{"id":288,"date":"2023-06-21T13:22:52","date_gmt":"2023-06-21T13:22:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/logarithmic-equations\/"},"modified":"2023-07-04T04:24:28","modified_gmt":"2023-07-04T04:24:28","slug":"logarithmic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/logarithmic-equations\/","title":{"raw":"\u25aa   Logarithmic Equations","rendered":"\u25aa   Logarithmic Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve a logarithmic equation algebraically.<\/li>\r\n \t<li>Solve a logarithmic equation graphically.<\/li>\r\n \t<li>Use the one-to-one property of logarithms to solve a logarithmic equation.<\/li>\r\n \t<li>Solve a radioactive decay problem.<\/li>\r\n<\/ul>\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 equal 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\r\nFor 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 as a single log and then apply the definition of logs to solve for [latex]x[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\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{Convert to exponential form}.\\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 both sides by 6}.\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations<\/h3>\r\nFor 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],\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{ if and only if }{b}^{c}=S[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Algebra to Solve a Logarithmic Equation<\/h3>\r\nSolve [latex]2\\mathrm{ln}x+3=7[\/latex].\r\n\r\n[reveal-answer q=\"977891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"977891\"]\r\n\r\n[latex]\\begin{array}{l}2\\mathrm{ln}x+3=7\\hfill &amp; \\hfill \\\\ \\text{}2\\mathrm{ln}x=4\\hfill &amp; \\text{Subtract 3 from both sides}.\\hfill \\\\ \\text{}\\mathrm{ln}x=2\\hfill &amp; \\text{Divide both sides 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<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nAs was the case when using the properties and rules of exponents and logarithms to rewrite expressions containing them, there can be more than one good way to solve a logarithmic equation. It is good practice to follow the examples given for each of the situations in this section, but you should think about alternative ways to creatively and correctly apply the properties and rules.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]6+\\mathrm{ln}x=10[\/latex].\r\n\r\n[reveal-answer q=\"183568\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"183568\"]\r\n\r\n[latex]x={e}^{4}[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]129911[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Algebra Before and After Using the Definition of the Natural Logarithm<\/h3>\r\nSolve [latex]2\\mathrm{ln}\\left(6x\\right)=7[\/latex].\r\n\r\n[reveal-answer q=\"231886\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"231886\"]\r\n\r\n[latex]\\begin{array}{l}2\\mathrm{ln}\\left(6x\\right)=7\\hfill &amp; \\hfill \\\\ \\text{}\\mathrm{ln}\\left(6x\\right)=\\frac{7}{2}\\hfill &amp; \\text{Divide both sides 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 both sides by 6}.\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]2\\mathrm{ln}\\left(x+1\\right)=10[\/latex].\r\n\r\n[reveal-answer q=\"62905\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"62905\"]\r\n\r\n[latex]x={e}^{5}-1[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]14406[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using a Graph to Understand the Solution to a Logarithmic Equation<\/h3>\r\nSolve [latex]\\mathrm{ln}x=3[\/latex].\r\n\r\n[reveal-answer q=\"960461\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"960461\"]\r\n\r\n[latex]\\begin{array}{l}\\mathrm{ln}x=3\\hfill &amp; \\hfill \\\\ x={e}^{3}\\hfill &amp; \\text{Use the definition of }\\mathrm{ln}\\text{.}\\hfill \\end{array}[\/latex]\r\n\r\nBelow is a\u00a0graph of the equation. On the graph the <em>x<\/em>-coordinate of the point where 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].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03172012\/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\" \/> The graphs of [latex]y=\\mathrm{ln}x[\/latex] and y\u00a0= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex] which is approximately (20.0855, 3).[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nUse a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[\/latex] to 2 decimal places.\r\n\r\n[reveal-answer q=\"889911\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"889911\"]\r\n\r\n[latex]x\\approx 9.97[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Using the One-to-One Property of Logarithms to Solve Logarithmic Equations<\/h2>\r\nAs 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],\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/p>\r\nFor example,\r\n<p style=\"text-align: center;\">[latex]\\text{If }{\\mathrm{log}}_{2}\\left(x - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right),\\text{then }x - 1=8[\/latex]<\/p>\r\nSo if [latex]x - 1=8[\/latex], then we can solve for <em>x\u00a0<\/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.\r\n\r\nFor example, consider the equation [latex]\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)[\/latex]. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm and then apply the one-to-one property to solve for <em>x<\/em>:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\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}.\\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]<\/p>\r\nTo 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].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\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]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Using the One-to-One Property of Logarithms to Solve Logarithmic Equations<\/h3>\r\nFor any algebraic expressions <em>S<\/em> and <em>T<\/em> and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/p>\r\nNote, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an equation containing logarithms, solve it using the one-to-one property<\/h3>\r\n<ol>\r\n \t<li>Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation is of 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 to each other.<\/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>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation Using the One-to-One Property of Logarithms<\/h3>\r\nSolve [latex]\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}\\left(2x+3\\right)[\/latex].\r\n\r\n[reveal-answer q=\"957758\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"957758\"]\r\n\r\n[latex]\\begin{array}{l}\\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<h4>Analysis of the Solution<\/h4>\r\nThere 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 function is still positive.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}1[\/latex].\r\n\r\n[reveal-answer q=\"807762\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"807762\"]\r\n\r\n[latex]x=1[\/latex] or [latex]x=\u20131[\/latex][\/hidden-answer]\r\n\r\n[ohm_question]129918[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve a logarithmic equation algebraically.<\/li>\n<li>Solve a logarithmic equation graphically.<\/li>\n<li>Use the one-to-one property of logarithms to solve a logarithmic equation.<\/li>\n<li>Solve a radioactive decay problem.<\/li>\n<\/ul>\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 equal 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>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 as a single log and then apply the definition of logs to solve for [latex]x[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\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{Convert to exponential form}.\\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 both sides by 6}.\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations<\/h3>\n<p>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<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}\\left(S\\right)=c\\text{ if and only if }{b}^{c}=S[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Algebra to Solve a Logarithmic Equation<\/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=\"q977891\">Show Solution<\/span><\/p>\n<div id=\"q977891\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}2\\mathrm{ln}x+3=7\\hfill & \\hfill \\\\ \\text{}2\\mathrm{ln}x=4\\hfill & \\text{Subtract 3 from both sides}.\\hfill \\\\ \\text{}\\mathrm{ln}x=2\\hfill & \\text{Divide both sides 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<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>As was the case when using the properties and rules of exponents and logarithms to rewrite expressions containing them, there can be more than one good way to solve a logarithmic equation. It is good practice to follow the examples given for each of the situations in this section, but you should think about alternative ways to creatively and correctly apply the properties and rules.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]6+\\mathrm{ln}x=10[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q183568\">Show Solution<\/span><\/p>\n<div id=\"q183568\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x={e}^{4}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm129911\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129911&theme=oea&iframe_resize_id=ohm129911&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using Algebra Before and After Using the Definition of the Natural Logarithm<\/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=\"q231886\">Show Solution<\/span><\/p>\n<div id=\"q231886\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}2\\mathrm{ln}\\left(6x\\right)=7\\hfill & \\hfill \\\\ \\text{}\\mathrm{ln}\\left(6x\\right)=\\frac{7}{2}\\hfill & \\text{Divide both sides 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 both sides by 6}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]2\\mathrm{ln}\\left(x+1\\right)=10[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q62905\">Show Solution<\/span><\/p>\n<div id=\"q62905\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x={e}^{5}-1[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm14406\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14406&theme=oea&iframe_resize_id=ohm14406&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using a Graph to Understand the Solution to a Logarithmic Equation<\/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=\"q960461\">Show Solution<\/span><\/p>\n<div id=\"q960461\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}\\mathrm{ln}x=3\\hfill & \\hfill \\\\ x={e}^{3}\\hfill & \\text{Use the definition of }\\mathrm{ln}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<p>Below is a\u00a0graph of the equation. On the graph the <em>x<\/em>-coordinate of the point where the two graphs intersect is close to 20. In other words [latex]{e}^{3}\\approx 20[\/latex]. A calculator gives a better approximation: [latex]{e}^{3}\\approx 20.0855[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03172012\/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\" \/><\/p>\n<p class=\"wp-caption-text\">The graphs of [latex]y=\\mathrm{ln}x[\/latex] and y\u00a0= 3 cross at the point [latex]\\left(e^3,3\\right)[\/latex] which is approximately (20.0855, 3).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Use a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[\/latex] to 2 decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q889911\">Show Solution<\/span><\/p>\n<div id=\"q889911\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x\\approx 9.97[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Using the One-to-One Property of Logarithms to Solve Logarithmic Equations<\/h2>\n<p>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<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/p>\n<p>For example,<\/p>\n<p style=\"text-align: center;\">[latex]\\text{If }{\\mathrm{log}}_{2}\\left(x - 1\\right)={\\mathrm{log}}_{2}\\left(8\\right),\\text{then }x - 1=8[\/latex]<\/p>\n<p>So if [latex]x - 1=8[\/latex], then we can solve for <em>x\u00a0<\/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<p>For example, consider the equation [latex]\\mathrm{log}\\left(3x - 2\\right)-\\mathrm{log}\\left(2\\right)=\\mathrm{log}\\left(x+4\\right)[\/latex]. To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm and then apply the one-to-one property to solve for <em>x<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\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}.\\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>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<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\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]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Using the One-to-One Property of Logarithms to Solve Logarithmic Equations<\/h3>\n<p>For any algebraic expressions <em>S<\/em> and <em>T<\/em> and any positive real number <em>b<\/em>, where [latex]b\\ne 1[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}S={\\mathrm{log}}_{b}T\\text{ if and only if }S=T[\/latex]<\/p>\n<p>Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an equation containing logarithms, solve it using the one-to-one property<\/h3>\n<ol>\n<li>Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation is of 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 to each other.<\/li>\n<li>Solve the resulting equation, <em>S<\/em> =\u00a0<em>T<\/em>, for the unknown.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation Using the One-to-One Property of Logarithms<\/h3>\n<p>Solve [latex]\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}\\left(2x+3\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q957758\">Show Solution<\/span><\/p>\n<div id=\"q957758\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}\\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<h4>Analysis of the Solution<\/h4>\n<p>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 function is still positive.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]\\mathrm{ln}\\left({x}^{2}\\right)=\\mathrm{ln}1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q807762\">Show Solution<\/span><\/p>\n<div id=\"q807762\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=1[\/latex] or [latex]x=\u20131[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm129918\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129918&theme=oea&iframe_resize_id=ohm129918&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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-288\">\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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Question ID 14406. <strong>Authored by<\/strong>: Sousa, James. <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>: IMathAS Community License CC-BY + GPL<\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/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\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 122911. <strong>Authored 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>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t 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