{"id":267,"date":"2023-06-21T13:22:50","date_gmt":"2023-06-21T13:22:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/convert-between-logarithmic-and-exponential-form\/"},"modified":"2023-07-22T07:24:22","modified_gmt":"2023-07-22T07:24:22","slug":"convert-between-logarithmic-and-exponential-form","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/convert-between-logarithmic-and-exponential-form\/","title":{"raw":"\u25aa   Converting Between Logarithmic and Exponential Form","rendered":"\u25aa   Converting Between Logarithmic and Exponential Form"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Convert from logarithmic to exponential form.<\/li>\r\n \t<li>Convert from exponential to logarithmic form.<\/li>\r\n \t<li>Solve real-world problems of logarithmic functions.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is [latex]{10}^{x}=500[\/latex] where <em>x<\/em>\u00a0represents the difference in magnitudes on the <strong>Richter Scale<\/strong>. How would we solve for\u00a0<em>x<\/em>?\r\n\r\nWe have not yet learned a method for solving exponential equations algebraically. None of the algebraic tools discussed so far is sufficient to solve [latex]{10}^{x}=500[\/latex]. We know that [latex]{10}^{2}=100[\/latex] and [latex]{10}^{3}=1000[\/latex], so it is clear that <em>x<\/em>\u00a0must be some value between 2 and 3 since [latex]y={10}^{x}[\/latex] is increasing. We can examine a graph\u00a0to better estimate the solution.\r\n\r\n<img class=\"aligncenter size-full wp-image-3089\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16195659\/CNX_Precalc_Figure_04_03_0022.jpg\" alt=\"Graph of the intersections of the equations y=10^x and y=500.\" width=\"487\" height=\"477\" \/>\r\n\r\nEstimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph above\u00a0passes the horizontal line test. The exponential function [latex]y={b}^{x}[\/latex] is <strong>one-to-one<\/strong>, so its inverse, [latex]x={b}^{y}[\/latex] is also a function. As is the case with all inverse functions, we simply interchange <em>x<\/em>\u00a0and <em>y<\/em>\u00a0and solve for <em>y<\/em>\u00a0to find the inverse function. To represent <em>y<\/em>\u00a0as a function of <em>x<\/em>, we use a logarithmic function of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. The base <em>b<\/em>\u00a0<strong>logarithm<\/strong> of a number is the exponent by which we must raise <em>b<\/em>\u00a0to get that number.\r\n<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nUnderstanding what a logarithm is requires understanding what an exponent is. A logarithm is an exponent.\r\n\r\nRead the paragraphs and boxes below carefully, perhaps more than once or twice, to gain the understanding of the inverse relationship between logarithms and exponents. Keep in mind that the inverse of a function effectively \"undoes\" what the other does.\r\n\r\nYou can use the definition of the logarithm given below to solve certain equations involving exponents and logarithms.\r\n\r\n<\/div>\r\nWe read a logarithmic expression as, \"The logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is equal to <em>y<\/em>,\" or, simplified, \"log base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is <em>y<\/em>.\" We can also say, \"<em>b<\/em>\u00a0raised to the power of <em>y<\/em>\u00a0is <em>x<\/em>,\" because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since [latex]{2}^{5}=32[\/latex], we can write [latex]{\\mathrm{log}}_{2}32=5[\/latex]. We read this as \"log base 2 of 32 is 5.\"\r\n\r\nWe can express the relationship between logarithmic form and its corresponding exponential form as follows:\r\n\r\n[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b&gt;0,b\\ne 1[\/latex]\r\n\r\nNote that the base <em>b<\/em>\u00a0is always positive.\r\n\r\n<img class=\"aligncenter size-full wp-image-3090\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16195822\/CNX_Precalc_Figure_04_03_0042.jpg\" alt=\"Think b to the y equals x.\" width=\"487\" height=\"83\" \/>\r\n\r\nBecause a logarithm is a function, it is most correctly written as [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] using parentheses to denote function evaluation just as we would with [latex]f\\left(x\\right)[\/latex]. However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses as [latex]{\\mathrm{log}}_{b}x[\/latex]. Note that many calculators require parentheses around the <em>x<\/em>.\r\n\r\nWe can illustrate the notation of logarithms as follows:\r\n\r\n<img class=\"aligncenter size-full wp-image-3092\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16200035\/CNX_Precalc_Figure_04_03_0032.jpg\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"487\" height=\"101\" \/>\r\n\r\nNotice that when comparing the logarithm function and the exponential function, the input and the output are switched. This means [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and [latex]y={b}^{x}[\/latex] are inverse functions.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Definition of the Logarithmic Function<\/h3>\r\nA <strong>logarithm<\/strong> base <em>b<\/em>\u00a0of a positive number <em>x<\/em>\u00a0satisfies the following definition:\r\n\r\nFor [latex]x&gt;0,b&gt;0,b\\ne 1[\/latex],\r\n\r\n[latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equal to }{b}^{y}=x[\/latex], where\r\n<ul>\r\n \t<li>we read [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] as, \"the logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>\" or the \"log base <em>b<\/em>\u00a0of <em>x<\/em>.\"<\/li>\r\n \t<li>the logarithm <em>y<\/em>\u00a0is the exponent to which <em>b<\/em>\u00a0must be raised to get <em>x<\/em>.<\/li>\r\n \t<li>if no base [latex]b[\/latex] is indicated, the base of the logarithm is assumed to be [latex]10[\/latex].<\/li>\r\n<\/ul>\r\nAlso, since the logarithmic and exponential functions switch the <em>x<\/em>\u00a0and <em>y<\/em>\u00a0values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,\r\n<ul>\r\n \t<li>the domain of the logarithm function with base [latex]b \\text{ is} \\left(0,\\infty \\right)[\/latex].<\/li>\r\n \t<li>the range of the logarithm function with base [latex]b \\text{ is} \\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h4>Can we take the logarithm of a negative number?<\/h4>\r\n<em>No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an equation in logarithmic form [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex], convert it to exponential form<\/h3>\r\n<ol>\r\n \t<li>Examine the equation [latex]y={\\mathrm{log}}_{b}x[\/latex] and identify <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>.<\/li>\r\n \t<li>Rewrite [latex]{\\mathrm{log}}_{b}x=y[\/latex] as [latex]{b}^{y}=x[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Converting from Logarithmic Form to Exponential Form<\/h3>\r\nWrite the following logarithmic equations in exponential form.\r\n<ol>\r\n \t<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"642511\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"642511\"]\r\n\r\nFirst, identify the values of <em>b<\/em>,\u00a0<em>y<\/em>, and\u00a0<em>x<\/em>. Then, write the equation in the form [latex]{b}^{y}=x[\/latex].\r\n<ol>\r\n \t<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] Here, [latex]b=6,y=\\frac{1}{2},\\text{and } x=\\sqrt{6}[\/latex]. Therefore, the equation [latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] is equal to [latex]{6}^{\\frac{1}{2}}=\\sqrt{6}[\/latex].<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] Here, <em>b\u00a0<\/em>= 3, <em>y\u00a0<\/em>= 2, and <em>x\u00a0<\/em>= 9. Therefore, the equation [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] is equal to [latex]{3}^{2}=9[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite the following logarithmic equations in exponential form.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex]<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"200815\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"200815\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex] is equal to [latex]{10}^{6}=1,000,000[\/latex]<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex] is equal to [latex]{5}^{2}=25[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n[ohm_question]29661[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Convert from Exponential to Logarithmic Form<\/h2>\r\nTo convert from exponential to logarithmic form, we follow the same steps in reverse. We identify the base <em>b<\/em>, exponent <em>x<\/em>, and output <em>y<\/em>. Then we write [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Converting from Exponential Form to Logarithmic Form<\/h3>\r\nWrite the following exponential equations in logarithmic form.\r\n<ol>\r\n \t<li>[latex]{2}^{3}=8[\/latex]<\/li>\r\n \t<li>[latex]{5}^{2}=25[\/latex]<\/li>\r\n \t<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"583658\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"583658\"]\r\n\r\nFirst, identify the values of <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>. Then, write the equation in the form [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].\r\n<ol>\r\n \t<li>[latex]{2}^{3}=8[\/latex] Here, <em>b\u00a0<\/em>= 2, <em>x\u00a0<\/em>= 3, and <em>y\u00a0<\/em>= 8. Therefore, the equation [latex]{2}^{3}=8[\/latex] is equal to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/li>\r\n \t<li>[latex]{5}^{2}=25[\/latex] Here, <em>b\u00a0<\/em>= 5, <em>x\u00a0<\/em>= 2, and <em>y\u00a0<\/em>= 25. Therefore, the equation [latex]{5}^{2}=25[\/latex] is equal to [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex].<\/li>\r\n \t<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] Here, <em>b\u00a0<\/em>= 10, <em>x\u00a0<\/em>= \u20134, and [latex]y=\\frac{1}{10,000}[\/latex]. Therefore, the equation [latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] is equal to [latex]{\\text{log}}_{10}\\left(\\frac{1}{10,000}\\right)=-4[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite the following exponential equations in logarithmic form.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{3}^{2}=9[\/latex]<\/li>\r\n \t<li>[latex]{5}^{3}=125[\/latex]<\/li>\r\n \t<li>[latex]{2}^{-1}=\\frac{1}{2}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"767260\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"767260\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]{3}^{2}=9[\/latex] is equal to [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\r\n \t<li>[latex]{5}^{3}=125[\/latex] is equal to [latex]{\\mathrm{log}}_{5}\\left(125\\right)=3[\/latex]<\/li>\r\n \t<li>[latex]{2}^{-1}=\\frac{1}{2}[\/latex] is equal to [latex]{\\text{log}}_{2}\\left(\\frac{1}{2}\\right)=-1[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n[ohm_question]29668[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Real-World Applications<\/h2>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Rewriting and Solving a Real-World Exponential Model<\/h3>\r\nThe amount of energy released from one earthquake was [latex]500[\/latex] times greater than the amount of energy released from another. The equation <span style=\"text-wrap: nowrap;\">[latex]10^x=500[\/latex]<\/span>\u00a0represents this situation, where <span style=\"text-wrap: nowrap;\">[latex]x[\/latex]<\/span>\u00a0is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?\r\n\r\n[reveal-answer q=\"361449\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"361449\"]\r\n\r\nWe begin by rewriting the exponential equation in logarithmic form.\r\n<p style=\"padding-left: 30px;\">[latex]10^x=500[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]log(500)=x[\/latex]\u00a0 \u00a0 \u00a0[latex]\\leftarrow[\/latex] use the definition of the common logarithm<\/p>\r\nNext, we evaluate the logarithm using a calculator:\r\n<ul>\r\n \t<li>Press \"log\"<\/li>\r\n \t<li>Enter 500, followed by \")\"<\/li>\r\n \t<li>Press \"Enter\"<\/li>\r\n \t<li>To the nearest thousandth, [latex]log(500) \\approx 2.699[\/latex]<\/li>\r\n<\/ul>\r\nSo, the difference in magnitudes was about 2.699.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nThe amount of energy released from one earthquake was <span style=\"text-wrap: nowrap;\">[latex]8,500[\/latex]<\/span>\u00a0times greater than the amount of energy released from another. The equation <span style=\"text-wrap: nowrap;\">[latex]10^x=8500[\/latex]<\/span>\u00a0represents this situation, where <span style=\"text-wrap: nowrap;\">[latex]x[\/latex]<\/span>\u00a0is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?\r\n\r\n[reveal-answer q=\"406177\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"406177\"]\r\n\r\nSince [latex]10^x=8500[\/latex], [latex]x=log(8500) \\approx 3.929[\/latex].\r\n\r\nSo, the difference in magnitudes was about 3.929.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Exposure Index <em>EI<\/em><\/h3>\r\nThe exposure index <em>EI\u00a0<\/em>for a 35-millimeter camera is a measurement of the amount of light that hits the film. It is determined by the equation [latex]EI = log_2(\\frac{f^2}{t})[\/latex], where [latex]f[\/latex] is the \u201cf-stop\u201d setting on the camera, and [latex]t[\/latex] is the exposure time in seconds. Suppose the f-stop setting is 8 and the desired exposure time is 2 seconds. What will the resulting exposure index be?\r\n\r\n[reveal-answer q=\"372356\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"372356\"]\r\n\r\nSince\u00a0the f-stop setting is 8 and the desired exposure time is 2 seconds, [latex]f=8[\/latex] and [latex]t=2[\/latex]. So, [latex]EI = log_2(\\frac {f^2}{t}) = log_2(\\frac{8^2}{2}) = log_2(\\frac {64}{2}) = log_2(32) = 5[\/latex].\r\n\r\nThus, the result exposure index\u00a0<em>EI\u00a0<\/em>is 5.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nRefer to the previous example. Suppose the light meter on a camera indicates an\u00a0<em>EI\u00a0<\/em>of [latex]-2[\/latex],\u00a0and the desired exposure time is 16 seconds. What should the f-stop setting be? (* Hint: f-stop is a positive number.)\r\n\r\n[reveal-answer q=\"906743\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"906743\"]\r\n\r\nSince\u00a0<em>EI <\/em>is [latex]-2[\/latex] and the desired exposure time is 16 seconds, [latex]EI=-2[\/latex] and [latex]t=16[\/latex]. So, [latex]-2=log_2(\\frac{f^2}{16})[\/latex].\r\n\r\nRewrite the logarithmic equation in exponential form:\r\n<p style=\"padding-left: 30px;\">[latex]\\frac {f^2}{16} = 2^{-2} = \\frac {1}{2^2}[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]\\frac {f^2}{16} = \\frac {1}{4}[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]4f^2 =16[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]f^2 = \\frac {16}{4} = 4[\/latex]<\/p>\r\n<p style=\"padding-left: 30px;\">[latex]f = \\sqrt 4 = 2[\/latex]<\/p>\r\nSo, the f-stop setting should be 2.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nIf 1 in [latex]x[\/latex] person dies as a result of doing some given activity each year, the safety index for that activity is [latex]SI = log x[\/latex]. For example, according to a statistic in the US, 1 in 5,300 dies each year due to car crashes. Then the safety index for car crash is [latex]log(5300) \\approx 3.7[\/latex]. If 1 in 2,000,000 is killed by lighting, what is the safety index for lightning?[footnote]Source: https:\/\/janav.wordpress.com\/2013\/09\/29\/logarithms-in-real-life\/[\/footnote]\r\n\r\n[reveal-answer q=\"866414\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"866414\"]\r\n\r\nSince\u00a01 in 2,000,000 is killed by lighting, [latex]x=2000000[\/latex]. So, [latex]SI = log(2000000) \\approx 6.3[\/latex].\r\n\r\nTherefore, the safety index for lightning is 6.3.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Convert from logarithmic to exponential form.<\/li>\n<li>Convert from exponential to logarithmic form.<\/li>\n<li>Solve real-world problems of logarithmic functions.<\/li>\n<\/ul>\n<\/div>\n<p>In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is [latex]{10}^{x}=500[\/latex] where <em>x<\/em>\u00a0represents the difference in magnitudes on the <strong>Richter Scale<\/strong>. How would we solve for\u00a0<em>x<\/em>?<\/p>\n<p>We have not yet learned a method for solving exponential equations algebraically. None of the algebraic tools discussed so far is sufficient to solve [latex]{10}^{x}=500[\/latex]. We know that [latex]{10}^{2}=100[\/latex] and [latex]{10}^{3}=1000[\/latex], so it is clear that <em>x<\/em>\u00a0must be some value between 2 and 3 since [latex]y={10}^{x}[\/latex] is increasing. We can examine a graph\u00a0to better estimate the solution.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3089\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16195659\/CNX_Precalc_Figure_04_03_0022.jpg\" alt=\"Graph of the intersections of the equations y=10^x and y=500.\" width=\"487\" height=\"477\" \/><\/p>\n<p>Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph above\u00a0passes the horizontal line test. The exponential function [latex]y={b}^{x}[\/latex] is <strong>one-to-one<\/strong>, so its inverse, [latex]x={b}^{y}[\/latex] is also a function. As is the case with all inverse functions, we simply interchange <em>x<\/em>\u00a0and <em>y<\/em>\u00a0and solve for <em>y<\/em>\u00a0to find the inverse function. To represent <em>y<\/em>\u00a0as a function of <em>x<\/em>, we use a logarithmic function of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. The base <em>b<\/em>\u00a0<strong>logarithm<\/strong> of a number is the exponent by which we must raise <em>b<\/em>\u00a0to get that number.<\/p>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Understanding what a logarithm is requires understanding what an exponent is. A logarithm is an exponent.<\/p>\n<p>Read the paragraphs and boxes below carefully, perhaps more than once or twice, to gain the understanding of the inverse relationship between logarithms and exponents. Keep in mind that the inverse of a function effectively &#8220;undoes&#8221; what the other does.<\/p>\n<p>You can use the definition of the logarithm given below to solve certain equations involving exponents and logarithms.<\/p>\n<\/div>\n<p>We read a logarithmic expression as, &#8220;The logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is equal to <em>y<\/em>,&#8221; or, simplified, &#8220;log base <em>b<\/em>\u00a0of <em>x<\/em>\u00a0is <em>y<\/em>.&#8221; We can also say, &#8220;<em>b<\/em>\u00a0raised to the power of <em>y<\/em>\u00a0is <em>x<\/em>,&#8221; because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since [latex]{2}^{5}=32[\/latex], we can write [latex]{\\mathrm{log}}_{2}32=5[\/latex]. We read this as &#8220;log base 2 of 32 is 5.&#8221;<\/p>\n<p>We can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\n<p>[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b>0,b\\ne 1[\/latex]<\/p>\n<p>Note that the base <em>b<\/em>\u00a0is always positive.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3090\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16195822\/CNX_Precalc_Figure_04_03_0042.jpg\" alt=\"Think b to the y equals x.\" width=\"487\" height=\"83\" \/><\/p>\n<p>Because a logarithm is a function, it is most correctly written as [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] using parentheses to denote function evaluation just as we would with [latex]f\\left(x\\right)[\/latex]. However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses as [latex]{\\mathrm{log}}_{b}x[\/latex]. Note that many calculators require parentheses around the <em>x<\/em>.<\/p>\n<p>We can illustrate the notation of logarithms as follows:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-3092\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/01\/16200035\/CNX_Precalc_Figure_04_03_0032.jpg\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"487\" height=\"101\" \/><\/p>\n<p>Notice that when comparing the logarithm function and the exponential function, the input and the output are switched. This means [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] and [latex]y={b}^{x}[\/latex] are inverse functions.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Definition of the Logarithmic Function<\/h3>\n<p>A <strong>logarithm<\/strong> base <em>b<\/em>\u00a0of a positive number <em>x<\/em>\u00a0satisfies the following definition:<\/p>\n<p>For [latex]x>0,b>0,b\\ne 1[\/latex],<\/p>\n<p>[latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equal to }{b}^{y}=x[\/latex], where<\/p>\n<ul>\n<li>we read [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] as, &#8220;the logarithm with base <em>b<\/em>\u00a0of <em>x<\/em>&#8221; or the &#8220;log base <em>b<\/em>\u00a0of <em>x<\/em>.&#8221;<\/li>\n<li>the logarithm <em>y<\/em>\u00a0is the exponent to which <em>b<\/em>\u00a0must be raised to get <em>x<\/em>.<\/li>\n<li>if no base [latex]b[\/latex] is indicated, the base of the logarithm is assumed to be [latex]10[\/latex].<\/li>\n<\/ul>\n<p>Also, since the logarithmic and exponential functions switch the <em>x<\/em>\u00a0and <em>y<\/em>\u00a0values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,<\/p>\n<ul>\n<li>the domain of the logarithm function with base [latex]b \\text{ is} \\left(0,\\infty \\right)[\/latex].<\/li>\n<li>the range of the logarithm function with base [latex]b \\text{ is} \\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h4>Can we take the logarithm of a negative number?<\/h4>\n<p><em>No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an equation in logarithmic form [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex], convert it to exponential form<\/h3>\n<ol>\n<li>Examine the equation [latex]y={\\mathrm{log}}_{b}x[\/latex] and identify <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>.<\/li>\n<li>Rewrite [latex]{\\mathrm{log}}_{b}x=y[\/latex] as [latex]{b}^{y}=x[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Converting from Logarithmic Form to Exponential Form<\/h3>\n<p>Write the following logarithmic equations in exponential form.<\/p>\n<ol>\n<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q642511\">Show Solution<\/span><\/p>\n<div id=\"q642511\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, identify the values of <em>b<\/em>,\u00a0<em>y<\/em>, and\u00a0<em>x<\/em>. Then, write the equation in the form [latex]{b}^{y}=x[\/latex].<\/p>\n<ol>\n<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] Here, [latex]b=6,y=\\frac{1}{2},\\text{and } x=\\sqrt{6}[\/latex]. Therefore, the equation [latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex] is equal to [latex]{6}^{\\frac{1}{2}}=\\sqrt{6}[\/latex].<\/li>\n<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] Here, <em>b\u00a0<\/em>= 3, <em>y\u00a0<\/em>= 2, and <em>x\u00a0<\/em>= 9. Therefore, the equation [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex] is equal to [latex]{3}^{2}=9[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the following logarithmic equations in exponential form.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q200815\">Show Solution<\/span><\/p>\n<div id=\"q200815\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex] is equal to [latex]{10}^{6}=1,000,000[\/latex]<\/li>\n<li>[latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex] is equal to [latex]{5}^{2}=25[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm29661\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29661&theme=oea&iframe_resize_id=ohm29661&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Convert from Exponential to Logarithmic Form<\/h2>\n<p>To convert from exponential to logarithmic form, we follow the same steps in reverse. We identify the base <em>b<\/em>, exponent <em>x<\/em>, and output <em>y<\/em>. Then we write [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Converting from Exponential Form to Logarithmic Form<\/h3>\n<p>Write the following exponential equations in logarithmic form.<\/p>\n<ol>\n<li>[latex]{2}^{3}=8[\/latex]<\/li>\n<li>[latex]{5}^{2}=25[\/latex]<\/li>\n<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q583658\">Show Solution<\/span><\/p>\n<div id=\"q583658\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, identify the values of <em>b<\/em>, <em>y<\/em>, and <em>x<\/em>. Then, write the equation in the form [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\n<ol>\n<li>[latex]{2}^{3}=8[\/latex] Here, <em>b\u00a0<\/em>= 2, <em>x\u00a0<\/em>= 3, and <em>y\u00a0<\/em>= 8. Therefore, the equation [latex]{2}^{3}=8[\/latex] is equal to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/li>\n<li>[latex]{5}^{2}=25[\/latex] Here, <em>b\u00a0<\/em>= 5, <em>x\u00a0<\/em>= 2, and <em>y\u00a0<\/em>= 25. Therefore, the equation [latex]{5}^{2}=25[\/latex] is equal to [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex].<\/li>\n<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] Here, <em>b\u00a0<\/em>= 10, <em>x\u00a0<\/em>= \u20134, and [latex]y=\\frac{1}{10,000}[\/latex]. Therefore, the equation [latex]{10}^{-4}=\\frac{1}{10,000}[\/latex] is equal to [latex]{\\text{log}}_{10}\\left(\\frac{1}{10,000}\\right)=-4[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the following exponential equations in logarithmic form.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{3}^{2}=9[\/latex]<\/li>\n<li>[latex]{5}^{3}=125[\/latex]<\/li>\n<li>[latex]{2}^{-1}=\\frac{1}{2}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q767260\">Show Solution<\/span><\/p>\n<div id=\"q767260\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]{3}^{2}=9[\/latex] is equal to [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\n<li>[latex]{5}^{3}=125[\/latex] is equal to [latex]{\\mathrm{log}}_{5}\\left(125\\right)=3[\/latex]<\/li>\n<li>[latex]{2}^{-1}=\\frac{1}{2}[\/latex] is equal to [latex]{\\text{log}}_{2}\\left(\\frac{1}{2}\\right)=-1[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm29668\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29668&theme=oea&iframe_resize_id=ohm29668&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Real-World Applications<\/h2>\n<div class=\"textbox exercises\">\n<h3>Example: Rewriting and Solving a Real-World Exponential Model<\/h3>\n<p>The amount of energy released from one earthquake was [latex]500[\/latex] times greater than the amount of energy released from another. The equation <span style=\"text-wrap: nowrap;\">[latex]10^x=500[\/latex]<\/span>\u00a0represents this situation, where <span style=\"text-wrap: nowrap;\">[latex]x[\/latex]<\/span>\u00a0is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q361449\">Show Solution<\/span><\/p>\n<div id=\"q361449\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by rewriting the exponential equation in logarithmic form.<\/p>\n<p style=\"padding-left: 30px;\">[latex]10^x=500[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]log(500)=x[\/latex]\u00a0 \u00a0 \u00a0[latex]\\leftarrow[\/latex] use the definition of the common logarithm<\/p>\n<p>Next, we evaluate the logarithm using a calculator:<\/p>\n<ul>\n<li>Press &#8220;log&#8221;<\/li>\n<li>Enter 500, followed by &#8220;)&#8221;<\/li>\n<li>Press &#8220;Enter&#8221;<\/li>\n<li>To the nearest thousandth, [latex]log(500) \\approx 2.699[\/latex]<\/li>\n<\/ul>\n<p>So, the difference in magnitudes was about 2.699.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>The amount of energy released from one earthquake was <span style=\"text-wrap: nowrap;\">[latex]8,500[\/latex]<\/span>\u00a0times greater than the amount of energy released from another. The equation <span style=\"text-wrap: nowrap;\">[latex]10^x=8500[\/latex]<\/span>\u00a0represents this situation, where <span style=\"text-wrap: nowrap;\">[latex]x[\/latex]<\/span>\u00a0is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q406177\">Show Solution<\/span><\/p>\n<div id=\"q406177\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since [latex]10^x=8500[\/latex], [latex]x=log(8500) \\approx 3.929[\/latex].<\/p>\n<p>So, the difference in magnitudes was about 3.929.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Exposure Index <em>EI<\/em><\/h3>\n<p>The exposure index <em>EI\u00a0<\/em>for a 35-millimeter camera is a measurement of the amount of light that hits the film. It is determined by the equation [latex]EI = log_2(\\frac{f^2}{t})[\/latex], where [latex]f[\/latex] is the \u201cf-stop\u201d setting on the camera, and [latex]t[\/latex] is the exposure time in seconds. Suppose the f-stop setting is 8 and the desired exposure time is 2 seconds. What will the resulting exposure index be?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q372356\">Show Solution<\/span><\/p>\n<div id=\"q372356\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since\u00a0the f-stop setting is 8 and the desired exposure time is 2 seconds, [latex]f=8[\/latex] and [latex]t=2[\/latex]. So, [latex]EI = log_2(\\frac {f^2}{t}) = log_2(\\frac{8^2}{2}) = log_2(\\frac {64}{2}) = log_2(32) = 5[\/latex].<\/p>\n<p>Thus, the result exposure index\u00a0<em>EI\u00a0<\/em>is 5.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Refer to the previous example. Suppose the light meter on a camera indicates an\u00a0<em>EI\u00a0<\/em>of [latex]-2[\/latex],\u00a0and the desired exposure time is 16 seconds. What should the f-stop setting be? (* Hint: f-stop is a positive number.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q906743\">Show Solution<\/span><\/p>\n<div id=\"q906743\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since\u00a0<em>EI <\/em>is [latex]-2[\/latex] and the desired exposure time is 16 seconds, [latex]EI=-2[\/latex] and [latex]t=16[\/latex]. So, [latex]-2=log_2(\\frac{f^2}{16})[\/latex].<\/p>\n<p>Rewrite the logarithmic equation in exponential form:<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac {f^2}{16} = 2^{-2} = \\frac {1}{2^2}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]\\frac {f^2}{16} = \\frac {1}{4}[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]4f^2 =16[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]f^2 = \\frac {16}{4} = 4[\/latex]<\/p>\n<p style=\"padding-left: 30px;\">[latex]f = \\sqrt 4 = 2[\/latex]<\/p>\n<p>So, the f-stop setting should be 2.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>If 1 in [latex]x[\/latex] person dies as a result of doing some given activity each year, the safety index for that activity is [latex]SI = log x[\/latex]. For example, according to a statistic in the US, 1 in 5,300 dies each year due to car crashes. Then the safety index for car crash is [latex]log(5300) \\approx 3.7[\/latex]. If 1 in 2,000,000 is killed by lighting, what is the safety index for lightning?<a class=\"footnote\" title=\"Source: https:\/\/janav.wordpress.com\/2013\/09\/29\/logarithms-in-real-life\/\" id=\"return-footnote-267-1\" href=\"#footnote-267-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q866414\">Show Answer<\/span><\/p>\n<div id=\"q866414\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since\u00a01 in 2,000,000 is killed by lighting, [latex]x=2000000[\/latex]. So, [latex]SI = log(2000000) \\approx 6.3[\/latex].<\/p>\n<p>Therefore, the safety index for lightning is 6.3.<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-267\">\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 29668, 29661. <strong>Authored by<\/strong>:  McClure,Caren. <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><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-267-1\">Source: https:\/\/janav.wordpress.com\/2013\/09\/29\/logarithms-in-real-life\/ <a href=\"#return-footnote-267-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":395986,"menu_order":29,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Question ID 29668, 29661\",\"author\":\" McClure,Caren\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"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 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