\/commons.wikimedia.org\/w\/index.php?curid=1567023<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Lemon\",\"author\":\"Andru00e9 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