{"id":2393,"date":"2022-05-27T21:57:35","date_gmt":"2022-05-27T21:57:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=2393"},"modified":"2025-12-07T22:41:13","modified_gmt":"2025-12-07T22:41:13","slug":"6-3-the-inverse-of-a-logarithmic-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/6-3-the-inverse-of-a-logarithmic-function\/","title":{"raw":"6.3: The Inverse of a Logarithmic Function","rendered":"6.3: The Inverse of a Logarithmic Function"},"content":{"raw":"<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Objectives<\/h1>\r\n<ul>\r\n \t<li>Graph the inverse function of a logarithmic function<\/li>\r\n \t<li>Find the equation of the inverse function of a logarithmic function<\/li>\r\n \t<li><span class=\"TextRun SCXW18598214 BCX9\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW18598214 BCX9\">Use compositions of function to show equivalency: [latex]f\\left(f^{-1}(x)\\right)=f^{-1}\\left(f^(x)\\right)=x[\/latex]<\/span><\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Graphing the Inverse Function<\/h2>\r\nOne way to graph the inverse of a logarithmic function is by creating and using its inverse table. For example, to graph the inverse function of [latex]f(x)=log_2{x}[\/latex], we start by creating a table of values for the function (Table 1).\r\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 50%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 276.188px; text-align: center;\">[latex]x[\/latex]<\/th>\r\n<th style=\"width: 202.812px; text-align: center;\">[latex]y=\\log_2{x}[\/latex]<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 276.188px; text-align: center;\">[latex]\\dfrac{1}{8}[\/latex]<\/td>\r\n<td style=\"width: 202.812px; text-align: center;\">[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 276.188px; text-align: center;\">[latex]\\dfrac{1}{4}[\/latex]<\/td>\r\n<td style=\"width: 202.812px; text-align: center;\">[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 276.188px; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/td>\r\n<td style=\"width: 202.812px; text-align: center;\">[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 276.188px; text-align: center;\">[latex]1[\/latex]<\/td>\r\n<td style=\"width: 202.812px; text-align: center;\">[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 276.188px; text-align: center;\">[latex]2[\/latex]<\/td>\r\n<td style=\"width: 202.812px; text-align: center;\">[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 276.188px; text-align: center;\">[latex]4[\/latex]<\/td>\r\n<td style=\"width: 202.812px; text-align: center;\">[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 276.188px; text-align: center;\">[latex]8[\/latex]<\/td>\r\n<td style=\"width: 202.812px; text-align: center;\">[latex]3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 276.188px;\" colspan=\"2\">Table 1. Table of values for\u00a0[latex]f(x)=\\log_2{x}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: left;\">To create an inverse table, we switch the values in the [latex]x[\/latex] and [latex]y[\/latex] columns so that the inputs become the values of [latex]y[\/latex] and the outputs become the values of [latex]x[\/latex]. The table after switching the values of the [latex]x[\/latex] and [latex]y[\/latex] columns is the inverse table (Table 2). Since\u00a0[latex]f(x)=\\log_2{x}[\/latex] is a one-to-one function, its inverse will also be a one-to-one function.<\/p>\r\n\r\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 50%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<th style=\"width: 2.40964%; text-align: center;\">[latex]x[\/latex]<\/th>\r\n<th style=\"width: 2.40964%; text-align: center;\">[latex]y=f^{-1}(x)[\/latex]<\/th>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]-3[\/latex]<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]\\dfrac{1}{8}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]-2[\/latex]<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]\\dfrac{1}{4}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]-1[\/latex]<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]0[\/latex]<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]1[\/latex]<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]2[\/latex]<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]3[\/latex]<\/td>\r\n<td style=\"width: 2.40964%; text-align: center;\">[latex]8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 2.40964%;\" colspan=\"2\">Table 2. Inverse table<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nLooking at the values in table 2, notice that as [latex]x[\/latex] increases by 1, the inverse function value increases by a multiple of 2. This means that the inverse of a logarithmic function is an exponential function. In this case, the inverse of\u00a0[latex]f(x)=\\log_2{x}[\/latex] is [latex]f^{-1}(x)=2^x[\/latex].\r\n\r\nFigure 1 shows the graphs of the function [latex]f(x)=\\log_2{x}[\/latex] (the blue curve) and its inverse (the green curve) graphed usings the values in Tables 1 and 2.\u00a0 Notice that the graph of the inverse function is a reflection of the graph of the original function with respect to the line [latex]y=x[\/latex] (the red line).\r\n\r\n[caption id=\"attachment_2941\" align=\"aligncenter\" width=\"408\"]<img class=\"wp-image-2941\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/22145601\/6-3-InverseFunctionNew-300x300.png\" alt=\"A green curve rising quickly is the reflection of a blue curve that increases gradually to the right, reflected across the line \u2060y = x.\" width=\"408\" height=\"408\" \/> Figure 1. The graphs of the function [latex]f(x)=\\log_2{x}[\/latex] and its inverse function.[\/caption]Notice that the inverse is also a function because it passes the vertical line test. The inverse is a one-to-one mapping. Therefore, the inverse is a function.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nUse Desmos to graph [latex]f(x)=2\\log_2{(x-3)}+1[\/latex], then use reflection to graph its inverse function.\r\n<h4>Solution<\/h4>\r\nCopying and pasting the function into Desmos yields:\r\n<p style=\"text-align: center;\"><img class=\"aligncenter wp-image-3036 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T105304.306-300x300.png\" alt=\"A blue curve rising gradually to the right with a vertical asymptote at x=3.\" width=\"300\" height=\"300\" \/><\/p>\r\nNow we draw in the line [latex]y=x[\/latex] and reflect every point on the graph of [latex]y=f(x)[\/latex] across the line [latex]y=x[\/latex] to get the inverse. Every point [latex](x,\\;y)[\/latex] is reflected to [latex](y,\\;x)[\/latex]. The vertical asymptote [latex]x=3[\/latex] reflects to a horizontal asymptote [latex]y=3[\/latex].\r\n<p style=\"text-align: center;\"><img class=\"aligncenter wp-image-3037 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T110817.876-300x300.png\" alt=\"A green curve rising gradually is the reflection of a blue curve that increases gradually to the right, reflected across the line \u2060y = x. A point (x,y) on the blue curve corresponds to the point (y,x) on the green curve. \" width=\"300\" height=\"300\" \/><\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nUse Desmos to graph [latex]f(x)=\\log_2{(x+4)}+3[\/latex], then use reflection to graph its inverse function.\r\n\r\n[reveal-answer q=\"hjm118\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm118\"]\r\n<p style=\"text-align: center;\"><img class=\"aligncenter wp-image-3038 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T111805.998-300x300.png\" alt=\"A green curve rising quickly is the reflection of a blue curve that increases gradually to the right, reflected across the line \u2060y = x. The two curves intersect at two points on the line of symmetry.\" width=\"300\" height=\"300\" \/><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>The Inverse Function Equation<\/h2>\r\nWe already know that to find the equation of an inverse function for a given function, we set [latex]y=f(x)[\/latex], switch [latex]x[\/latex] and [latex]y[\/latex], then solve for [latex]y[\/latex]. For example, given the logarithmic function\u00a0[latex]f(x)=\\log_2{x}[\/latex], we first set [latex]y=f(x)[\/latex]:\r\n<p style=\"text-align: center;\">[latex]y=\\log_2{x}[\/latex]<\/p>\r\nThen we switch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\r\n<p style=\"text-align: center;\">[latex]x=\\log_2y[\/latex]<\/p>\r\nNow we solve for [latex]y[\/latex] by writing this logarithmic equation as an equivalent exponential equation. Based on the definition of logarithm, the base of the logarithm, 2, becomes the base of the exponential:\r\n<p style=\"text-align: center;\">[latex]y=2^x[\/latex]<\/p>\r\nTherefore, the inverse function of the logarithmic function\u00a0[latex]f(x)=\\log_2{x}[\/latex] is:\r\n<p style=\"text-align: center;\">[latex]f^{-1}(x)=2^{x}[\/latex]<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nDetermine the inverse function of [latex]g(x)=\\log_5x[\/latex].\r\n<h4>Solution<\/h4>\r\nFirst set [latex]y=g(x)[\/latex]:\u00a0\u00a0[latex]y=\\log_5x[\/latex]\r\n\r\nSwitch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\u00a0 [latex]x=\\log_5y[\/latex]\r\n\r\nUse the definition of logarithm to solve for [latex]y[\/latex]:\u00a0 [latex]y=5^x[\/latex]\r\n\r\nWrite in function notation:\u00a0 [latex]g^{-1}(x)=5^x[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nDetermine the inverse function of [latex]h(x)=\\log_7x[\/latex].\r\n\r\n[reveal-answer q=\"hjm291\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm291\"]\r\n\r\n[latex]h^{-1}(x)=7^x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 3<\/h3>\r\nDetermine the inverse function of the logarithmic function\u00a0[latex]f(x)=\\log_4{(x+5)}[\/latex].\r\n<h4>Solution<\/h4>\r\nFirst set [latex]y=f(x)[\/latex]:\u00a0\u00a0[latex]y=\\log_4{(x+5)}[\/latex]\r\n\r\nSwitch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\u00a0 [latex]x=\\log_4{(y+5)}[\/latex]\r\n\r\nUse the definition of logarithm to solve for [latex]y[\/latex]:\u00a0 [latex]y+5=4^x[\/latex]\r\n\r\nSolve for [latex]y[\/latex] by subtracting 5 from both sides:\u00a0\u00a0[latex]y=4^x-5[\/latex]\r\n\r\nWrite in function notation:\u00a0[latex]f^{-1}(x)=4^x-5[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nDetermine the inverse function of the logarithmic function\u00a0[latex]g(x)=\\log_3{(x-6)}[\/latex].\r\n\r\n[reveal-answer q=\"hjm926\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm926\"][latex]g^{-1}(x)=3^x+6[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nFor a logarithmic function that is in the form [latex]f(x)=a\\log_b{x-h}+k[\/latex], we must first isolate the logarithm, before using the definition of a logarithm to solve for [latex]y[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Example 4<\/h3>\r\nDetermine the inverse function of the logarithmic function\u00a0[latex]f(x)=4\\log_3{(x+2)}[\/latex].\r\n<h4>Solution<\/h4>\r\nFirst set [latex]y=f(x)[\/latex]:[latex]y=4\\log_3{(x+2)}[\/latex]\r\n\r\nSwitch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\u00a0 [latex]x=4\\log_3{(y+2)}[\/latex]\r\n\r\nIsolate the logarithm by dividing both sides by 4:\u00a0[latex]\\frac{1}{4}x=\\log_3{(y+2)}[\/latex]\r\n\r\nUse the definition of logarithm to solve for [latex]y[\/latex]:\u00a0 [latex]y+2=3^{\\frac{x}{4}}[\/latex]\r\n\r\nSolve for [latex]y[\/latex] by subtracting 2 from both sides:\u00a0[latex]y=3^{\\frac{x}{4}}-2[\/latex]\r\n\r\nWrite in function notation:\u00a0 [latex]f^{-1}(x)=3^{\\frac{x}{4}}-2[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 4<\/h3>\r\nDetermine the inverse function of the logarithmic function\u00a0[latex]f(x)=5\\log_7{(x-3)}[\/latex].\r\n\r\n[reveal-answer q=\"hjm985\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm985\"]\r\n\r\n[latex]f^{-1}(x)=7^{\\frac{x}{5}}+3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 5<\/h3>\r\nDetermine the inverse function of the logarithmic function\u00a0[latex]f(x)=\\log_5{(x-3)}+2[\/latex].\r\n<h4>Solution<\/h4>\r\nFirst set [latex]y=f(x)[\/latex]:\u00a0[latex]y=\\log_5{(x-3)}+2[\/latex]\r\n\r\nSwitch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\u00a0[latex]x=\\log_5{(y-3)}+2[\/latex]\r\n\r\nIsolate the logarithm by subtracting 2 from both sides:\u00a0\u00a0[latex]x-2=\\log_5{(y-3)}[\/latex]\r\n\r\nUse the definition of logarithm to solve for [latex]y[\/latex]:\u00a0 [latex]y-3=5^{x-2}[\/latex]\r\n\r\nSolve for [latex]y[\/latex] by adding 3 to both sides:\u00a0\u00a0[latex]y=5^{x-2}+3[\/latex]\r\n\r\nWrite in function notation:\u00a0[latex]f^{-1}(x)=5^{x-2}+3[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 6<\/h3>\r\nDetermine the inverse function of the logarithmic function [latex]f(x)=\\dfrac{2}{5}\\log_3{(8x)}-4[\/latex].\r\n<h4>Solution<\/h4>\r\nFirst set [latex]y=f(x)[\/latex]:\u00a0 [latex]y=\\dfrac{2}{5}\\log_3{(8x)}-4[\/latex]\r\n\r\nSwitch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\u00a0 [latex]x=\\dfrac{2}{5}\\log_3{(8y)}-4[\/latex]\r\n\r\nIsolate the logarithm by adding 4 to both sides:\u00a0[latex]x+4=\\dfrac{2}{5}\\log_3{(8y)}[\/latex]\r\n\r\nIsolate the logarithm by multiplying both sides by [latex]\\dfrac{5}{2}[\/latex]:\u00a0\u00a0[latex]\\dfrac{5}{2}(x+4)=\\log_3{(8y)}[\/latex]\r\n\r\nUse the definition of logarithm to solve for [latex]y[\/latex]:\u00a0 [latex]8y=3^{\\frac{5}{2}(x+4)}[\/latex]\r\n\r\nSolve for [latex]y[\/latex] by dividing both sides by 8:\u00a0[latex]y=\\frac{1}{8}\\cdot3^{\\frac{5}{2}(x+4)}[\/latex]\r\n\r\nWrite in function notation:\u00a0[latex]f^{-1}(x)=\\frac{1}{8}\\cdot3^{\\frac{5}{2}(x+4)}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 5<\/h3>\r\nDetermine the inverse function of the logarithmic function:\r\n<ol>\r\n \t<li>[latex]f(x)=\\log_2{(x-9)}+1[\/latex]<\/li>\r\n \t<li>[latex]g(x)=\\dfrac{3}{4}\\log_5{x}-2[\/latex]<\/li>\r\n \t<li>[latex]h(x)=\\dfrac{1}{3}\\log_2{(x+5)}+8[\/latex]<\/li>\r\n \t<li>[latex]F(x)=6\\log_{10}{(x-3)}+6[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"hjm417\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm417\"]\r\n<ol>\r\n \t<li>[latex]f^{-1}(x)=2^{x-1}+9[\/latex]<\/li>\r\n \t<li>[latex]g^{-1}=5^{\\frac{4}{3}(x+2)}[\/latex]<\/li>\r\n \t<li>[latex]h^{-1}=2^{3(x-8)}-5[\/latex]<\/li>\r\n \t<li>[latex]F^{-1}(x)=10^{\\frac{x-6}{6}}+3[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Composition of Inverse Functions<\/h2>\r\nIt is important to remember that an inverse function 'undoes' what was 'done' by the original function. Logarithmic functions and inverse functions are inverses of one another, so if we apply one function then apply its inverse, we should get back to where we started.\r\n\r\nConsider the function [latex]f(x)=\\log_4x[\/latex] and its inverse function [latex]f^{-1}(x)=4^x[\/latex]. Now let's perform a composition of functions:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(f\\circ f^{-1}\\right)(x)&amp;=f\\left(f^{-1}(x)\\right)\\\\&amp;=f\\left(4^x\\right)\\\\&amp;=\\log_4{4^x}\\end{aligned}[\/latex]<\/p>\r\nTo simplify [latex]\\log_4{4^x}[\/latex], let's set [latex]a=\\log_4{4^x}[\/latex]. Then using the definition of logarithms, we can write an equivalent exponential equation:\r\n<p style=\"text-align: center;\">[latex]4^a=4^x[\/latex]<\/p>\r\nSince the bases are identical, the exponents must be equal:\r\n<p style=\"text-align: center;\">[latex]a=x[\/latex]<\/p>\r\nConsequently,\u00a0[latex]\\log_4{4^x}=x[\/latex] and, therefore, [latex]\\left(f\\circ f^{-1}\\right)(x)=x[\/latex].\r\n\r\nIf we perform the composition in the opposite direction we get:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(f^{-1}\\circ f\\right)(x)&amp;=f^{-1}\\left(f(x)\\right)\\\\&amp;=f^{-1}\\left(\\log_4x\\right)\\\\&amp;=4^{\\log_4x}\\end{aligned}[\/latex]<\/p>\r\nTo simplify [latex]4^{\\log_4x}[\/latex], let's set [latex]a=4^{\\log_4x}[\/latex]. Then using the definition of logarithms, we can write an equivalent exponential equation:\r\n<p style=\"text-align: center;\">[latex]\\log_4a=\\log_4x[\/latex]<\/p>\r\nSince logarithms are one-to-one functions, [latex]a=x[\/latex].\r\n\r\nConsequently,\u00a0[latex]4^{\\log_4x}=x[\/latex] and, therefore, [latex]\\left(f^{-1}\\circ f\\right)(x)=x[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Example 7<\/h3>\r\nUse composition of functions to show that [latex]f(x)=\\log_2{(x+1)}[\/latex] and [latex]g(x)=2^x-1[\/latex] are inverse functions.\r\n<h4>Solution<\/h4>\r\nIt doesn't matter the order in which we compose the functions.\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(f\\circ g\\right)(x)&amp;=f\\left(2^x-1\\right)\\\\&amp;=\\log_2{(2^x-1+1)}\\\\&amp;=\\log_2{2^x}\\end{aligned}[\/latex]<\/p>\r\nTo simplify [latex]\\log_2{2^x}[\/latex], let [latex]a=\\log_2{2^x}[\/latex], then using the definition of logarithms:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}2^a&amp;=2^x\\\\a&amp;=x\\end{aligned}[\/latex]<\/p>\r\nConsequently,\u00a0[latex]\\left(f\\circ g\\right)(x)=x[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(g\\circ f\\right)(x)&amp;=g\\left(\\log_2{(x+1)}\\right)\\\\&amp;=2^{\\log_2{(x+1)}}-1\\end{aligned}[\/latex]<\/p>\r\nTo simplify [latex]2^{\\log_2{(x+1)}}[\/latex], let [latex]a=2^{\\log_2{(x+1)}}[\/latex], then using the definition of logarithms:\r\n<p style=\"text-align: center;\">[latex]\\log_2{a}=\\log_2{(x+1)}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]a=x+1[\/latex]<\/p>\r\nHence,\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(g\\circ f\\right)(x)&amp;=g\\left(\\log_2{(x+1)}\\right)\\\\&amp;=2^{\\log_2{(x+1)}}-1\\\\&amp;=(x+1)-1\\\\&amp;=x\\end{aligned}[\/latex]<\/p>\r\nConsequently,\u00a0[latex]\\left(g\\circ f\\right)(x)=x[\/latex]\r\n\r\n&nbsp;\r\n\r\nTherefore, [latex]f(x)=\\log_2{(x+1)}[\/latex] and [latex]g(x)=2^x-1[\/latex] are inverse functions.\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 6<\/h3>\r\nUse composition of functions to show that [latex]f(x)=\\log_3{(x+4)}[\/latex] and [latex]g(x)=3^x-4[\/latex] are inverse functions.\r\n\r\n[reveal-answer q=\"hjm562\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm562\"]\r\n\r\n[latex]\\left(f\\circ g\\right)(x)=\\left(g\\circ f\\right)(x)=x[\/latex] so [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are inverse functions.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<script>\r\nwindow.embeddedChatbotConfig = {\r\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\r\ndomain: \"www.chatbase.co\"\r\n}\r\n<\/script>\r\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" chatbotId=\"ejVb5sgc1-w972OOCgl5x\" domain=\"www.chatbase.co\" defer>\r\n<\/script>\r\n\r\n<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>","rendered":"<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n<div class=\"textbox learning-objectives\">\n<h1>Learning Objectives<\/h1>\n<ul>\n<li>Graph the inverse function of a logarithmic function<\/li>\n<li>Find the equation of the inverse function of a logarithmic function<\/li>\n<li><span class=\"TextRun SCXW18598214 BCX9\" lang=\"EN-US\" xml:lang=\"EN-US\" data-contrast=\"none\"><span class=\"NormalTextRun SCXW18598214 BCX9\">Use compositions of function to show equivalency: [latex]f\\left(f^{-1}(x)\\right)=f^{-1}\\left(f^(x)\\right)=x[\/latex]<\/span><\/span><\/li>\n<\/ul>\n<\/div>\n<h2>Graphing the Inverse Function<\/h2>\n<p>One way to graph the inverse of a logarithmic function is by creating and using its inverse table. For example, to graph the inverse function of [latex]f(x)=log_2{x}[\/latex], we start by creating a table of values for the function (Table 1).<\/p>\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 50%;\">\n<tbody>\n<tr>\n<th style=\"width: 276.188px; text-align: center;\">[latex]x[\/latex]<\/th>\n<th style=\"width: 202.812px; text-align: center;\">[latex]y=\\log_2{x}[\/latex]<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 276.188px; text-align: center;\">[latex]\\dfrac{1}{8}[\/latex]<\/td>\n<td style=\"width: 202.812px; text-align: center;\">[latex]-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 276.188px; text-align: center;\">[latex]\\dfrac{1}{4}[\/latex]<\/td>\n<td style=\"width: 202.812px; text-align: center;\">[latex]-2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 276.188px; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/td>\n<td style=\"width: 202.812px; text-align: center;\">[latex]-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 276.188px; text-align: center;\">[latex]1[\/latex]<\/td>\n<td style=\"width: 202.812px; text-align: center;\">[latex]0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 276.188px; text-align: center;\">[latex]2[\/latex]<\/td>\n<td style=\"width: 202.812px; text-align: center;\">[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 276.188px; text-align: center;\">[latex]4[\/latex]<\/td>\n<td style=\"width: 202.812px; text-align: center;\">[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 276.188px; text-align: center;\">[latex]8[\/latex]<\/td>\n<td style=\"width: 202.812px; text-align: center;\">[latex]3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 276.188px;\" colspan=\"2\">Table 1. Table of values for\u00a0[latex]f(x)=\\log_2{x}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: left;\">To create an inverse table, we switch the values in the [latex]x[\/latex] and [latex]y[\/latex] columns so that the inputs become the values of [latex]y[\/latex] and the outputs become the values of [latex]x[\/latex]. The table after switching the values of the [latex]x[\/latex] and [latex]y[\/latex] columns is the inverse table (Table 2). Since\u00a0[latex]f(x)=\\log_2{x}[\/latex] is a one-to-one function, its inverse will also be a one-to-one function.<\/p>\n<table class=\"aligncenter\" style=\"border-collapse: collapse; width: 50%;\">\n<tbody>\n<tr>\n<th style=\"width: 2.40964%; text-align: center;\">[latex]x[\/latex]<\/th>\n<th style=\"width: 2.40964%; text-align: center;\">[latex]y=f^{-1}(x)[\/latex]<\/th>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]-3[\/latex]<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]\\dfrac{1}{8}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]-2[\/latex]<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]\\dfrac{1}{4}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]-1[\/latex]<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]\\dfrac{1}{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]0[\/latex]<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]1[\/latex]<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]2[\/latex]<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]3[\/latex]<\/td>\n<td style=\"width: 2.40964%; text-align: center;\">[latex]8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 2.40964%;\" colspan=\"2\">Table 2. Inverse table<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Looking at the values in table 2, notice that as [latex]x[\/latex] increases by 1, the inverse function value increases by a multiple of 2. This means that the inverse of a logarithmic function is an exponential function. In this case, the inverse of\u00a0[latex]f(x)=\\log_2{x}[\/latex] is [latex]f^{-1}(x)=2^x[\/latex].<\/p>\n<p>Figure 1 shows the graphs of the function [latex]f(x)=\\log_2{x}[\/latex] (the blue curve) and its inverse (the green curve) graphed usings the values in Tables 1 and 2.\u00a0 Notice that the graph of the inverse function is a reflection of the graph of the original function with respect to the line [latex]y=x[\/latex] (the red line).<\/p>\n<div id=\"attachment_2941\" style=\"width: 418px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2941\" class=\"wp-image-2941\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/05\/22145601\/6-3-InverseFunctionNew-300x300.png\" alt=\"A green curve rising quickly is the reflection of a blue curve that increases gradually to the right, reflected across the line \u2060y = x.\" width=\"408\" height=\"408\" \/><\/p>\n<p id=\"caption-attachment-2941\" class=\"wp-caption-text\">Figure 1. The graphs of the function [latex]f(x)=\\log_2{x}[\/latex] and its inverse function.<\/p>\n<\/div>\n<p>Notice that the inverse is also a function because it passes the vertical line test. The inverse is a one-to-one mapping. Therefore, the inverse is a function.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>Use Desmos to graph [latex]f(x)=2\\log_2{(x-3)}+1[\/latex], then use reflection to graph its inverse function.<\/p>\n<h4>Solution<\/h4>\n<p>Copying and pasting the function into Desmos yields:<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3036 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T105304.306-300x300.png\" alt=\"A blue curve rising gradually to the right with a vertical asymptote at x=3.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T105304.306-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T105304.306-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T105304.306-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T105304.306-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T105304.306-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T105304.306-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T105304.306.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Now we draw in the line [latex]y=x[\/latex] and reflect every point on the graph of [latex]y=f(x)[\/latex] across the line [latex]y=x[\/latex] to get the inverse. Every point [latex](x,\\;y)[\/latex] is reflected to [latex](y,\\;x)[\/latex]. The vertical asymptote [latex]x=3[\/latex] reflects to a horizontal asymptote [latex]y=3[\/latex].<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3037 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T110817.876-300x300.png\" alt=\"A green curve rising gradually is the reflection of a blue curve that increases gradually to the right, reflected across the line \u2060y = x. A point (x,y) on the blue curve corresponds to the point (y,x) on the green curve.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T110817.876-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T110817.876-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T110817.876-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T110817.876-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T110817.876-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T110817.876-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T110817.876.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>Use Desmos to graph [latex]f(x)=\\log_2{(x+4)}+3[\/latex], then use reflection to graph its inverse function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm118\">Show Answer<\/span><\/p>\n<div id=\"qhjm118\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3038 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T111805.998-300x300.png\" alt=\"A green curve rising quickly is the reflection of a blue curve that increases gradually to the right, reflected across the line \u2060y = x. The two curves intersect at two points on the line of symmetry.\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T111805.998-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T111805.998-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T111805.998-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T111805.998-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T111805.998-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T111805.998-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/05\/desmos-graph-2022-06-27T111805.998.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>The Inverse Function Equation<\/h2>\n<p>We already know that to find the equation of an inverse function for a given function, we set [latex]y=f(x)[\/latex], switch [latex]x[\/latex] and [latex]y[\/latex], then solve for [latex]y[\/latex]. For example, given the logarithmic function\u00a0[latex]f(x)=\\log_2{x}[\/latex], we first set [latex]y=f(x)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]y=\\log_2{x}[\/latex]<\/p>\n<p>Then we switch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:<\/p>\n<p style=\"text-align: center;\">[latex]x=\\log_2y[\/latex]<\/p>\n<p>Now we solve for [latex]y[\/latex] by writing this logarithmic equation as an equivalent exponential equation. Based on the definition of logarithm, the base of the logarithm, 2, becomes the base of the exponential:<\/p>\n<p style=\"text-align: center;\">[latex]y=2^x[\/latex]<\/p>\n<p>Therefore, the inverse function of the logarithmic function\u00a0[latex]f(x)=\\log_2{x}[\/latex] is:<\/p>\n<p style=\"text-align: center;\">[latex]f^{-1}(x)=2^{x}[\/latex]<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Determine the inverse function of [latex]g(x)=\\log_5x[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>First set [latex]y=g(x)[\/latex]:\u00a0\u00a0[latex]y=\\log_5x[\/latex]<\/p>\n<p>Switch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\u00a0 [latex]x=\\log_5y[\/latex]<\/p>\n<p>Use the definition of logarithm to solve for [latex]y[\/latex]:\u00a0 [latex]y=5^x[\/latex]<\/p>\n<p>Write in function notation:\u00a0 [latex]g^{-1}(x)=5^x[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>Determine the inverse function of [latex]h(x)=\\log_7x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm291\">Show Answer<\/span><\/p>\n<div id=\"qhjm291\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]h^{-1}(x)=7^x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 3<\/h3>\n<p>Determine the inverse function of the logarithmic function\u00a0[latex]f(x)=\\log_4{(x+5)}[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>First set [latex]y=f(x)[\/latex]:\u00a0\u00a0[latex]y=\\log_4{(x+5)}[\/latex]<\/p>\n<p>Switch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\u00a0 [latex]x=\\log_4{(y+5)}[\/latex]<\/p>\n<p>Use the definition of logarithm to solve for [latex]y[\/latex]:\u00a0 [latex]y+5=4^x[\/latex]<\/p>\n<p>Solve for [latex]y[\/latex] by subtracting 5 from both sides:\u00a0\u00a0[latex]y=4^x-5[\/latex]<\/p>\n<p>Write in function notation:\u00a0[latex]f^{-1}(x)=4^x-5[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Determine the inverse function of the logarithmic function\u00a0[latex]g(x)=\\log_3{(x-6)}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm926\">Show Answer<\/span><\/p>\n<div id=\"qhjm926\" class=\"hidden-answer\" style=\"display: none\">[latex]g^{-1}(x)=3^x+6[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>For a logarithmic function that is in the form [latex]f(x)=a\\log_b{x-h}+k[\/latex], we must first isolate the logarithm, before using the definition of a logarithm to solve for [latex]y[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4<\/h3>\n<p>Determine the inverse function of the logarithmic function\u00a0[latex]f(x)=4\\log_3{(x+2)}[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>First set [latex]y=f(x)[\/latex]:[latex]y=4\\log_3{(x+2)}[\/latex]<\/p>\n<p>Switch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\u00a0 [latex]x=4\\log_3{(y+2)}[\/latex]<\/p>\n<p>Isolate the logarithm by dividing both sides by 4:\u00a0[latex]\\frac{1}{4}x=\\log_3{(y+2)}[\/latex]<\/p>\n<p>Use the definition of logarithm to solve for [latex]y[\/latex]:\u00a0 [latex]y+2=3^{\\frac{x}{4}}[\/latex]<\/p>\n<p>Solve for [latex]y[\/latex] by subtracting 2 from both sides:\u00a0[latex]y=3^{\\frac{x}{4}}-2[\/latex]<\/p>\n<p>Write in function notation:\u00a0 [latex]f^{-1}(x)=3^{\\frac{x}{4}}-2[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 4<\/h3>\n<p>Determine the inverse function of the logarithmic function\u00a0[latex]f(x)=5\\log_7{(x-3)}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm985\">Show Answer<\/span><\/p>\n<div id=\"qhjm985\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f^{-1}(x)=7^{\\frac{x}{5}}+3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 5<\/h3>\n<p>Determine the inverse function of the logarithmic function\u00a0[latex]f(x)=\\log_5{(x-3)}+2[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>First set [latex]y=f(x)[\/latex]:\u00a0[latex]y=\\log_5{(x-3)}+2[\/latex]<\/p>\n<p>Switch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\u00a0[latex]x=\\log_5{(y-3)}+2[\/latex]<\/p>\n<p>Isolate the logarithm by subtracting 2 from both sides:\u00a0\u00a0[latex]x-2=\\log_5{(y-3)}[\/latex]<\/p>\n<p>Use the definition of logarithm to solve for [latex]y[\/latex]:\u00a0 [latex]y-3=5^{x-2}[\/latex]<\/p>\n<p>Solve for [latex]y[\/latex] by adding 3 to both sides:\u00a0\u00a0[latex]y=5^{x-2}+3[\/latex]<\/p>\n<p>Write in function notation:\u00a0[latex]f^{-1}(x)=5^{x-2}+3[\/latex]<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 6<\/h3>\n<p>Determine the inverse function of the logarithmic function [latex]f(x)=\\dfrac{2}{5}\\log_3{(8x)}-4[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>First set [latex]y=f(x)[\/latex]:\u00a0 [latex]y=\\dfrac{2}{5}\\log_3{(8x)}-4[\/latex]<\/p>\n<p>Switch [latex]x[\/latex] and [latex]y[\/latex] to get the inverse:\u00a0 [latex]x=\\dfrac{2}{5}\\log_3{(8y)}-4[\/latex]<\/p>\n<p>Isolate the logarithm by adding 4 to both sides:\u00a0[latex]x+4=\\dfrac{2}{5}\\log_3{(8y)}[\/latex]<\/p>\n<p>Isolate the logarithm by multiplying both sides by [latex]\\dfrac{5}{2}[\/latex]:\u00a0\u00a0[latex]\\dfrac{5}{2}(x+4)=\\log_3{(8y)}[\/latex]<\/p>\n<p>Use the definition of logarithm to solve for [latex]y[\/latex]:\u00a0 [latex]8y=3^{\\frac{5}{2}(x+4)}[\/latex]<\/p>\n<p>Solve for [latex]y[\/latex] by dividing both sides by 8:\u00a0[latex]y=\\frac{1}{8}\\cdot3^{\\frac{5}{2}(x+4)}[\/latex]<\/p>\n<p>Write in function notation:\u00a0[latex]f^{-1}(x)=\\frac{1}{8}\\cdot3^{\\frac{5}{2}(x+4)}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 5<\/h3>\n<p>Determine the inverse function of the logarithmic function:<\/p>\n<ol>\n<li>[latex]f(x)=\\log_2{(x-9)}+1[\/latex]<\/li>\n<li>[latex]g(x)=\\dfrac{3}{4}\\log_5{x}-2[\/latex]<\/li>\n<li>[latex]h(x)=\\dfrac{1}{3}\\log_2{(x+5)}+8[\/latex]<\/li>\n<li>[latex]F(x)=6\\log_{10}{(x-3)}+6[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm417\">Show Answer<\/span><\/p>\n<div id=\"qhjm417\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]f^{-1}(x)=2^{x-1}+9[\/latex]<\/li>\n<li>[latex]g^{-1}=5^{\\frac{4}{3}(x+2)}[\/latex]<\/li>\n<li>[latex]h^{-1}=2^{3(x-8)}-5[\/latex]<\/li>\n<li>[latex]F^{-1}(x)=10^{\\frac{x-6}{6}}+3[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>Composition of Inverse Functions<\/h2>\n<p>It is important to remember that an inverse function &#8216;undoes&#8217; what was &#8216;done&#8217; by the original function. Logarithmic functions and inverse functions are inverses of one another, so if we apply one function then apply its inverse, we should get back to where we started.<\/p>\n<p>Consider the function [latex]f(x)=\\log_4x[\/latex] and its inverse function [latex]f^{-1}(x)=4^x[\/latex]. Now let&#8217;s perform a composition of functions:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(f\\circ f^{-1}\\right)(x)&=f\\left(f^{-1}(x)\\right)\\\\&=f\\left(4^x\\right)\\\\&=\\log_4{4^x}\\end{aligned}[\/latex]<\/p>\n<p>To simplify [latex]\\log_4{4^x}[\/latex], let&#8217;s set [latex]a=\\log_4{4^x}[\/latex]. Then using the definition of logarithms, we can write an equivalent exponential equation:<\/p>\n<p style=\"text-align: center;\">[latex]4^a=4^x[\/latex]<\/p>\n<p>Since the bases are identical, the exponents must be equal:<\/p>\n<p style=\"text-align: center;\">[latex]a=x[\/latex]<\/p>\n<p>Consequently,\u00a0[latex]\\log_4{4^x}=x[\/latex] and, therefore, [latex]\\left(f\\circ f^{-1}\\right)(x)=x[\/latex].<\/p>\n<p>If we perform the composition in the opposite direction we get:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(f^{-1}\\circ f\\right)(x)&=f^{-1}\\left(f(x)\\right)\\\\&=f^{-1}\\left(\\log_4x\\right)\\\\&=4^{\\log_4x}\\end{aligned}[\/latex]<\/p>\n<p>To simplify [latex]4^{\\log_4x}[\/latex], let&#8217;s set [latex]a=4^{\\log_4x}[\/latex]. Then using the definition of logarithms, we can write an equivalent exponential equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\log_4a=\\log_4x[\/latex]<\/p>\n<p>Since logarithms are one-to-one functions, [latex]a=x[\/latex].<\/p>\n<p>Consequently,\u00a0[latex]4^{\\log_4x}=x[\/latex] and, therefore, [latex]\\left(f^{-1}\\circ f\\right)(x)=x[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 7<\/h3>\n<p>Use composition of functions to show that [latex]f(x)=\\log_2{(x+1)}[\/latex] and [latex]g(x)=2^x-1[\/latex] are inverse functions.<\/p>\n<h4>Solution<\/h4>\n<p>It doesn&#8217;t matter the order in which we compose the functions.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(f\\circ g\\right)(x)&=f\\left(2^x-1\\right)\\\\&=\\log_2{(2^x-1+1)}\\\\&=\\log_2{2^x}\\end{aligned}[\/latex]<\/p>\n<p>To simplify [latex]\\log_2{2^x}[\/latex], let [latex]a=\\log_2{2^x}[\/latex], then using the definition of logarithms:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}2^a&=2^x\\\\a&=x\\end{aligned}[\/latex]<\/p>\n<p>Consequently,\u00a0[latex]\\left(f\\circ g\\right)(x)=x[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(g\\circ f\\right)(x)&=g\\left(\\log_2{(x+1)}\\right)\\\\&=2^{\\log_2{(x+1)}}-1\\end{aligned}[\/latex]<\/p>\n<p>To simplify [latex]2^{\\log_2{(x+1)}}[\/latex], let [latex]a=2^{\\log_2{(x+1)}}[\/latex], then using the definition of logarithms:<\/p>\n<p style=\"text-align: center;\">[latex]\\log_2{a}=\\log_2{(x+1)}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]a=x+1[\/latex]<\/p>\n<p>Hence,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\\left(g\\circ f\\right)(x)&=g\\left(\\log_2{(x+1)}\\right)\\\\&=2^{\\log_2{(x+1)}}-1\\\\&=(x+1)-1\\\\&=x\\end{aligned}[\/latex]<\/p>\n<p>Consequently,\u00a0[latex]\\left(g\\circ f\\right)(x)=x[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Therefore, [latex]f(x)=\\log_2{(x+1)}[\/latex] and [latex]g(x)=2^x-1[\/latex] are inverse functions.<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 6<\/h3>\n<p>Use composition of functions to show that [latex]f(x)=\\log_3{(x+4)}[\/latex] and [latex]g(x)=3^x-4[\/latex] are inverse functions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm562\">Show Answer<\/span><\/p>\n<div id=\"qhjm562\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(f\\circ g\\right)(x)=\\left(g\\circ f\\right)(x)=x[\/latex] so [latex]f(x)[\/latex] and [latex]g(x)[\/latex] are inverse functions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><script>\nwindow.embeddedChatbotConfig = {\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\ndomain: \"www.chatbase.co\"\n}\n<\/script><br \/>\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" defer=\"defer\">\n<\/script><\/p>\n<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\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-2393\">\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>The Inverse of a Logarithmic Function. <strong>Authored by<\/strong>: Leo Chang and Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>All graphs created using desmos graphing calculator. <strong>Authored by<\/strong>: Leo Chang and Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.desmos.com\/calculator\">http:\/\/www.desmos.com\/calculator<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>All examples and Try Its: hjm417; hjm985; hjm926; hjm291; hjm118; hjm562. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Composition of Inverse Functions. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Vally University. <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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":422608,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"The Inverse of a Logarithmic Function\",\"author\":\"Leo Chang and Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"All graphs created using desmos graphing calculator\",\"author\":\"Leo Chang and Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"www.desmos.com\/calculator\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"All examples and Try Its: hjm417; hjm985; hjm926; hjm291; hjm118; hjm562\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Composition of Inverse Functions\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Vally University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2393","chapter","type-chapter","status-publish","hentry"],"part":2310,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2393","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/users\/422608"}],"version-history":[{"count":45,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2393\/revisions"}],"predecessor-version":[{"id":4707,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2393\/revisions\/4707"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/parts\/2310"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapters\/2393\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/media?parent=2393"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2393"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/contributor?post=2393"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-json\/wp\/v2\/license?post=2393"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}