{"id":5342,"date":"2021-10-13T18:24:27","date_gmt":"2021-10-13T18:24:27","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/introduction-exponential-and-logarithmic-equations\/"},"modified":"2023-07-24T15:31:11","modified_gmt":"2023-07-24T15:31:11","slug":"introduction-exponential-and-logarithmic-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/introduction-exponential-and-logarithmic-equations\/","title":{"raw":"Solving Exponential and Logarithmic Equations","rendered":"Solving Exponential and Logarithmic Equations"},"content":{"raw":"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.\r\n\r\nUncontrolled 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 and logarithmic equations.\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dd id=\"fs-id1165137838640\"><\/dd>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve an exponential equation using the definition of a logarithm.<\/li>\r\n \t<li>Solve an exponential equation graphically.<\/li>\r\n \t<li>Solve a logarithmic equation graphically.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Exponential Equations<\/h2>\r\nThe first technique we will introduce for solving exponential equations involves using the definition of the logarithm.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an equation of the form [latex]y=A{e}^{kt}[\/latex], solve for [latex]t[\/latex]<\/h3>\r\n<ol>\r\n \t<li>Isolate the exponential expression, that is, the base with its exponent should be isolated to one side of the equation.<\/li>\r\n \t<li>Change from exponential form of the equation to logarithmic form.<\/li>\r\n \t<li>Use your calculator to find the approximate solution.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation of the Form [latex]y=A{e}^{kt}[\/latex]<\/h3>\r\nSolve [latex]100=20{e}^{2t}[\/latex].\r\n\r\n[reveal-answer q=\"7965\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"7965\"]\r\n\r\n[latex]\\begin{array}{l}100\\hfill &amp; =20{e}^{2t}\\hfill &amp; \\hfill \\\\ 5\\hfill &amp; ={e}^{2t}\\hfill &amp; \\text{Divide by the coefficient 20}\\text{.}\\hfill \\\\ \\mathrm{ln}5\\hfill &amp; =\\mathrm{ln}{{e}^{2t}}\\hfill &amp; \\text{Take ln of both sides.}\\hfill \\\\ \\mathrm{ln}5\\hfill &amp; =2t\\hfill &amp; \\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&nbsp;\r\n<h4>Analysis of the Solution<\/h4>\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, then we use a calculator.\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]3{e}^{0.5t}=11[\/latex].\r\n\r\n[reveal-answer q=\"585330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"585330\"]\r\n\r\n[latex]t=2\\mathrm{ln}\\left(\\frac{11}{3}\\right)[\/latex] or [latex]\\mathrm{ln}{\\left(\\frac{11}{3}\\right)}^{2}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98596&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation That Can Be Simplified to the Form [latex]y=A{e}^{kt}[\/latex]<\/h3>\r\nSolve [latex]4{e}^{2x}+5=12[\/latex].\r\n\r\n[reveal-answer q=\"480495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"480495\"]\r\n\r\n[latex]\\begin{array}{l}4{e}^{2x}+5=12\\hfill &amp; \\hfill \\\\ 4{e}^{2x}=7\\hfill &amp; \\text{Subtract 5 from both sides}.\\hfill \\\\ {e}^{2x}=\\frac{7}{4}\\hfill &amp; \\text{Divide both sides by 4}.\\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 class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]3+{e}^{2t}=7{e}^{2t}[\/latex].\r\n\r\n[reveal-answer q=\"326491\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"326491\"]\r\n\r\n[latex]t=\\mathrm{ln}\\left(\\frac{1}{\\sqrt{2}}\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(2\\right)[\/latex][\/hidden-answer]\r\n<iframe id=\"mom20\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129891&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"350\"><\/iframe>\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 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<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 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<iframe id=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129911&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/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\"]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<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14406&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/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\"]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&nbsp;\r\n<h2>Change-of-Base Formula for Logarithms<\/h2>\r\nSome calculators can only evaluate common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[\/latex], we use the\u00a0<strong>change-of-base formula<\/strong> to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.\r\n\r\nTo derive the change-of-base formula, we use the <strong>one-to-one<\/strong> property and <strong>power rule for logarithms<\/strong>.\r\n\r\nGiven any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1 [\/latex] and [latex]b\\ne 1[\/latex], we show\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex]<\/p>\r\nLet [latex]y={\\mathrm{log}}_{b}M[\/latex]. Converting to exponential form, we obtain [latex]{b}^{y}=M[\/latex]. It follows that:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{n}\\left({b}^{y}\\right)\\hfill &amp; ={\\mathrm{log}}_{n}M\\hfill &amp; \\text{Apply the one-to-one property}.\\hfill \\\\ y{\\mathrm{log}}_{n}b\\hfill &amp; ={\\mathrm{log}}_{n}M \\hfill &amp; \\text{Apply the power rule for logarithms}.\\hfill \\\\ y\\hfill &amp; =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill &amp; \\text{Isolate }y.\\hfill \\\\ {\\mathrm{log}}_{b}M\\hfill &amp; =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill &amp; \\text{Substitute for }y.\\hfill \\end{array}[\/latex]<\/p>\r\nFor example, to evaluate [latex]{\\mathrm{log}}_{5}36[\/latex] using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{5}36\\hfill &amp; =\\frac{\\mathrm{log}\\left(36\\right)}{\\mathrm{log}\\left(5\\right)}\\hfill &amp; \\text{Apply the change of base formula using base 10}\\text{.}\\hfill \\\\ \\hfill &amp; \\approx 2.2266\\text{ }\\hfill &amp; \\text{Use a calculator to evaluate to 4 decimal places}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: The Change-of-Base Formula<\/h3>\r\nThe <strong>change-of-base formula<\/strong> can be used to evaluate a logarithm with any base.\r\n\r\nFor any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1 [\/latex] and [latex]b\\ne 1[\/latex],\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex].<\/p>\r\nIt follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{log}M}{\\mathrm{log}b}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Can we change common logarithms to natural logarithms?<\/strong>\r\n\r\n<em>Yes. Remember that [latex]\\mathrm{log}9[\/latex] means [latex]{\\text{log}}_{\\text{10}}\\text{9}[\/latex]. So, [latex]\\mathrm{log}9=\\frac{\\mathrm{ln}9}{\\mathrm{ln}10}[\/latex].<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Change-of-Base Formula with a Calculator<\/h3>\r\nEvaluate [latex]{\\mathrm{log}}_{2}\\left(10\\right)[\/latex] using the change-of-base formula with a calculator.\r\n\r\n[reveal-answer q=\"448676\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"448676\"]\r\n\r\nAccording to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm which is the log base <i>e<\/i>.\r\n\r\n[latex]\\begin{array}{l}{\\mathrm{log}}_{2}10=\\frac{\\mathrm{ln}10}{\\mathrm{ln}2}\\hfill &amp; \\text{Apply the change of base formula using base }e.\\hfill \\\\ \\approx 3.3219\\hfill &amp; \\text{Use a calculator to evaluate to 4 decimal places}.\\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\nEvaluate [latex]{\\mathrm{log}}_{5}\\left(100\\right)[\/latex] using the change-of-base formula.\r\n\r\n[reveal-answer q=\"732930\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"732930\"]\r\n\r\n[latex]\\frac{\\mathrm{ln}100}{\\mathrm{ln}5}\\approx \\frac{4.6051}{1.6094}=2.861[\/latex][\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35015&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe>\r\n<h2><\/h2>\r\n<ul>\r\n \t<li>We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula.<\/li>\r\n \t<li>The change-of-base formula is often used to rewrite a logarithm with a base other than 10 or [latex]e[\/latex]\u00a0as the quotient of natural or common logs. A calculator can then be used to evaluate it.<\/li>\r\n<\/ul>\r\n<dl id=\"fs-id1165135397912\" class=\"definition\">\r\n \t<dd id=\"fs-id1165135397918\"><\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt>\r\n<dl id=\"fs-id1165137838635\" class=\"definition\">\r\n \t<dt><\/dt>\r\n<\/dl>\r\n<\/dt>\r\n<\/dl>\r\n<\/div>\r\n<h2>Using a Graph to Approximate a Solution to an Exponential Equation<\/h2>\r\nGraphing can help you confirm or find the solution to an exponential equation.\u00a0For example,[latex]42=1.2{\\left(5\\right)}^{x}+2.8[\/latex] can be solved to find the specific value for x that makes it a true statement. Graphing [latex]y=4[\/latex] along with [latex]y=2^{x}[\/latex] in the same window, the point(s) of intersection if any represent the solutions of the equation.\r\n\r\nTo use a calculator to solve this, press <strong>[Y=]<\/strong> and enter [latex]1.2(5)x+2.8 [\/latex] next to <strong>Y1=<\/strong>. Then enter 42 next to <strong>Y2=<\/strong>. For a window, use the values \u20133 to 3 for[latex] x[\/latex] and \u20135 to 55 for[latex]y[\/latex].Press <strong>[GRAPH]<\/strong>. The graphs should intersect somewhere near[latex]x=2[\/latex].\r\n\r\nFor a better approximation, press <strong>[2ND]<\/strong> then <strong>[CALC]<\/strong>. Select <strong>[5: intersect]<\/strong> and press <strong>[ENTER]<\/strong> three times. The x-coordinate of the point of intersection is displayed as 2.1661943. (Your answer may be different if you use a different window or use a different value for Guess?) To the nearest thousandth,x\u22482.166.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]4=7.85{\\left(1.15\\right)}^{x}-2.27[\/latex] graphically. Round to the nearest thousandth.\r\n\r\n[reveal-answer q=\"407425\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"407425\"]\r\n\r\n[latex]x\\approx -1.608[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Using a Graph to Approximate a Solution to a Logarithmic Equation<\/h2>\r\n<h3>How To: Given a logarithmic equation, use a graphing calculator to approximate solutions<\/h3>\r\n<ol>\r\n \t<li>Press <strong>[Y=]<\/strong>. Enter the given logarithmic equation or equations as <strong>Y<sub>1<\/sub>=<\/strong> and, if needed, <strong>Y<sub>2<\/sub>=<\/strong>.<\/li>\r\n \t<li>Press <strong>[GRAPH]<\/strong> to observe the graphs of the curves and use <strong>[WINDOW]<\/strong> to find an appropriate view of the graphs, including their point(s) of intersection.<\/li>\r\n \t<li>To find the value of <em>x<\/em>, we compute the point of intersection. Press <strong>[2ND] <\/strong>then <strong>[CALC]<\/strong>. Select \"intersect\" and press <strong>[ENTER]<\/strong> three times. The point of intersection gives the value of <i>x\u00a0<\/i>for the point(s) of intersection.<\/li>\r\n<\/ol>\r\n<h3>Example: Approximating the Solution of a Logarithmic Equation<\/h3>\r\nSolve [latex]4\\mathrm{ln}\\left(x\\right)+1=-2\\mathrm{ln}\\left(x - 1\\right)[\/latex] graphically. Round to the nearest thousandth.\r\n\r\n[reveal-answer q=\"435068\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"435068\"]\r\n\r\nPress <strong>[Y=]<\/strong> and enter [latex]4\\mathrm{ln}\\left(x\\right)+1[\/latex] next to <strong>Y<sub>1<\/sub><\/strong>=. Then enter [latex]-2\\mathrm{ln}\\left(x - 1\\right)[\/latex] next to <strong>Y<sub>2<\/sub>=<\/strong>. For a window, use the values 0 to 5 for <em>x<\/em>\u00a0and \u201310 to 10 for <em>y<\/em>. Press <strong>[GRAPH]<\/strong>. The graphs should intersect somewhere a little to the right of <em>x\u00a0<\/em>= 1.\r\n\r\nFor a better approximation, press <strong>[2ND] <\/strong>then <strong>[CALC]<\/strong>. Select <strong>[5: intersect]<\/strong> and press <strong>[ENTER]<\/strong> three times. The <em>x<\/em>-coordinate of the point of intersection is displayed as 1.3385297. (Your answer may be different if you use a different window or use a different value for <strong>Guess?<\/strong>) So, to the nearest thousandth, [latex]x\\approx 1.339[\/latex].\r\n\r\n[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]5\\mathrm{log}\\left(x+2\\right)=4-\\mathrm{log}\\left(x\\right)[\/latex] graphically. Round to the nearest thousandth.\r\n\r\n[reveal-answer q=\"280798\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"280798\"]\r\n\r\n[latex]x\\approx 3.049[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<span style=\"font-size: 1.15em;\">Key Concepts<\/span>\r\n<ul>\r\n \t<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex], where <em>S<\/em>\u00a0is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation [latex]{b}^{c}=S[\/latex] and solve for the unknown.<\/li>\r\n \t<li>A graphing calculator may be used to approximate solutions to some exponential and logarithmic equations.<\/li>\r\n \t<li>We can also use graphing to solve equations of the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex]. We graph both equations [latex]y={\\mathrm{log}}_{b}\\left(S\\right)[\/latex] and [latex]y=c[\/latex] on the same coordinate plane and identify the solution as the <em>x-<\/em>value of the point of intersecting.<\/li>\r\n \t<li>Combining the skills learned in this and previous sections, we can solve equations that model real world situations whether the unknown is in an exponent or in the argument of a logarithm.<\/li>\r\n<\/ul>\r\nGlossary\r\n\r\n<strong>change-of-base formula <\/strong>\r\n<p style=\"padding-left: 30px;\">a formula for converting a logarithm with any base to a quotient of logarithms with any other base<\/p>\r\n\r\n<table summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 254.004px;\">Definition of a logarithm<\/td>\r\n<td style=\"width: 991.523px;\">For any algebraic expression <em>S<\/em> and positive real numbers <em>b<\/em>\u00a0and <em>c<\/em>, where [latex]b\\ne 1[\/latex], [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex] if and only if [latex]{b}^{c}=S[\/latex].<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 238.145px;\">The Change-of-Base Formula<\/td>\r\n<td style=\"width: 978.125px;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\text{ }n&gt;0,n\\ne 1,b\\ne 1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>","rendered":"<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>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 and logarithmic equations.<\/p>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt>\n<\/dt>\n<dd id=\"fs-id1165137838640\"><\/dd>\n<\/dl>\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve an exponential equation using the definition of a logarithm.<\/li>\n<li>Solve an exponential equation graphically.<\/li>\n<li>Solve a logarithmic equation graphically.<\/li>\n<\/ul>\n<\/div>\n<h2>Exponential Equations<\/h2>\n<p>The first technique we will introduce for solving exponential equations involves using the definition of the logarithm.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an equation of the form [latex]y=A{e}^{kt}[\/latex], solve for [latex]t[\/latex]<\/h3>\n<ol>\n<li>Isolate the exponential expression, that is, the base with its exponent should be isolated to one side of the equation.<\/li>\n<li>Change from exponential form of the equation to logarithmic form.<\/li>\n<li>Use your calculator to find the approximate solution.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation of the Form [latex]y=A{e}^{kt}[\/latex]<\/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=\"q7965\">Show Solution<\/span><\/p>\n<div id=\"q7965\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}100\\hfill & =20{e}^{2t}\\hfill & \\hfill \\\\ 5\\hfill & ={e}^{2t}\\hfill & \\text{Divide by the coefficient 20}\\text{.}\\hfill \\\\ \\mathrm{ln}5\\hfill & =\\mathrm{ln}{{e}^{2t}}\\hfill & \\text{Take ln of both sides.}\\hfill \\\\ \\mathrm{ln}5\\hfill & =2t\\hfill & \\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<p>&nbsp;<\/p>\n<h4>Analysis of the Solution<\/h4>\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, then we use a calculator.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]3{e}^{0.5t}=11[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q585330\">Show Solution<\/span><\/p>\n<div id=\"q585330\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]t=2\\mathrm{ln}\\left(\\frac{11}{3}\\right)[\/latex] or [latex]\\mathrm{ln}{\\left(\\frac{11}{3}\\right)}^{2}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom10\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=98596&amp;theme=oea&amp;iframe_resize_id=mom10\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation That Can Be Simplified to the Form [latex]y=A{e}^{kt}[\/latex]<\/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=\"q480495\">Show Solution<\/span><\/p>\n<div id=\"q480495\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{array}{l}4{e}^{2x}+5=12\\hfill & \\hfill \\\\ 4{e}^{2x}=7\\hfill & \\text{Subtract 5 from both sides}.\\hfill \\\\ {e}^{2x}=\\frac{7}{4}\\hfill & \\text{Divide both sides by 4}.\\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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]3+{e}^{2t}=7{e}^{2t}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q326491\">Show Solution<\/span><\/p>\n<div id=\"q326491\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]t=\\mathrm{ln}\\left(\\frac{1}{\\sqrt{2}}\\right)=-\\frac{1}{2}\\mathrm{ln}\\left(2\\right)[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom20\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129891&amp;theme=oea&amp;iframe_resize_id=mom20\" width=\"100%\" height=\"350\"><\/iframe><\/p>\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<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 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=\"mom6\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=129911&amp;theme=oea&amp;iframe_resize_id=mom6\" width=\"100%\" height=\"250\"><\/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=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=14406&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/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<p>&nbsp;<\/p>\n<h2>Change-of-Base Formula for Logarithms<\/h2>\n<p>Some calculators can only evaluate common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[\/latex], we use the\u00a0<strong>change-of-base formula<\/strong> to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.<\/p>\n<p>To derive the change-of-base formula, we use the <strong>one-to-one<\/strong> property and <strong>power rule for logarithms<\/strong>.<\/p>\n<p>Given any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1[\/latex] and [latex]b\\ne 1[\/latex], we show<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex]<\/p>\n<p>Let [latex]y={\\mathrm{log}}_{b}M[\/latex]. Converting to exponential form, we obtain [latex]{b}^{y}=M[\/latex]. It follows that:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{n}\\left({b}^{y}\\right)\\hfill & ={\\mathrm{log}}_{n}M\\hfill & \\text{Apply the one-to-one property}.\\hfill \\\\ y{\\mathrm{log}}_{n}b\\hfill & ={\\mathrm{log}}_{n}M \\hfill & \\text{Apply the power rule for logarithms}.\\hfill \\\\ y\\hfill & =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill & \\text{Isolate }y.\\hfill \\\\ {\\mathrm{log}}_{b}M\\hfill & =\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\hfill & \\text{Substitute for }y.\\hfill \\end{array}[\/latex]<\/p>\n<p>For example, to evaluate [latex]{\\mathrm{log}}_{5}36[\/latex] using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{\\mathrm{log}}_{5}36\\hfill & =\\frac{\\mathrm{log}\\left(36\\right)}{\\mathrm{log}\\left(5\\right)}\\hfill & \\text{Apply the change of base formula using base 10}\\text{.}\\hfill \\\\ \\hfill & \\approx 2.2266\\text{ }\\hfill & \\text{Use a calculator to evaluate to 4 decimal places}\\text{.}\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Change-of-Base Formula<\/h3>\n<p>The <strong>change-of-base formula<\/strong> can be used to evaluate a logarithm with any base.<\/p>\n<p>For any positive real numbers <em>M<\/em>, <em>b<\/em>, and <em>n<\/em>, where [latex]n\\ne 1[\/latex] and [latex]b\\ne 1[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}[\/latex].<\/p>\n<p>It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{ln}M}{\\mathrm{ln}b}[\/latex]<\/p>\n<p style=\"text-align: center;\">and<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{b}M=\\frac{\\mathrm{log}M}{\\mathrm{log}b}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Can we change common logarithms to natural logarithms?<\/strong><\/p>\n<p><em>Yes. Remember that [latex]\\mathrm{log}9[\/latex] means [latex]{\\text{log}}_{\\text{10}}\\text{9}[\/latex]. So, [latex]\\mathrm{log}9=\\frac{\\mathrm{ln}9}{\\mathrm{ln}10}[\/latex].<\/em><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Change-of-Base Formula with a Calculator<\/h3>\n<p>Evaluate [latex]{\\mathrm{log}}_{2}\\left(10\\right)[\/latex] using the change-of-base formula with a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q448676\">Show Solution<\/span><\/p>\n<div id=\"q448676\" class=\"hidden-answer\" style=\"display: none\">\n<p>According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm which is the log base <i>e<\/i>.<\/p>\n<p>[latex]\\begin{array}{l}{\\mathrm{log}}_{2}10=\\frac{\\mathrm{ln}10}{\\mathrm{ln}2}\\hfill & \\text{Apply the change of base formula using base }e.\\hfill \\\\ \\approx 3.3219\\hfill & \\text{Use a calculator to evaluate to 4 decimal places}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]{\\mathrm{log}}_{5}\\left(100\\right)[\/latex] using the change-of-base formula.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q732930\">Show Solution<\/span><\/p>\n<div id=\"q732930\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\mathrm{ln}100}{\\mathrm{ln}5}\\approx \\frac{4.6051}{1.6094}=2.861[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=35015&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<h2><\/h2>\n<ul>\n<li>We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula.<\/li>\n<li>The change-of-base formula is often used to rewrite a logarithm with a base other than 10 or [latex]e[\/latex]\u00a0as the quotient of natural or common logs. A calculator can then be used to evaluate it.<\/li>\n<\/ul>\n<dl id=\"fs-id1165135397912\" class=\"definition\">\n<dd id=\"fs-id1165135397918\"><\/dd>\n<\/dl>\n<dl id=\"fs-id1165137838635\" class=\"definition\">\n<dt>\n<\/dt>\n<dt><\/dt>\n<\/dl>\n<\/div>\n<h2>Using a Graph to Approximate a Solution to an Exponential Equation<\/h2>\n<p>Graphing can help you confirm or find the solution to an exponential equation.\u00a0For example,[latex]42=1.2{\\left(5\\right)}^{x}+2.8[\/latex] can be solved to find the specific value for x that makes it a true statement. Graphing [latex]y=4[\/latex] along with [latex]y=2^{x}[\/latex] in the same window, the point(s) of intersection if any represent the solutions of the equation.<\/p>\n<p>To use a calculator to solve this, press <strong>[Y=]<\/strong> and enter [latex]1.2(5)x+2.8[\/latex] next to <strong>Y1=<\/strong>. Then enter 42 next to <strong>Y2=<\/strong>. For a window, use the values \u20133 to 3 for[latex]x[\/latex] and \u20135 to 55 for[latex]y[\/latex].Press <strong>[GRAPH]<\/strong>. The graphs should intersect somewhere near[latex]x=2[\/latex].<\/p>\n<p>For a better approximation, press <strong>[2ND]<\/strong> then <strong>[CALC]<\/strong>. Select <strong>[5: intersect]<\/strong> and press <strong>[ENTER]<\/strong> three times. The x-coordinate of the point of intersection is displayed as 2.1661943. (Your answer may be different if you use a different window or use a different value for Guess?) To the nearest thousandth,x\u22482.166.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]4=7.85{\\left(1.15\\right)}^{x}-2.27[\/latex] graphically. Round to the nearest thousandth.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q407425\">Show Solution<\/span><\/p>\n<div id=\"q407425\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x\\approx -1.608[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Using a Graph to Approximate a Solution to a Logarithmic Equation<\/h2>\n<h3>How To: Given a logarithmic equation, use a graphing calculator to approximate solutions<\/h3>\n<ol>\n<li>Press <strong>[Y=]<\/strong>. Enter the given logarithmic equation or equations as <strong>Y<sub>1<\/sub>=<\/strong> and, if needed, <strong>Y<sub>2<\/sub>=<\/strong>.<\/li>\n<li>Press <strong>[GRAPH]<\/strong> to observe the graphs of the curves and use <strong>[WINDOW]<\/strong> to find an appropriate view of the graphs, including their point(s) of intersection.<\/li>\n<li>To find the value of <em>x<\/em>, we compute the point of intersection. Press <strong>[2ND] <\/strong>then <strong>[CALC]<\/strong>. Select &#8220;intersect&#8221; and press <strong>[ENTER]<\/strong> three times. The point of intersection gives the value of <i>x\u00a0<\/i>for the point(s) of intersection.<\/li>\n<\/ol>\n<h3>Example: Approximating the Solution of a Logarithmic Equation<\/h3>\n<p>Solve [latex]4\\mathrm{ln}\\left(x\\right)+1=-2\\mathrm{ln}\\left(x - 1\\right)[\/latex] graphically. Round to the nearest thousandth.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q435068\">Show Solution<\/span><\/p>\n<div id=\"q435068\" class=\"hidden-answer\" style=\"display: none\">\n<p>Press <strong>[Y=]<\/strong> and enter [latex]4\\mathrm{ln}\\left(x\\right)+1[\/latex] next to <strong>Y<sub>1<\/sub><\/strong>=. Then enter [latex]-2\\mathrm{ln}\\left(x - 1\\right)[\/latex] next to <strong>Y<sub>2<\/sub>=<\/strong>. For a window, use the values 0 to 5 for <em>x<\/em>\u00a0and \u201310 to 10 for <em>y<\/em>. Press <strong>[GRAPH]<\/strong>. The graphs should intersect somewhere a little to the right of <em>x\u00a0<\/em>= 1.<\/p>\n<p>For a better approximation, press <strong>[2ND] <\/strong>then <strong>[CALC]<\/strong>. Select <strong>[5: intersect]<\/strong> and press <strong>[ENTER]<\/strong> three times. The <em>x<\/em>-coordinate of the point of intersection is displayed as 1.3385297. (Your answer may be different if you use a different window or use a different value for <strong>Guess?<\/strong>) So, to the nearest thousandth, [latex]x\\approx 1.339[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]5\\mathrm{log}\\left(x+2\\right)=4-\\mathrm{log}\\left(x\\right)[\/latex] graphically. Round to the nearest thousandth.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q280798\">Show Solution<\/span><\/p>\n<div id=\"q280798\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x\\approx 3.049[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p><span style=\"font-size: 1.15em;\">Key Concepts<\/span><\/p>\n<ul>\n<li>When given an equation of the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex], where <em>S<\/em>\u00a0is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation [latex]{b}^{c}=S[\/latex] and solve for the unknown.<\/li>\n<li>A graphing calculator may be used to approximate solutions to some exponential and logarithmic equations.<\/li>\n<li>We can also use graphing to solve equations of the form [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex]. We graph both equations [latex]y={\\mathrm{log}}_{b}\\left(S\\right)[\/latex] and [latex]y=c[\/latex] on the same coordinate plane and identify the solution as the <em>x-<\/em>value of the point of intersecting.<\/li>\n<li>Combining the skills learned in this and previous sections, we can solve equations that model real world situations whether the unknown is in an exponent or in the argument of a logarithm.<\/li>\n<\/ul>\n<p>Glossary<\/p>\n<p><strong>change-of-base formula <\/strong><\/p>\n<p style=\"padding-left: 30px;\">a formula for converting a logarithm with any base to a quotient of logarithms with any other base<\/p>\n<table summary=\"...\">\n<tbody>\n<tr>\n<td style=\"width: 254.004px;\">Definition of a logarithm<\/td>\n<td style=\"width: 991.523px;\">For any algebraic expression <em>S<\/em> and positive real numbers <em>b<\/em>\u00a0and <em>c<\/em>, where [latex]b\\ne 1[\/latex], [latex]{\\mathrm{log}}_{b}\\left(S\\right)=c[\/latex] if and only if [latex]{b}^{c}=S[\/latex].<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 238.145px;\">The Change-of-Base Formula<\/td>\n<td style=\"width: 978.125px;\">[latex]{\\mathrm{log}}_{b}M\\text{=}\\frac{{\\mathrm{log}}_{n}M}{{\\mathrm{log}}_{n}b}\\text{ }n>0,n\\ne 1,b\\ne 1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\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-5342\">\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":167848,"menu_order":8,"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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