{"id":5282,"date":"2018-05-17T02:11:23","date_gmt":"2018-05-17T02:11:23","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/coreq-collegealgebra\/?post_type=chapter&#038;p=5282"},"modified":"2018-05-21T17:47:25","modified_gmt":"2018-05-21T17:47:25","slug":"logarithmic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/logarithmic-equations\/","title":{"raw":"Logarithmic Equations*","rendered":"Logarithmic Equations*"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Solve a logarithmic equation with algebra<\/li>\r\n \t<li>Solve a logarithmic equation with a graph<\/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\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\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 in compact form and then apply the definition of logs to solve for <em>x<\/em>:\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{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]<\/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\"]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}.\\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<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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"183568\"][latex]x={e}^{4}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129911&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\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\"]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 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\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"62905\"][latex]x={e}^{5}-1[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=14406&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\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\"]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 the natural logarithm}\\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 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].\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\" data-media-type=\"image\/jpg\" \/> 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<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\" rel=\"noopener\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n\r\n[reveal-answer q=\"889911\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"889911\"][latex]x\\approx 9.97[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Use 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<\/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 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]<\/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 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>\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\"]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 functions 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<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\" rel=\"noopener\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n[reveal-answer q=\"807762\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"807762\"][latex]x=1[\/latex] or [latex]x=\u20131[\/latex][\/hidden-answer]\r\n<iframe id=\"mom100\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129918&amp;theme=oea&amp;iframe_resize_id=mom100\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Solve a logarithmic equation with algebra<\/li>\n<li>Solve a logarithmic equation with a graph<\/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<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>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<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{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]<\/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\">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}.\\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<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\">Solution<\/span><\/p>\n<div id=\"q183568\" class=\"hidden-answer\" style=\"display: none\">[latex]x={e}^{4}[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom6\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129911&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/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\">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 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<\/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\">Solution<\/span><\/p>\n<div id=\"q62905\" class=\"hidden-answer\" style=\"display: none\">[latex]x={e}^{5}-1[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=14406&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/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\">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 the natural logarithm}\\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 at which the two graphs intersect is close to 20. In other words [latex]{e}^{3}\\approx 20[\/latex]. A calculator gives a better approximation: [latex]{e}^{3}\\approx 20.0855[\/latex].<\/p>\n<div 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\" data-media-type=\"image\/jpg\" \/><\/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.<br \/>\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q889911\">Solution<\/span><\/p>\n<div id=\"q889911\" class=\"hidden-answer\" style=\"display: none\">[latex]x\\approx 9.97[\/latex]<\/div>\n<\/div>\n<\/div>\n<h2>Use 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<\/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 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>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 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>\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\">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 functions 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].<br \/>\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q807762\">Solution<\/span><\/p>\n<div id=\"q807762\" class=\"hidden-answer\" style=\"display: none\">[latex]x=1[\/latex] or [latex]x=\u20131[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom100\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=129918&amp;theme=oea&amp;iframe_resize_id=mom100\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/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-5282\">\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>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>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 2637, 2620, 2638. <strong>Authored by<\/strong>: Greg Langkamp. <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>Question ID 98554, 98555, 98596. <strong>Authored by<\/strong>: Michael Jenck. <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>Question ID 14406. <strong>Authored by<\/strong>: James Sousa. <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>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 <\/section>","protected":false},"author":160,"menu_order":6,"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\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 2637, 2620, 2638\",\"author\":\"Greg Langkamp\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 98554, 98555, 98596\",\"author\":\"Michael 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