{"id":3518,"date":"2016-08-05T05:30:12","date_gmt":"2016-08-05T05:30:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=3518"},"modified":"2016-09-27T00:23:27","modified_gmt":"2016-09-27T00:23:27","slug":"introduction-to-exponential-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tallahassee-intermediatealgebra\/chapter\/introduction-to-exponential-functions\/","title":{"raw":"Logarithmic Functions","rendered":"Logarithmic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Composite and InverseFunctions\r\n<ul>\r\n \t<li>Define a composite function<\/li>\r\n \t<li>Define an inverse function<\/li>\r\n \t<li>Use compositions of functions to verify inverses algebraically<\/li>\r\n \t<li>Identify an inverse algebraically<\/li>\r\n \t<li>Identify the domain and range of inverse functions with tables<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Logarithmic Functions\r\n<ul>\r\n \t<li>Define a logarithmic function as the inverse\u00a0of an exponential function<\/li>\r\n \t<li>Convert between logarithmic and exponential forms<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Evaluate Logarithms\r\n<ul>\r\n \t<li>Mentally evaluate logarithms<\/li>\r\n \t<li>Define natural logarithm, evaluate natural logarithms with a calculator<\/li>\r\n \t<li>Define common logarithm, evaluate common logarithms mentally and with a calculator<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Graphs of Logarithmic Functions\r\n<ul>\r\n \t<li>Identify the domain of a logarithmic function.<\/li>\r\n \t<li>Graph logarithmic functions.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<figure id=\"CNX_Precalc_Figure_04_03_001\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"alignleft\" width=\"269\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051922\/CNX_Precalc_Figure_04_03_0012.jpg\" alt=\"Photo of the aftermath of the earthquake in Japan with a focus on the Japanese flag.\" width=\"269\" height=\"180\" \/> <strong>Figure 1.\u00a0<\/strong>Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce)[\/caption]\r\n\r\n<\/figure>\r\n<p id=\"fs-id1165137557013\">In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes.[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary\" target=\"_blank\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary<\/a>. Accessed 3\/4\/2013.[\/footnote] One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings,[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary\" target=\"_blank\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary<\/a>. Accessed 3\/4\/2013.[\/footnote]\u00a0like those shown in the picture above. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/\" target=\"_blank\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/<\/a>. Accessed 3\/4\/2013.[\/footnote]\u00a0whereas the Japanese earthquake registered a 9.0.[footnote]<a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details\" target=\"_blank\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details<\/a>. Accessed 3\/4\/2013.[\/footnote]<\/p>\r\n<p id=\"fs-id1165137760714\">The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is [latex]{10}^{8 - 4}={10}^{4}=10,000[\/latex] times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.<\/p>\r\nOur first topic will be about inverse functions, logarithmic functions are the inverse of an exponential functions, and sometimes understanding this helps us make sense of what a logarithm is.\r\n<h2>Composite and Inverse Functions<\/h2>\r\n<p id=\"fs-id1165134094620\">Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25200743\/CNX_Precalc_Figure_01_04_0062.jpg\" alt=\"Explanation of C(T(5)), which is the cost for the temperature and T(5) is the temperature on day 5.\" width=\"487\" height=\"140\" data-media-type=\"image\/jpg\" \/> <b>Figure 1<\/b>[\/caption]\r\n<p id=\"fs-id1165134038788\">Using descriptive variables, we can notate these two functions. The function [latex]C\\left(T\\right)[\/latex] gives the cost [latex]C[\/latex] of heating a house for a given average daily temperature in [latex]T[\/latex] degrees Celsius. The function [latex]T\\left(d\\right)[\/latex] gives the average daily temperature on day [latex]d[\/latex] of the year. For any given day, [latex]\\text{Cost}=C\\left(T\\left(d\\right)\\right)[\/latex] means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature [latex]T\\left(d\\right)[\/latex]. For example, we could evaluate [latex]T\\left(5\\right)[\/latex] to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the <strong>cost function<\/strong> at that temperature. We would write [latex]C\\left(T\\left(5\\right)\\right)[\/latex].\u00a0By combining these two relationships into one function, we have performed function composition.<\/p>\r\nWe read the left-hand side as [latex]\"f[\/latex] composed with [latex]g[\/latex] at [latex]x,\"[\/latex] and the right-hand side as [latex]\"f[\/latex] of [latex]g[\/latex] of [latex]x.\"[\/latex] The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol [latex]\\circ [\/latex] is called the composition operator.\r\n\r\nIt is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside.\r\n\r\n&nbsp;\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25200744\/CNX_Precalc_Figure_01_04_0012.jpg\" alt=\"Explanation of the composite function. g(x), the output of g is the input of f. X is the input of g.\" width=\"487\" height=\"171\" data-media-type=\"image\/jpg\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUsing the functions provided, find [latex]f\\left(g\\left(x\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)[\/latex].\r\n\r\n[latex]f\\left(x\\right)=2x+1[\/latex]\r\n\r\n[latex]g\\left(x\\right)=3-x[\/latex]\r\n[reveal-answer q=\"337338\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"337338\"]\r\n\r\n[latex]f\\left(x\\right)=2x+1[\/latex]\r\n\r\n[latex]g\\left(x\\right)=3-x[\/latex]\r\n\r\nLet\u2019s begin by substituting [latex]g\\left(x\\right)[\/latex] into [latex]f\\left(x\\right)[\/latex].[latex]\\begin{array}f\\left(g\\left(x\\right)\\right)=2\\left(3-x\\right)+1\\hfill \\\\ \\text{ }=6 - 2x+1\\hfill \\\\ \\text{ }=7 - 2x\\hfill \\end{array}[\/latex]\r\nNow we can substitute [latex]f\\left(x\\right)[\/latex] into [latex]g\\left(x\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}g\\left(f\\left(x\\right)\\right)=3-\\left(2x+1\\right)\\hfill \\\\ \\text{ }=3 - 2x - 1\\hfill \\\\ \\text{ }=-2x+2\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video you will see another example of how to find the composition of two functions.\r\n\r\nhttps:\/\/youtu.be\/r_LssVS4NHk\r\n<h2>Inverse Functions<\/h2>\r\n<p id=\"fs-id1165137827441\">An <strong>inverse function<\/strong>\u00a0is a function for which the input of the original function becomes the output of the inverse function. This naturally leads to the output of the original function becoming the input of the inverse function. The reason we want to introduce inverse functions is because exponential and logarithmic functions are inverses of each other, and understanding this quality helps to make understanding logarithmic functions easier. And the reason we introduced composite functions is because you can verify, algebraically, whether two functions are inverses of each other by using a composition.<\/p>\r\n<p id=\"fs-id1165135528385\">Given a function [latex]f\\left(x\\right)[\/latex], we represent its inverse as [latex]{f}^{-1}\\left(x\\right)[\/latex], read as [latex]\"f[\/latex] inverse of [latex]x.\\text{\"}[\/latex] The raised [latex]-1[\/latex] is part of the notation. It is not an exponent; it does not imply a power of [latex]-1[\/latex] . In other words, [latex]{f}^{-1}\\left(x\\right)[\/latex] does <em data-effect=\"italics\">not<\/em> mean [latex]\\frac{1}{f\\left(x\\right)}[\/latex] because [latex]\\frac{1}{f\\left(x\\right)}[\/latex] is the reciprocal of [latex]f[\/latex] and not the inverse.<\/p>\r\n<p id=\"fs-id1165137724926\">Just as zero does not have a <strong>reciprocal<\/strong>, some functions do not have inverses.<\/p>\r\n\r\n<div id=\"fs-id1165137933105\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\r\n<h3 class=\"title\" data-type=\"title\">Inverse Function<\/h3>\r\n<p id=\"fs-id1165137473076\">For any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex].<\/p>\r\n<p id=\"fs-id1165137444821\">The notation [latex]{f}^{-1}[\/latex] is read [latex]\\text{\"}f[\/latex] inverse.\" Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x.\"[\/latex]\r\nKeep in mind that<\/p>\r\n\r\n<div id=\"fs-id1165137581324\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{f}^{-1}\\left(x\\right)\\ne \\frac{1}{f\\left(x\\right)}[\/latex]<\/div>\r\n<p id=\"fs-id1165135194095\">and not all functions have inverses.<\/p>\r\n\r\n<\/div>\r\nIn our first example we will identify an inverse function from ordered pairs.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf for a particular one-to-one function [latex]f\\left(2\\right)=4[\/latex] and [latex]f\\left(5\\right)=12[\/latex], what are the corresponding input and output values for the inverse function?\r\n[reveal-answer q=\"664782\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"664782\"]\r\n<p id=\"fs-id1165137737081\">The inverse function reverses the input and output quantities, so if<\/p>\r\n\r\n<div id=\"fs-id1165137462459\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}f\\left(2\\right)=4,\\text{ then }{f}^{-1}\\left(4\\right)=2;\\\\ f\\left(5\\right)=12,{\\text{ then f}}^{-1}\\left(12\\right)=5.\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137659464\">Alternatively, if we want to name the inverse function [latex]g[\/latex], then [latex]g\\left(4\\right)=2[\/latex] and [latex]g\\left(12\\right)=5[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Analysis of the Solution<\/h3>\r\n<div id=\"Example_01_07_01\" class=\"example\" data-type=\"example\">\r\n<div id=\"fs-id1165137656641\" class=\"exercise\" data-type=\"exercise\">\r\n<div id=\"fs-id1165135245520\" class=\"commentary\" data-type=\"commentary\">\r\n<p id=\"fs-id1165135508518\">Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed.<\/p>\r\n\r\n<table style=\"width: 30%;\" summary=\"For (x,f(x)) we have the values (2, 4) and (5, 12); for (x, g(x)), we have the values (4, 2) and (12, 5).\">\r\n<thead>\r\n<tr>\r\n<th data-align=\"center\">[latex]\\left(x,f\\left(x\\right)\\right)[\/latex]<\/th>\r\n<th data-align=\"center\">[latex]\\left(x,g\\left(x\\right)\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td data-align=\"center\">[latex]\\left(2,4\\right)[\/latex]<\/td>\r\n<td data-align=\"center\">[latex]\\left(4,2\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td data-align=\"center\">[latex]\\left(5,12\\right)[\/latex]<\/td>\r\n<td data-align=\"center\">[latex]\\left(12,5\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nIn the following video we show an example of finding corresponding input and output values given two ordered pairs from functions that are inverses.\r\n\r\nhttps:\/\/youtu.be\/IR_1L1mnpvw\r\n<div id=\"fs-id1165134357354\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\r\n<h3 id=\"fs-id1165135434077\">How To: Given two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex], test whether the functions are inverses of each other.<\/h3>\r\n<ol id=\"fs-id1165137452358\" data-number-style=\"arabic\">\r\n \t<li>Substitute\u00a0[latex]g(x)[\/latex]\u00a0into [latex]f(x)[\/latex]. The result must be x. [latex]f\\left(g(x)\\right)=x[\/latex]<\/li>\r\n \t<li>Substitute\u00a0[latex]f(x)[\/latex]\u00a0into [latex]g(x)[\/latex]. The result must be x. [latex]g\\left(f(x)\\right)=x[\/latex]<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">If\u00a0[latex]f(x)[\/latex] and\u00a0\u00a0[latex]g(x)[\/latex] are inverses, then\u00a0\u00a0[latex]f(x)=g^{-1}(x)[\/latex] and\u00a0[latex]g(x)=f^{-1}(x)[\/latex]<\/p>\r\n\r\n<\/div>\r\nIn our next example we will test inverse relationships algebraically.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf [latex]f\\left(x\\right)=x^2-3[\/latex], for [latex]x\\ge0[\/latex] and [latex]g\\left(x\\right)=\\sqrt{x+3}[\/latex], is g the inverse of f? \u00a0[latex]g={f}^{-1}?[\/latex]\r\n[reveal-answer q=\"598434\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"598434\"]\r\n\r\nSubstitute [latex]g(x)=\\sqrt{x+3}[\/latex] into [latex]f(x)[\/latex], this means the new variable in\u00a0[latex]f(x)[\/latex] is [latex]\\sqrt{x+3}[\/latex] so you will substitute that expression where you see x. \u00a0Using parentheses helps keep track of things.\r\n\r\n[latex]\\begin{array}{c}f\\left(\\sqrt{x+3}\\right)={(\\sqrt{x+3})}^2-3\\hfill\\\\=x+3-3\\\\=x\\hfill \\end{array}[\/latex]\r\n\r\nOur result implies that [latex]g(x)[\/latex] is indeed the inverse of\u00a0[latex]f(x)[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]g={f}^{-1}[\/latex],\u00a0for [latex]x\\ge0[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we use algebra to determine if two functions are inverses.\r\n\r\nhttps:\/\/youtu.be\/vObCvTOatfQ\r\n\r\nWe will show one more example of how to verify whether you have an inverse algebraically.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf [latex]f\\left(x\\right)=\\frac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\frac{1}{x}-2[\/latex], is g the inverse of f? \u00a0[latex]g={f}^{-1}?[\/latex]\r\n[reveal-answer q=\"56557\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"56557\"]\r\n\r\nSubstitute [latex]g(x)=\\frac{1}{x}-2[\/latex] into [latex]f(x)[\/latex], this means the new variable in\u00a0[latex]f(x)[\/latex] is [latex]\\frac{1}{x}-2[\/latex] so you will substitute that expression where you see x. \u00a0Using parentheses helps keep track of things.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c} f\\left(\\frac{1}{x}-2\\right)=\\frac{1}{\\left(\\frac{1}{x}-2\\right)+2}\\hfill\\\\=\\frac{1}{\\frac{1}{x}}\\hfill\\\\={ x }\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]g={f}^{-1}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe will show one more example of how to use algebra to determine whether two functions are inverses of each other.\r\n\r\nhttps:\/\/youtu.be\/hzehBtNmw08\r\n<h2>\u00a0Domain and Range of a Function and It's Inverse<\/h2>\r\nThe outputs of the function [latex]f[\/latex] are the inputs to [latex]{f}^{-1}[\/latex], so the range of [latex]f[\/latex] is also the domain of [latex]{f}^{-1}[\/latex]. Likewise, because the inputs to [latex]f[\/latex] are the outputs of [latex]{f}^{-1}[\/latex], the domain of [latex]f[\/latex] is the range of [latex]{f}^{-1}[\/latex]. We can visualize the situation.\r\n<div class=\"mceTemp\"><\/div>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25200955\/CNX_Precalc_Figure_01_07_0032.jpg\" alt=\"Domain and range of a function and its inverse.\" width=\"487\" height=\"143\" data-media-type=\"image\/jpg\" \/>\r\n\r\nDomain and range of a function and its inverse\r\n\r\nIn many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function [latex]f\\left(x\\right)={x}^{2}[\/latex] with its range limited to [latex]\\left[0,\\infty \\right)[\/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).\r\n<div class=\"textbox\">\r\n<h3 class=\"title\" data-type=\"title\">Domain and Range of Inverse Functions<\/h3>\r\n<p id=\"fs-id1165135319550\">The range of a function [latex]f\\left(x\\right)[\/latex] is the domain of the inverse function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165137673886\">The domain of [latex]f\\left(x\\right)[\/latex] is the range of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\nIn our last example we will define the domain and range of a function's inverse using a table of values, and evaluate the inverse at a specific value.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<p id=\"fs-id1165135435474\">A function [latex]f\\left(t\\right)[\/latex] is given\u00a0below, showing distance in miles that a car has traveled in [latex]t[\/latex] minutes.<\/p>\r\n\r\n<ol>\r\n \t<li>Define the domain and range of the function and it's inverse.<\/li>\r\n \t<li>Find and interpret [latex]{f}^{-1}\\left(70\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<table style=\"width: 30%;\" summary=\"Two rows and five columns. The first row is labeled\">\r\n<tbody>\r\n<tr>\r\n<td data-align=\"left\"><strong>[latex]t\\text{ (minutes)}[\/latex]<\/strong><\/td>\r\n<td data-align=\"left\">30<\/td>\r\n<td data-align=\"left\">50<\/td>\r\n<td data-align=\"left\">70<\/td>\r\n<td data-align=\"left\">90<\/td>\r\n<\/tr>\r\n<tr>\r\n<td data-align=\"left\"><strong>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex] <\/strong><\/td>\r\n<td data-align=\"left\">20<\/td>\r\n<td data-align=\"left\">40<\/td>\r\n<td data-align=\"left\">60<\/td>\r\n<td data-align=\"left\">70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"713219\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"713219\"]\r\n\r\n1.<span style=\"text-decoration: underline;\">Domain and Range of the Original Function<\/span>\r\n\r\nThe domain of this tabular function, [latex]f\\left(t\\right)[\/latex] , is all the input values, t in minutes: {30, 50, 70, 90}\r\n\r\nThe range of this tabular function,[latex]f\\left(t\\right)[\/latex], \u00a0is all the output values[latex]f\\left(t\\right)[\/latex] in miles: {20, 40, 60, 70}\r\n\r\n<span style=\"text-decoration: underline;\">Domain and Range of the Inverse Function<\/span>\r\n<p id=\"fs-id1165137640334\">The domain for the inverse will be the outputs from the original, so the domain of \u00a0[latex]{f}^{-1}(x)[\/latex] is the output values from\u00a0[latex]f\\left(t\\right)[\/latex]:\u00a0{20, 40, 60, 70}<\/p>\r\nThe range for the inverse will be the inputs from the original:\u00a0{30, 50, 70, 90}\r\n\r\nThis translates to putting in a number of miles and getting out how long it took to drive that far in minutes.\r\n\r\n2.\u00a0So in the expression [latex]{f}^{-1}\\left(70\\right)[\/latex], 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function [latex]f[\/latex], 90 minutes, so [latex]{f}^{-1}\\left(70\\right)=90[\/latex]. The interpretation of this is that, to drive 70 miles, it took 90 minutes.[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Define Logarithmic Functions<\/h2>\r\n<p id=\"fs-id1165135192781\">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 were 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>\r\nWe have not yet learned a method for solving exponential equations. 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[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051924\/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\" data-media-type=\"image\/jpg\" \/> <b>Figure 2<\/b>[\/caption]\r\n<p id=\"fs-id1165137662989\">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 data-effect=\"bold\">logarithm<\/strong> of a number is the exponent by which we must raise <em>b<\/em>\u00a0to get that number.<\/p>\r\n<p id=\"fs-id1165137404844\">We 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.\"<\/p>\r\n<p id=\"fs-id1165137597501\">We can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\r\n\r\n<div id=\"eip-604\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b&gt;0,b\\ne 1[\/latex]<\/div>\r\n<p id=\"fs-id1165137678993\">Note that the base <em>b<\/em>\u00a0is always positive.<span id=\"fs-id1165137696233\" data-type=\"media\" data-alt=\"\" data-display=\"block\">\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051926\/CNX_Precalc_Figure_04_03_0042.jpg\" alt=\"Think b to the y equals x.\" width=\"487\" height=\"83\" data-media-type=\"image\/jpg\" \/><\/span><\/p>\r\n<p id=\"fs-id1165137400957\">Because 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>\r\n<p id=\"fs-id1165137827516\">We can illustrate the notation of logarithms as follows:<span id=\"fs-id1165137771679\" data-type=\"media\" data-alt=\"\" data-display=\"block\">\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051928\/CNX_Precalc_Figure_04_03_0032.jpg\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"487\" height=\"101\" data-media-type=\"image\/jpg\" \/><\/span><\/p>\r\n<p id=\"fs-id1165137575165\">Notice that, 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>\r\n\r\n<div id=\"fs-id1165137472937\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\r\n<h3 class=\"title\" data-type=\"title\">Definition of the Logarithmic Function<\/h3>\r\n<p id=\"fs-id1165137704597\">A <strong>logarithm<\/strong> base <em>b<\/em>\u00a0of a positive number <em>x<\/em>\u00a0satisfies the following definition.<\/p>\r\n<p id=\"fs-id1165137584967\">For [latex]x&gt;0,b&gt;0,b\\ne 1[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165137433829\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equivalent to }{b}^{y}=x[\/latex]<\/div>\r\n<p id=\"fs-id1165137893373\">where,<\/p>\r\n\r\n<ul id=\"fs-id1165135530561\">\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<\/ul>\r\n<p id=\"fs-id1165137547773\">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>\r\n\r\n<ul id=\"fs-id1165137643167\">\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\nIn our first example we will convert logarithmic equations into exponential equations.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<p id=\"fs-id1165137580570\">Write the following logarithmic equations in exponential form.<\/p>\r\n\r\n<ol id=\"fs-id1165137705346\" data-number-style=\"lower-alpha\">\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]\r\n[reveal-answer q=\"161275\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"161275\"]\r\n<p id=\"fs-id1165137408172\">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>\r\n\r\n<ol id=\"fs-id1165137705659\" data-number-style=\"lower-alpha\">\r\n \t<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]\r\n<p id=\"fs-id1165137602796\">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 equivalent to [latex]{6}^{\\frac{1}{2}}=\\sqrt{6}[\/latex].<\/p>\r\n<\/li>\r\n \t<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]\r\n<p id=\"fs-id1165137698078\">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 equivalent to [latex]{3}^{2}=9[\/latex].<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn the following video we present more examples of rewriting logarithmic equations as exponential equations.\r\n\r\nhttps:\/\/youtu.be\/q9_s0wqhIXU\r\n<div id=\"fs-id1165137874700\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\r\n<h3 id=\"fs-id1165137806301\">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 id=\"fs-id1165137641669\" data-number-style=\"arabic\">\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 key-takeaways\">\r\n<h3>Think About It<\/h3>\r\n<p id=\"eip-id1549475\">Can we take the logarithm of a negative number? Re-read the definition of a logarithm and formulate an answer. \u00a0Think about the behavior of exponents. \u00a0You can use the textbox below to formulate your ideas before you look at an answer.<\/p>\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"162494\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"162494\"]\r\n<p id=\"fs-id1165137653864\">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.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Convert from exponential to logarithmic form<\/h2>\r\nTo convert from exponents to logarithms, 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<\/h3>\r\n<p id=\"fs-id1165137804412\">Write the following exponential equations in logarithmic form.<\/p>\r\n\r\n<ol id=\"fs-id1165135192287\" data-number-style=\"lower-alpha\">\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]\r\n[reveal-answer q=\"516026\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"516026\"]\r\n<p id=\"fs-id1165137474116\">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>\r\n\r\n<ol id=\"fs-id1165137573458\" data-number-style=\"lower-alpha\">\r\n \t<li>[latex]{2}^{3}=8[\/latex]\r\n<p id=\"fs-id1165137466396\">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 equivalent to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/p>\r\n<\/li>\r\n \t<li>[latex]{5}^{2}=25[\/latex]\r\n<p id=\"fs-id1165135193035\">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 equivalent to [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex].<\/p>\r\n<\/li>\r\n \t<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]\r\n<p id=\"fs-id1165135187822\">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 equivalent to [latex]{\\text{log}}_{10}\\left(\\frac{1}{10,000}\\right)=-4[\/latex].<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn our last video we show more examples of writing logarithmic equations as exponential equations.\r\n\r\nhttps:\/\/youtu.be\/9_GPPUWEJQQ\r\n<h2>Evaluate Logarithms<\/h2>\r\n<section id=\"fs-id1165137405741\" data-depth=\"1\">\r\n<p id=\"fs-id1165137422589\">Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider [latex]{\\mathrm{log}}_{2}8[\/latex]. We ask, \"To what exponent must 2\u00a0be raised in order to get 8?\" Because we already know [latex]{2}^{3}=8[\/latex], it follows that [latex]{\\mathrm{log}}_{2}8=3[\/latex].<\/p>\r\n<p id=\"fs-id1165137733822\">Now consider solving [latex]{\\mathrm{log}}_{7}49[\/latex] and [latex]{\\mathrm{log}}_{3}27[\/latex] mentally.<\/p>\r\n\r\n<ul id=\"fs-id1165137937690\">\r\n \t<li>We ask, \"To what exponent must 7 be raised in order to get 49?\" We know [latex]{7}^{2}=49[\/latex]. Therefore, [latex]{\\mathrm{log}}_{7}49=2[\/latex]<\/li>\r\n \t<li>We ask, \"To what exponent must 3 be raised in order to get 27?\" We know [latex]{3}^{3}=27[\/latex]. Therefore, [latex]{\\mathrm{log}}_{3}27=3[\/latex]<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137456358\">Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let\u2019s evaluate [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\frac{4}{9}[\/latex] mentally.<\/p>\r\n\r\n<ul id=\"fs-id1165137584208\">\r\n \t<li>We ask, \"To what exponent must [latex]\\frac{2}{3}[\/latex] be raised in order to get [latex]\\frac{4}{9}[\/latex]? \" We know [latex]{2}^{2}=4[\/latex] and [latex]{3}^{2}=9[\/latex], so [latex]{\\left(\\frac{2}{3}\\right)}^{2}=\\frac{4}{9}[\/latex]. Therefore, [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\left(\\frac{4}{9}\\right)=2[\/latex].<\/li>\r\n<\/ul>\r\nIn our first example we will evaluate logarithms mentally.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]y={\\mathrm{log}}_{4}\\left(64\\right)[\/latex] without using a calculator.\r\n[reveal-answer q=\"161686\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"161686\"]\r\n<p id=\"fs-id1165137611276\">First we rewrite the logarithm in exponential form: [latex]{4}^{y}=64[\/latex]. Next, we ask, \"To what exponent must 4 be raised in order to get 64?\"<\/p>\r\nWe know\r\n<div id=\"eip-id1165134583995\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{4}^{3}=64[\/latex]<\/div>\r\n<p id=\"fs-id1165137619013\">Therefore,<\/p>\r\n\r\n<div id=\"eip-id1165135606935\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\mathrm{log}{}_{4}\\left(64\\right)=3[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our first video we will show more examples of evaluating logarithms mentally, this helps you get familiar with what a logarithm represents.\r\n\r\nhttps:\/\/youtu.be\/dxj5J9OpWGA\r\n\r\nIn our next example we will evaluate the logarithm of a reciprocal.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nEvaluate [latex]y={\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)[\/latex] without using a calculator.\r\n[reveal-answer q=\"534439\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"534439\"]\r\n<p id=\"fs-id1165137638179\">First we rewrite the logarithm in exponential form: [latex]{3}^{y}=\\frac{1}{27}[\/latex]. Next, we ask, \"To what exponent must 3 be raised in order to get [latex]\\frac{1}{27}[\/latex]\"?<\/p>\r\n<p id=\"fs-id1165137552085\">We know [latex]{3}^{3}=27[\/latex], but what must we do to get the reciprocal, [latex]\\frac{1}{27}[\/latex]? Recall from working with exponents that [latex]{b}^{-a}=\\frac{1}{{b}^{a}}[\/latex]. We use this information to write<\/p>\r\n\r\n<div id=\"eip-id1165137550550\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{3}^{-3}=\\frac{1}{{3}^{3}}\\hfill \\\\ =\\frac{1}{27}\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137585807\">Therefore, [latex]{\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)=-3[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3 id=\"fs-id1165137453770\">How To: Given a logarithm of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], evaluate it mentally.<\/h3>\r\n<ol id=\"fs-id1165134079724\" data-number-style=\"arabic\">\r\n \t<li>Rewrite the argument <em>x<\/em>\u00a0as a power of <em>b<\/em>: [latex]{b}^{y}=x[\/latex].<\/li>\r\n \t<li>Use previous knowledge of powers of <em>b<\/em>\u00a0identify <em>y<\/em>\u00a0by asking, \"To what exponent should <em>b<\/em>\u00a0be raised in order to get <em>x<\/em>?\"<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1165137661970\"><\/p>\r\n\r\n<\/div>\r\n<h2>\u00a0Natural logarithms<\/h2>\r\n<span style=\"line-height: 1.5;\">The most frequently used base for logarithms is <\/span><em style=\"line-height: 1.5;\">e<\/em><span style=\"line-height: 1.5;\">. Base <\/span><em style=\"line-height: 1.5;\">e<\/em><span style=\"line-height: 1.5;\">\u00a0logarithms are important in calculus and some scientific applications; they are called <\/span><strong style=\"line-height: 1.5;\" data-effect=\"bold\">natural logarithms<\/strong><span style=\"line-height: 1.5;\">. The base <\/span><em style=\"line-height: 1.5;\">e<\/em><span style=\"line-height: 1.5;\">\u00a0logarithm, [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex], has its own notation, [latex]\\mathrm{ln}\\left(x\\right)[\/latex].<\/span>\r\n<p id=\"fs-id1165137473872\">Most values of [latex]\\mathrm{ln}\\left(x\\right)[\/latex] can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, [latex]\\mathrm{ln}1=0[\/latex]. For other natural logarithms, we can use the [latex]\\mathrm{ln}[\/latex] key that can be found on most scientific calculators. We can also find the natural logarithm of any power of <em>e<\/em>\u00a0using the inverse property of logarithms.<\/p>\r\n\r\n<div id=\"fs-id1165137452317\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\r\n<h3 class=\"title\" data-type=\"title\">A General Note: Definition of the Natural Logarithm<\/h3>\r\n<p id=\"fs-id1165137579241\">A <strong>natural logarithm<\/strong> is a logarithm with base <em>e<\/em>. We write [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] simply as [latex]\\mathrm{ln}\\left(x\\right)[\/latex]. The natural logarithm of a positive number <em>x<\/em>\u00a0satisfies the following definition.<\/p>\r\n<p id=\"fs-id1165135613642\">For [latex]x&gt;0[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1165137580230\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]y=\\mathrm{ln}\\left(x\\right)\\text{ is equivalent to }{e}^{y}=x[\/latex]<\/div>\r\n<p id=\"fs-id1165137658264\">We read [latex]\\mathrm{ln}\\left(x\\right)[\/latex] as, \"the logarithm with base <em>e<\/em>\u00a0of <em>x<\/em>\" or \"the natural logarithm of <em>x<\/em>.\"<\/p>\r\n<p id=\"fs-id1165137566720\">The logarithm <em>y<\/em>\u00a0is the exponent to which <em>e<\/em>\u00a0must be raised to get <em>x<\/em>.<\/p>\r\n<p id=\"fs-id1165137705251\">Since the functions [latex]y=e{}^{x}[\/latex] and [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] are inverse functions, [latex]\\mathrm{ln}\\left({e}^{x}\\right)=x[\/latex] for all <em>x<\/em>\u00a0and [latex]e{}^{\\mathrm{ln}\\left(x\\right)}=x[\/latex] for <em>x\u00a0<\/em>&gt; 0.<\/p>\r\n\r\n<\/div>\r\nIn the next\u00a0example, we will evaluate a natural logarithm using a calculator.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nEvaluate [latex]y=\\mathrm{ln}\\left(500\\right)[\/latex] to four decimal places using a calculator.\r\n[reveal-answer q=\"957920\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"957920\"]\r\n<ul id=\"fs-id1165137563770\">\r\n \t<li>Press <strong data-effect=\"bold\">[LN]<\/strong>.<\/li>\r\n \t<li>Enter 500, followed by <strong data-effect=\"bold\">[ ) ]<\/strong>.<\/li>\r\n \t<li>Press <strong data-effect=\"bold\">[ENTER]<\/strong>.<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137645024\">Rounding to four decimal places, [latex]\\mathrm{ln}\\left(500\\right)\\approx 6.2146[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next video, we show more examples of how to evaluate natural logarithms using a calculator.\r\n\r\nhttps:\/\/youtu.be\/Rpounu3epSc\r\n<h2>Common logarithms<\/h2>\r\nSometimes we may see a logarithm written without a base. In this case, we assume that the base is 10. In other words, the expression [latex]{\\mathrm{log}}_{}[\/latex] means [latex]{\\mathrm{log}}_{10}[\/latex] \u00a0We call a base-10 logarithm a <strong data-effect=\"bold\">common logarithm<\/strong>. Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars and the pH of acids and bases also use common logarithms.\r\n<div class=\"textbox\">\r\n<h3>Definition of Common Logarithm: Log is an exponent<\/h3>\r\nA common logarithm is a logarithm with base 10. \u00a0We write\u00a0[latex]{\\mathrm{log}}_{10}(x)[\/latex] \u00a0simpliy as\u00a0[latex]{\\mathrm{log}}_{}(x)[\/latex]. \u00a0The common logarithm of a positive number, x, satisfies the following definition:\r\n\r\nFor [latex]x\\gt0[\/latex]\r\n<p style=\"text-align: center;\">[latex]y={\\mathrm{log}}_{}(x)[\/latex] is equivalent to [latex]10^y=x[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We read [latex]{\\mathrm{log}}_{}(x)[\/latex] as \" the logarithm with base 10 of x\" or \"log base 10 of x\".<\/p>\r\n<p style=\"text-align: left;\">The logarithm y is the exponent to which 10 must be raised to get x.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nEvaluate [latex]{\\mathrm{log}}_{}(1000)[\/latex] without using a calculator.\r\n[reveal-answer q=\"80362\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"80362\"]We know 10^3=1000, therefore\r\n\r\n[latex]{\\mathrm{log}}_{}(1000)=3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nEvaluate [latex]y={\\mathrm{log}}_{}(321)[\/latex] to four decimal places using a calculator.\r\n[reveal-answer q=\"782139\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"782139\"]\r\n<ul id=\"fs-id1165137786486\">\r\n \t<li>Press <strong data-effect=\"bold\">[LOG]<\/strong>.<\/li>\r\n \t<li>Enter 321<em data-effect=\"italics\">,<\/em> followed by <strong data-effect=\"bold\">[ ) ]<\/strong>.<\/li>\r\n \t<li>Press <strong data-effect=\"bold\">[ENTER]<\/strong>.<\/li>\r\n<\/ul>\r\nRounding to four decimal places,\u00a0[latex]{\\mathrm{log}}_{}(321)\\approx2.5065[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our last example we will use a logarithm to find the difference in magnitude of two different earthquakes.\r\n\r\n<\/section>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. The equation [latex]10^x=500[\/latex] represents this situation, where x is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?\r\n[reveal-answer q=\"735383\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"735383\"]We begin by rewriting the exponential equation in logarithmic form.\r\n<p style=\"text-align: center;\">[latex]10^x=500[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{}(500)=x[\/latex]<\/p>\r\n<p id=\"fs-id1165137419444\">Next we evaluate the logarithm using a calculator:<\/p>\r\n\r\n<ul id=\"fs-id1165137736356\">\r\n \t<li>Press <strong data-effect=\"bold\">[LOG]<\/strong>.<\/li>\r\n \t<li>Enter<span style=\"font-size: 14px; line-height: normal; white-space: nowrap;\">\u00a0500\u00a0<\/span>followed by <strong data-effect=\"bold\">[ ) ]<\/strong>.<\/li>\r\n \t<li>Press <strong data-effect=\"bold\">[ENTER]<\/strong>.<\/li>\r\n \t<li>To the nearest thousandth,\u00a0[latex]{\\mathrm{log}}_{}(500)\\approx2.699[\/latex]<span id=\"MathJax-Element-202-Frame\" class=\"MathJax\" style=\"box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: 0px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; color: #333333; font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-variant: normal; orphans: auto; widows: 1; -webkit-text-stroke-width: 0px; position: relative; background-color: #ededed;\" tabindex=\"0\" data-mathml=\"&lt;math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot; display=&quot;inline&quot;&gt;&lt;semantics&gt;&lt;mrow&gt;&lt;mrow&gt;&lt;mtext&gt;&amp;#x2009;&lt;\/mtext&gt;&lt;mi&gt;log&lt;\/mi&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mn&gt;500&lt;\/mn&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;mo&gt;&amp;#x2248;&lt;\/mo&gt;&lt;mn&gt;2.699.&lt;\/mn&gt;&lt;\/mrow&gt;&lt;\/mrow&gt;&lt;annotation-xml encoding=&quot;MathML-Content&quot;&gt;&lt;mrow&gt;&lt;mtext&gt;\u2009&lt;\/mtext&gt;&lt;mi&gt;log&lt;\/mi&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mn&gt;500&lt;\/mn&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;mo&gt;\u2248&lt;\/mo&gt;&lt;mn&gt;2.699.&lt;\/mn&gt;&lt;\/mrow&gt;&lt;\/annotation-xml&gt;&lt;\/semantics&gt;&lt;\/math&gt;\"><span id=\"MathJax-Span-2627\" class=\"math\"><\/span><\/span><\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Graphs of Logarithmic Functions<\/h2>\r\n<p id=\"fs-id1165137748716\">Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined.<\/p>\r\n<p id=\"fs-id1165137758495\">Recall that the exponential function is defined as [latex]y={b}^{x}[\/latex] for any real number <em>x<\/em>\u00a0and constant [latex]b&gt;0[\/latex], [latex]b\\ne 1[\/latex], where<\/p>\r\n\r\n<ul id=\"fs-id1165137736024\">\r\n \t<li>The domain of <em>y<\/em>\u00a0is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n \t<li>The range of <em>y<\/em>\u00a0is [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165135641666\">In the last section we learned that the logarithmic function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the inverse of the exponential function [latex]y={b}^{x}[\/latex]. So, as inverse functions:<\/p>\r\n\r\n<ul id=\"fs-id1165137656096\">\r\n \t<li>The domain of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the range of [latex]y={b}^{x}[\/latex]:[latex]\\left(0,\\infty \\right)[\/latex].<\/li>\r\n \t<li>The range of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the domain of [latex]y={b}^{x}[\/latex]: [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1165137423048\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\r\n<h3 id=\"fs-id1165135173951\">How To: Given a logarithmic function, identify the domain.<strong>\r\n<\/strong><\/h3>\r\n<ol id=\"fs-id1165137823224\" data-number-style=\"arabic\">\r\n \t<li>Set up an inequality showing the argument greater than zero.<\/li>\r\n \t<li>Solve for <em>x<\/em>.<\/li>\r\n \t<li>Write the domain in interval notation.<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn our first example we will show how to identify the domain of a logarithmic function.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWhat is the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex]?\r\n[reveal-answer q=\"370398\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"370398\"]\r\n<p id=\"fs-id1165137693442\">The logarithmic function is defined only when the input is positive, so this function is defined when [latex]x+3&gt;0[\/latex]. Solving this inequality,<\/p>\r\n\r\n<div id=\"eip-id1165135381135\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}x+3&gt;0\\hfill &amp; \\text{The input must be positive}.\\hfill \\\\ x&gt;-3\\hfill &amp; \\text{Subtract 3}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137638183\">The domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex] is [latex]\\left(-3,\\infty \\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is another example of how to identify the domain of a logarithmic function.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWhat is the domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex]?\r\n[reveal-answer q=\"275313\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"275313\"]\r\n<p id=\"fs-id1165137780875\">The logarithmic function is defined only when the input is positive, so this function is defined when [latex]5 - 2x&gt;0[\/latex]. Solving this inequality,<\/p>\r\n\r\n<div id=\"eip-id1165135470032\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}5 - 2x&gt;0\\hfill &amp; \\text{The input must be positive}.\\hfill \\\\ -2x&gt;-5\\hfill &amp; \\text{Subtract }5.\\hfill \\\\ x&lt;\\frac{5}{2}\\hfill &amp; \\text{Divide by }-2\\text{ and switch the inequality}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137656879\">The domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex] is [latex]\\left(-\\infty ,\\frac{5}{2}\\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>\u00a0Graph logarithmic functions<\/h2>\r\n<p id=\"fs-id1165135194555\">Creating a graphical representation of most functions\u00a0gives us another layer of insight for predicting future events. How do logarithmic graphs give us insight into situations? Because every logarithmic function is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation. In other words, logarithms give the <em data-effect=\"italics\">cause<\/em> for an <em data-effect=\"italics\">effect<\/em>.<\/p>\r\n<p id=\"fs-id1165137603580\">To illustrate, suppose we invest $2500 in an account that offers an annual interest rate of 5%, compounded continuously. We already know that the balance in our account for any year <em>t<\/em>\u00a0can be found with the equation [latex]A=2500{e}^{0.05t}[\/latex].<\/p>\r\nBut what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation. By graphing the model, we can see the output (year) for any input (account balance). For instance, what if we wanted to know how many years it would take for our initial investment to double? Figure 1\u00a0shows this point on the logarithmic graph.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"900\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051930\/CNX_Precalc_Figure_04_04_0012.jpg\" alt=\"A graph titled,\" width=\"900\" height=\"459\" data-media-type=\"image\/jpg\" \/> <b>Figure 1<\/b>[\/caption]\r\n<p id=\"fs-id1165134104063\">Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] along with all its transformations: shifts, stretches, compressions, and reflections.<\/p>\r\n<p id=\"fs-id1165137679088\">We begin with the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. Because every logarithmic function of this form is the inverse of an exponential function with the form [latex]y={b}^{x}[\/latex], their graphs will be reflections of each other across the line [latex]y=x[\/latex]. To illustrate this, we can observe the relationship between the input and output values of [latex]y={2}^{x}[\/latex] and its equivalent [latex]x={\\mathrm{log}}_{2}\\left(y\\right)[\/latex] in the table below.<\/p>\r\n\r\n<table style=\"width: 70%;\" summary=\"Three rows and eight columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>\u20133<\/td>\r\n<td>\u20132<\/td>\r\n<td>\u20131<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]{2}^{x}=y[\/latex]<\/strong><\/td>\r\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]{\\mathrm{log}}_{2}\\left(y\\right)=x[\/latex]<\/strong><\/td>\r\n<td>\u20133<\/td>\r\n<td>\u20132<\/td>\r\n<td>\u20131<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165135509175\">Using the inputs and outputs from the table above, we can build another table to observe the relationship between points on the graphs of the inverse functions [latex]f\\left(x\\right)={2}^{x}[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/p>\r\n\r\n<table style=\"width: 70%;\" summary=\"Two rows and eight columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\r\n<td>[latex]\\left(-3,\\frac{1}{8}\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(-2,\\frac{1}{4}\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(-1,\\frac{1}{2}\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(0,1\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(1,2\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(2,4\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(3,8\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex]<\/strong><\/td>\r\n<td>[latex]\\left(\\frac{1}{8},-3\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(\\frac{1}{4},-2\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(\\frac{1}{2},-1\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(1,0\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(2,1\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(4,2\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(8,3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165137761335\">As we\u2019d expect, the <em data-effect=\"italics\">x<\/em>- and <em data-effect=\"italics\">y<\/em>-coordinates are reversed for the inverse functions. The figure below\u00a0shows the graph of <em>f<\/em>\u00a0and <em>g<\/em>.<\/p>\r\n\r\n<figure class=\"small\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051932\/CNX_Precalc_Figure_04_04_0022.jpg\" alt=\"Graph of two functions, f(x)=2^x and g(x)=log_2(x), with the line y=x denoting the axis of symmetry.\" data-media-type=\"image\/jpg\" \/><\/figure>\r\n<p style=\"text-align: center;\"><strong>\u00a0<\/strong>Notice that the graphs of [latex]f\\left(x\\right)={2}^{x}[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex] are reflections about the line <em>y\u00a0<\/em>= <em>x<\/em>.<\/p>\r\n<p id=\"fs-id1165137406913\">Observe the following from the graph:<\/p>\r\n\r\n<ul id=\"fs-id1165137408405\">\r\n \t<li>[latex]f\\left(x\\right)={2}^{x}[\/latex] has a <em data-effect=\"italics\">y<\/em>-intercept at [latex]\\left(0,1\\right)[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex] has an <em data-effect=\"italics\">x<\/em>-intercept at [latex]\\left(1,0\\right)[\/latex].<\/li>\r\n \t<li>The domain of [latex]f\\left(x\\right)={2}^{x}[\/latex], [latex]\\left(-\\infty ,\\infty \\right)[\/latex], is the same as the range of [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/li>\r\n \t<li>The range of [latex]f\\left(x\\right)={2}^{x}[\/latex], [latex]\\left(0,\\infty \\right)[\/latex], is the same as the domain of [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/li>\r\n<\/ul>\r\n<h3 class=\"title\" data-type=\"title\">A General Note: Characteristics of the Graph of the Parent Function, <em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = log<sub><em data-effect=\"italics\">b<\/em><\/sub>(<em data-effect=\"italics\">x<\/em>)<\/h3>\r\n<p id=\"fs-id1165135520250\">For any real number <em>x<\/em>\u00a0and constant <em>b\u00a0<\/em>&gt; 0, [latex]b\\ne 1[\/latex], we can see the following characteristics in the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]:<\/p>\r\n\r\n<ul id=\"fs-id1165137400150\">\r\n \t<li>one-to-one function<\/li>\r\n \t<li>vertical asymptote: <em>x\u00a0<\/em>= 0<\/li>\r\n \t<li>domain: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\r\n \t<li>range: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]<\/li>\r\n \t<li><em data-effect=\"italics\">x-<\/em>intercept: [latex]\\left(1,0\\right)[\/latex] and key point [latex]\\left(b,1\\right)[\/latex]<\/li>\r\n \t<li><em data-effect=\"italics\">y<\/em>-intercept: none<\/li>\r\n \t<li>increasing if [latex]b&gt;1[\/latex]<\/li>\r\n \t<li>decreasing if 0 &lt; <em>b\u00a0<\/em>&lt; 1<\/li>\r\n<\/ul>\r\n<figure id=\"CNX_Precalc_Figure_04_04_003\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"824\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051934\/CNX_Precalc_Figure_04_04_003G2.jpg\" alt=\"Two graphs of the function f(x)=log_b(x) with points (1,0) and (b, 1). The first graph shows the line when b&gt;1, and the second graph shows the line when 0&lt;b&lt;1.\" width=\"824\" height=\"367\" data-media-type=\"image\/jpg\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\n<\/figure>Figure 3\u00a0shows how changing the base <em>b<\/em>\u00a0in [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (<em data-effect=\"italics\">Note:<\/em> recall that the function [latex]\\mathrm{ln}\\left(x\\right)[\/latex] has base [latex]e\\approx \\text{2}.\\text{718.)}[\/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\/121\/2016\/08\/05051936\/CNX_Precalc_Figure_04_04_0042.jpg\" alt=\"Graph of three equations: y=log_2(x) in blue, y=ln(x) in orange, and y=log(x) in red. The y-axis is the asymptote.\" width=\"487\" height=\"363\" data-media-type=\"image\/jpg\" \/> <strong>Figure 4.\u00a0<\/strong>The graphs of three logarithmic functions with different bases, all greater than 1.[\/caption]\r\n\r\nIn our first example we will graph a logarithmic function of the form\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nGraph [latex]f\\left(x\\right)={\\mathrm{log}}_{5}\\left(x\\right)[\/latex]. State the domain, range.\r\n[reveal-answer q=\"486007\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"486007\"]\r\n<p id=\"fs-id1165137501970\">Before graphing, identify the behavior and key points for the graph.<\/p>\r\n\r\n<ul id=\"fs-id1165135497154\">\r\n \t<li>Since <em>b\u00a0<\/em>= 5 is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical line\u00a0<em>x\u00a0<\/em>= 0, and the right tail will increase slowly without bound.<\/li>\r\n \t<li>The <em data-effect=\"italics\">x<\/em>-intercept is [latex]\\left(1,0\\right)[\/latex].<\/li>\r\n \t<li>The key point [latex]\\left(5,1\\right)[\/latex] is on the graph.<\/li>\r\n \t<li>We plot and label the points, and draw a smooth curve through the points.<\/li>\r\n<\/ul>\r\n<figure id=\"CNX_Precalc_Figure_04_04_005\" class=\"small\"><span id=\"fs-id1165135508394\" data-type=\"media\" data-alt=\"Graph of f(x)=log_5(x) with labeled points at (1, 0) and (5, 1). The y-axis is the asymptote.\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051938\/CNX_Precalc_Figure_04_04_0052.jpg\" alt=\"Graph of f(x)=log_5(x) with labeled points at (1, 0) and (5, 1). The y-axis is the asymptote.\" data-media-type=\"image\/jpg\" \/><\/span><\/figure>\r\n<p id=\"fs-id1165135697920\" style=\"text-align: center;\"><strong>Figure 5.\u00a0<\/strong>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3 id=\"fs-id1165137805513\">How To: Given a logarithmic function with the form [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], graph the function.<\/h3>\r\n<ol id=\"fs-id1165135435529\" data-number-style=\"arabic\">\r\n \t<li>Plot the <em data-effect=\"italics\">x-<\/em>intercept, [latex]\\left(1,0\\right)[\/latex].<\/li>\r\n \t<li>Plot the key point [latex]\\left(b,1\\right)[\/latex].<\/li>\r\n \t<li>Draw a smooth curve through the points.<\/li>\r\n \t<li>State the domain, [latex]\\left(0,\\infty \\right)[\/latex], the range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ol>\r\n<h2>Summary<\/h2>\r\nThe inverse of a function can be defined for one-to-one functions. \u00a0If a function is not one-to-one, it can be possible to restrict it's domain to make it so. The domain of a function will become the range of it's inverse. \u00a0The range of a function will become the domain of it's inverse. \u00a0Inverses can be verified using tabular data as well as algebraically.\r\n\r\nThe base <em>b<\/em>\u00a0<strong data-effect=\"bold\">logarithm<\/strong> of a number is the exponent by which we must raise <em>b<\/em>\u00a0to get that number. Logarithmic functions are the inverse of Exponential functions, and it is often easier to understand them through this lens.\u00a0We can never take the logarithm of a negative number, therefore\u00a0[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is defined for [latex]b&gt;0[\/latex].\r\n\r\nKnowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally because the logarithm is an exponent. \u00a0Logarithms most commonly sue base 10, and often use base\u00a0<em>e.\u00a0<\/em>Logarithms can also be evaluated with most kinds of calculator.\r\n\r\nTo define the domain of a logarithmic function algebraically, set the argument greater than zero and solve. To plot a logarithmic function, it is easiest to find and plot the x-intercept, and the key point[latex]\\left(b,1\\right)[\/latex].\r\n<h2><\/h2>\r\n<h2>Earthquake!<\/h2>\r\n[caption id=\"attachment_3656\" align=\"alignleft\" width=\"200\"]<img class=\"size-medium wp-image-3656\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05230800\/AlaskaQuake-Tire-200x300.jpg\" alt=\"Image of A 2x6 plank driven through a ten-ply tire by the tsunami in Whittier.\" width=\"200\" height=\"300\" \/> A 2x6 plank driven through a ten-ply tire by the tsunami in Whittier, AK.[\/caption]\r\n\r\nJoan's grandpa comes over for dinner every Sunday. \u00a0Grandpa loves to talk, but it's really hard to understand him. Also, as he gets older his hearing has been getting worse. Usually Joan tries to find an excuse to be out after dinner on Sundays, but since she had food poisoning on Saturday night, this week she wasn't feeling like leaving the house.\r\n\r\nGrandpa was feeling extra chatty this week and since Joan was a captive audience, she got the full force of his attention. His stories started with the neighbor's dog who won't stop barking, then on to his arthritis.\r\n\r\n\"Joanie, do you know that I think I have arthritis in my ankles because of an earthquake injury?\" says Grandpa.\r\n\r\nGrandpa went on to tell Joan the story of when he lived in Anchorage, Alaska in the 60's and lived through the\u00a0Good Friday earthquake.\r\n\r\nThe <b>1964 Alaskan earthquake<\/b>, also known as the <b>Great Alaskan earthquake<\/b>, occurred at 5:36 p.m. AST on Good Friday, March 27.\u00a0Across south-central Alaska, ground fissures, collapsing structures, and earthquake-created tsunamis caused about 139 deaths, some of them being friends of Joan's grandpa.\r\n\r\nLasting four minutes and thirty-eight seconds, the magnitude 8.5 (Richter scale) [footnote]\"Facts About the 1964 Alaska Earthquake.\" LiveScience. Accessed August 18, 2016. http:\/\/www.livescience.com\/44412-1964-alaska-earthquake-facts.html.[\/footnote]megathrust earthquake was the most powerful recorded in North American history, and the second most powerful recorded in world history. Soil liquefaction, fissures, landslides, and other ground failures caused major structural damage in several communities and much damage to property.\u00a0Post-quake tsunamis severely affected numerous\u00a0Alaskan communities, as well as people and property in British Columbia, Washington, Oregon, and California.<span style=\"font-size: 13.3333px; line-height: 20px;\">\u00a0<\/span>Tsunamis also caused damage in Hawai'i and Japan. Evidence of motion directly related to the earthquake was reported from all over the world.\r\n\r\nSurviving the quake inspired Joan's grandpa to study Geology in college. \u00a0Joan had heard of the Richter scale before, but didn't really understand what it meant, so she asked her grandpa.\r\n\r\n[caption id=\"attachment_3654\" align=\"alignright\" width=\"190\"]<img class=\"size-medium wp-image-3654\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05230131\/CharlesRichter-190x300.jpg\" alt=\"Photo of Charles Richter\" width=\"190\" height=\"300\" \/> Charles Richter[\/caption]\r\n\r\nHe explained that the Richter scale magnitude is expressed in whole numbers and decimal fractions. For example, a magnitude 5.3 might be computed for a moderate earthquake, and a strong earthquake might be rated as magnitude 6.3. Because of the logarithmic basis of the scale, he said, each whole number increase in magnitude represents a tenfold increase in measured amplitude. For example, Grandpa explained, a magnitude 6.3 earthquake\u00a0corresponds to the release of about 31 times more energy than a magnitude 5.3.[footnote]\"Measuring the Size of an Earthquake.\" Measuring the Size of an Earthquake. Accessed August 05, 2016. http:\/\/earthquake.usgs.gov\/learn\/topics\/measure.php.[\/footnote]\r\n\r\nJoan really enjoyed the conversation she had with her grandpa. She had never asked him about his work as a geologist, and she could tell it made him really happy that she did. At least something good emerged out of\u00a0Joan's disastrous date and food poisoning fiasco from the night before.\r\n\r\n\"1964 Quake: The Great Alaska Earthquake\" is an eleven minute video created as part of USGS activities acknowledging the fifty year anniversary of the quake on March 27, 2014.\u00a0The video features USGS geologist George Plafker, who, in the 1960s, correctly interpreted the quake as a subduction zone event. This was a great leap forward in resolving key mechanisms of the developing theory of plate tectonics. The extreme loss of life and destruction from this earthquake and accompanying tsunamis inspired\u00a0the NOAA Tsunami Warning Centers and the USGS Earthquake Hazards Program.\r\n\r\nhttps:\/\/youtu.be\/lE2j10xyOgI","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Composite and InverseFunctions\n<ul>\n<li>Define a composite function<\/li>\n<li>Define an inverse function<\/li>\n<li>Use compositions of functions to verify inverses algebraically<\/li>\n<li>Identify an inverse algebraically<\/li>\n<li>Identify the domain and range of inverse functions with tables<\/li>\n<\/ul>\n<\/li>\n<li>Logarithmic Functions\n<ul>\n<li>Define a logarithmic function as the inverse\u00a0of an exponential function<\/li>\n<li>Convert between logarithmic and exponential forms<\/li>\n<\/ul>\n<\/li>\n<li>Evaluate Logarithms\n<ul>\n<li>Mentally evaluate logarithms<\/li>\n<li>Define natural logarithm, evaluate natural logarithms with a calculator<\/li>\n<li>Define common logarithm, evaluate common logarithms mentally and with a calculator<\/li>\n<\/ul>\n<\/li>\n<li>Graphs of Logarithmic Functions\n<ul>\n<li>Identify the domain of a logarithmic function.<\/li>\n<li>Graph logarithmic functions.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<figure id=\"CNX_Precalc_Figure_04_03_001\" class=\"small\">\n<div style=\"width: 279px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051922\/CNX_Precalc_Figure_04_03_0012.jpg\" alt=\"Photo of the aftermath of the earthquake in Japan with a focus on the Japanese flag.\" width=\"269\" height=\"180\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 1.\u00a0<\/strong>Devastation of March 11, 2011 earthquake in Honshu, Japan. (credit: Daniel Pierce)<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165137557013\">In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes.<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary. Accessed 3\/4\/2013.\" id=\"return-footnote-3518-1\" href=\"#footnote-3518-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings,<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary. Accessed 3\/4\/2013.\" id=\"return-footnote-3518-2\" href=\"#footnote-3518-2\" aria-label=\"Footnote 2\"><sup class=\"footnote\">[2]<\/sup><\/a>\u00a0like those shown in the picture above. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/. Accessed 3\/4\/2013.\" id=\"return-footnote-3518-3\" href=\"#footnote-3518-3\" aria-label=\"Footnote 3\"><sup class=\"footnote\">[3]<\/sup><\/a>\u00a0whereas the Japanese earthquake registered a 9.0.<a class=\"footnote\" title=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details. Accessed 3\/4\/2013.\" id=\"return-footnote-3518-4\" href=\"#footnote-3518-4\" aria-label=\"Footnote 4\"><sup class=\"footnote\">[4]<\/sup><\/a><\/p>\n<p id=\"fs-id1165137760714\">The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is [latex]{10}^{8 - 4}={10}^{4}=10,000[\/latex] times as great! In this lesson, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.<\/p>\n<p>Our first topic will be about inverse functions, logarithmic functions are the inverse of an exponential functions, and sometimes understanding this helps us make sense of what a logarithm is.<\/p>\n<h2>Composite and Inverse Functions<\/h2>\n<p id=\"fs-id1165134094620\">Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day.<\/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-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25200743\/CNX_Precalc_Figure_01_04_0062.jpg\" alt=\"Explanation of C(T(5)), which is the cost for the temperature and T(5) is the temperature on day 5.\" width=\"487\" height=\"140\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165134038788\">Using descriptive variables, we can notate these two functions. The function [latex]C\\left(T\\right)[\/latex] gives the cost [latex]C[\/latex] of heating a house for a given average daily temperature in [latex]T[\/latex] degrees Celsius. The function [latex]T\\left(d\\right)[\/latex] gives the average daily temperature on day [latex]d[\/latex] of the year. For any given day, [latex]\\text{Cost}=C\\left(T\\left(d\\right)\\right)[\/latex] means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature [latex]T\\left(d\\right)[\/latex]. For example, we could evaluate [latex]T\\left(5\\right)[\/latex] to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the <strong>cost function<\/strong> at that temperature. We would write [latex]C\\left(T\\left(5\\right)\\right)[\/latex].\u00a0By combining these two relationships into one function, we have performed function composition.<\/p>\n<p>We read the left-hand side as [latex]\"f[\/latex] composed with [latex]g[\/latex] at [latex]x,\"[\/latex] and the right-hand side as [latex]\"f[\/latex] of [latex]g[\/latex] of [latex]x.\"[\/latex] The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol [latex]\\circ[\/latex] is called the composition operator.<\/p>\n<p>It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside.<\/p>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25200744\/CNX_Precalc_Figure_01_04_0012.jpg\" alt=\"Explanation of the composite function. g(x), the output of g is the input of f. X is the input of g.\" width=\"487\" height=\"171\" data-media-type=\"image\/jpg\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Using the functions provided, find [latex]f\\left(g\\left(x\\right)\\right)[\/latex] and [latex]g\\left(f\\left(x\\right)\\right)[\/latex].<\/p>\n<p>[latex]f\\left(x\\right)=2x+1[\/latex]<\/p>\n<p>[latex]g\\left(x\\right)=3-x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q337338\">Show Answer<\/span><\/p>\n<div id=\"q337338\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(x\\right)=2x+1[\/latex]<\/p>\n<p>[latex]g\\left(x\\right)=3-x[\/latex]<\/p>\n<p>Let\u2019s begin by substituting [latex]g\\left(x\\right)[\/latex] into [latex]f\\left(x\\right)[\/latex].[latex]\\begin{array}f\\left(g\\left(x\\right)\\right)=2\\left(3-x\\right)+1\\hfill \\\\ \\text{ }=6 - 2x+1\\hfill \\\\ \\text{ }=7 - 2x\\hfill \\end{array}[\/latex]<br \/>\nNow we can substitute [latex]f\\left(x\\right)[\/latex] into [latex]g\\left(x\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}g\\left(f\\left(x\\right)\\right)=3-\\left(2x+1\\right)\\hfill \\\\ \\text{ }=3 - 2x - 1\\hfill \\\\ \\text{ }=-2x+2\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video you will see another example of how to find the composition of two functions.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Composition of Function\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/r_LssVS4NHk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Inverse Functions<\/h2>\n<p id=\"fs-id1165137827441\">An <strong>inverse function<\/strong>\u00a0is a function for which the input of the original function becomes the output of the inverse function. This naturally leads to the output of the original function becoming the input of the inverse function. The reason we want to introduce inverse functions is because exponential and logarithmic functions are inverses of each other, and understanding this quality helps to make understanding logarithmic functions easier. And the reason we introduced composite functions is because you can verify, algebraically, whether two functions are inverses of each other by using a composition.<\/p>\n<p id=\"fs-id1165135528385\">Given a function [latex]f\\left(x\\right)[\/latex], we represent its inverse as [latex]{f}^{-1}\\left(x\\right)[\/latex], read as [latex]\"f[\/latex] inverse of [latex]x.\\text{\"}[\/latex] The raised [latex]-1[\/latex] is part of the notation. It is not an exponent; it does not imply a power of [latex]-1[\/latex] . In other words, [latex]{f}^{-1}\\left(x\\right)[\/latex] does <em data-effect=\"italics\">not<\/em> mean [latex]\\frac{1}{f\\left(x\\right)}[\/latex] because [latex]\\frac{1}{f\\left(x\\right)}[\/latex] is the reciprocal of [latex]f[\/latex] and not the inverse.<\/p>\n<p id=\"fs-id1165137724926\">Just as zero does not have a <strong>reciprocal<\/strong>, some functions do not have inverses.<\/p>\n<div id=\"fs-id1165137933105\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\n<h3 class=\"title\" data-type=\"title\">Inverse Function<\/h3>\n<p id=\"fs-id1165137473076\">For any <strong>one-to-one function<\/strong> [latex]f\\left(x\\right)=y[\/latex], a function [latex]{f}^{-1}\\left(x\\right)[\/latex] is an <strong>inverse function<\/strong> of [latex]f[\/latex] if [latex]{f}^{-1}\\left(y\\right)=x[\/latex].<\/p>\n<p id=\"fs-id1165137444821\">The notation [latex]{f}^{-1}[\/latex] is read [latex]\\text{\"}f[\/latex] inverse.&#8221; Like any other function, we can use any variable name as the input for [latex]{f}^{-1}[\/latex], so we will often write [latex]{f}^{-1}\\left(x\\right)[\/latex], which we read as [latex]\"f[\/latex] inverse of [latex]x.\"[\/latex]<br \/>\nKeep in mind that<\/p>\n<div id=\"fs-id1165137581324\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{f}^{-1}\\left(x\\right)\\ne \\frac{1}{f\\left(x\\right)}[\/latex]<\/div>\n<p id=\"fs-id1165135194095\">and not all functions have inverses.<\/p>\n<\/div>\n<p>In our first example we will identify an inverse function from ordered pairs.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If for a particular one-to-one function [latex]f\\left(2\\right)=4[\/latex] and [latex]f\\left(5\\right)=12[\/latex], what are the corresponding input and output values for the inverse function?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q664782\">Show Answer<\/span><\/p>\n<div id=\"q664782\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137737081\">The inverse function reverses the input and output quantities, so if<\/p>\n<div id=\"fs-id1165137462459\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}f\\left(2\\right)=4,\\text{ then }{f}^{-1}\\left(4\\right)=2;\\\\ f\\left(5\\right)=12,{\\text{ then f}}^{-1}\\left(12\\right)=5.\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137659464\">Alternatively, if we want to name the inverse function [latex]g[\/latex], then [latex]g\\left(4\\right)=2[\/latex] and [latex]g\\left(12\\right)=5[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Analysis of the Solution<\/h3>\n<div id=\"Example_01_07_01\" class=\"example\" data-type=\"example\">\n<div id=\"fs-id1165137656641\" class=\"exercise\" data-type=\"exercise\">\n<div id=\"fs-id1165135245520\" class=\"commentary\" data-type=\"commentary\">\n<p id=\"fs-id1165135508518\">Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed.<\/p>\n<table style=\"width: 30%;\" summary=\"For (x,f(x)) we have the values (2, 4) and (5, 12); for (x, g(x)), we have the values (4, 2) and (12, 5).\">\n<thead>\n<tr>\n<th data-align=\"center\">[latex]\\left(x,f\\left(x\\right)\\right)[\/latex]<\/th>\n<th data-align=\"center\">[latex]\\left(x,g\\left(x\\right)\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td data-align=\"center\">[latex]\\left(2,4\\right)[\/latex]<\/td>\n<td data-align=\"center\">[latex]\\left(4,2\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td data-align=\"center\">[latex]\\left(5,12\\right)[\/latex]<\/td>\n<td data-align=\"center\">[latex]\\left(12,5\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show an example of finding corresponding input and output values given two ordered pairs from functions that are inverses.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Function and Inverse Function Values\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/IR_1L1mnpvw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div id=\"fs-id1165134357354\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\n<h3 id=\"fs-id1165135434077\">How To: Given two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex], test whether the functions are inverses of each other.<\/h3>\n<ol id=\"fs-id1165137452358\" data-number-style=\"arabic\">\n<li>Substitute\u00a0[latex]g(x)[\/latex]\u00a0into [latex]f(x)[\/latex]. The result must be x. [latex]f\\left(g(x)\\right)=x[\/latex]<\/li>\n<li>Substitute\u00a0[latex]f(x)[\/latex]\u00a0into [latex]g(x)[\/latex]. The result must be x. [latex]g\\left(f(x)\\right)=x[\/latex]<\/li>\n<\/ol>\n<p style=\"text-align: center;\">If\u00a0[latex]f(x)[\/latex] and\u00a0\u00a0[latex]g(x)[\/latex] are inverses, then\u00a0\u00a0[latex]f(x)=g^{-1}(x)[\/latex] and\u00a0[latex]g(x)=f^{-1}(x)[\/latex]<\/p>\n<\/div>\n<p>In our next example we will test inverse relationships algebraically.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If [latex]f\\left(x\\right)=x^2-3[\/latex], for [latex]x\\ge0[\/latex] and [latex]g\\left(x\\right)=\\sqrt{x+3}[\/latex], is g the inverse of f? \u00a0[latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q598434\">Show Answer<\/span><\/p>\n<div id=\"q598434\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute [latex]g(x)=\\sqrt{x+3}[\/latex] into [latex]f(x)[\/latex], this means the new variable in\u00a0[latex]f(x)[\/latex] is [latex]\\sqrt{x+3}[\/latex] so you will substitute that expression where you see x. \u00a0Using parentheses helps keep track of things.<\/p>\n<p>[latex]\\begin{array}{c}f\\left(\\sqrt{x+3}\\right)={(\\sqrt{x+3})}^2-3\\hfill\\\\=x+3-3\\\\=x\\hfill \\end{array}[\/latex]<\/p>\n<p>Our result implies that [latex]g(x)[\/latex] is indeed the inverse of\u00a0[latex]f(x)[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]g={f}^{-1}[\/latex],\u00a0for [latex]x\\ge0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we use algebra to determine if two functions are inverses.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 1: Determine if Two Functions Are Inverses\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/vObCvTOatfQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We will show one more example of how to verify whether you have an inverse algebraically.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If [latex]f\\left(x\\right)=\\frac{1}{x+2}[\/latex] and [latex]g\\left(x\\right)=\\frac{1}{x}-2[\/latex], is g the inverse of f? \u00a0[latex]g={f}^{-1}?[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q56557\">Show Answer<\/span><\/p>\n<div id=\"q56557\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute [latex]g(x)=\\frac{1}{x}-2[\/latex] into [latex]f(x)[\/latex], this means the new variable in\u00a0[latex]f(x)[\/latex] is [latex]\\frac{1}{x}-2[\/latex] so you will substitute that expression where you see x. \u00a0Using parentheses helps keep track of things.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c} f\\left(\\frac{1}{x}-2\\right)=\\frac{1}{\\left(\\frac{1}{x}-2\\right)+2}\\hfill\\\\=\\frac{1}{\\frac{1}{x}}\\hfill\\\\={ x }\\hfill \\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]g={f}^{-1}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We will show one more example of how to use algebra to determine whether two functions are inverses of each other.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex 2: Determine if Two Functions Are Inverses\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/hzehBtNmw08?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>\u00a0Domain and Range of a Function and It&#8217;s Inverse<\/h2>\n<p>The outputs of the function [latex]f[\/latex] are the inputs to [latex]{f}^{-1}[\/latex], so the range of [latex]f[\/latex] is also the domain of [latex]{f}^{-1}[\/latex]. Likewise, because the inputs to [latex]f[\/latex] are the outputs of [latex]{f}^{-1}[\/latex], the domain of [latex]f[\/latex] is the range of [latex]{f}^{-1}[\/latex]. We can visualize the situation.<\/p>\n<div class=\"mceTemp\"><\/div>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25200955\/CNX_Precalc_Figure_01_07_0032.jpg\" alt=\"Domain and range of a function and its inverse.\" width=\"487\" height=\"143\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>Domain and range of a function and its inverse<\/p>\n<p>In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function [latex]f\\left(x\\right)={x}^{2}[\/latex] with its range limited to [latex]\\left[0,\\infty \\right)[\/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).<\/p>\n<div class=\"textbox\">\n<h3 class=\"title\" data-type=\"title\">Domain and Range of Inverse Functions<\/h3>\n<p id=\"fs-id1165135319550\">The range of a function [latex]f\\left(x\\right)[\/latex] is the domain of the inverse function [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<p id=\"fs-id1165137673886\">The domain of [latex]f\\left(x\\right)[\/latex] is the range of [latex]{f}^{-1}\\left(x\\right)[\/latex].<\/p>\n<\/div>\n<p>In our last example we will define the domain and range of a function&#8217;s inverse using a table of values, and evaluate the inverse at a specific value.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p id=\"fs-id1165135435474\">A function [latex]f\\left(t\\right)[\/latex] is given\u00a0below, showing distance in miles that a car has traveled in [latex]t[\/latex] minutes.<\/p>\n<ol>\n<li>Define the domain and range of the function and it&#8217;s inverse.<\/li>\n<li>Find and interpret [latex]{f}^{-1}\\left(70\\right)[\/latex].<\/li>\n<\/ol>\n<table style=\"width: 30%;\" summary=\"Two rows and five columns. The first row is labeled\">\n<tbody>\n<tr>\n<td data-align=\"left\"><strong>[latex]t\\text{ (minutes)}[\/latex]<\/strong><\/td>\n<td data-align=\"left\">30<\/td>\n<td data-align=\"left\">50<\/td>\n<td data-align=\"left\">70<\/td>\n<td data-align=\"left\">90<\/td>\n<\/tr>\n<tr>\n<td data-align=\"left\"><strong>[latex]f\\left(t\\right)\\text{ (miles)}[\/latex] <\/strong><\/td>\n<td data-align=\"left\">20<\/td>\n<td data-align=\"left\">40<\/td>\n<td data-align=\"left\">60<\/td>\n<td data-align=\"left\">70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q713219\">Show Answer<\/span><\/p>\n<div id=\"q713219\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.<span style=\"text-decoration: underline;\">Domain and Range of the Original Function<\/span><\/p>\n<p>The domain of this tabular function, [latex]f\\left(t\\right)[\/latex] , is all the input values, t in minutes: {30, 50, 70, 90}<\/p>\n<p>The range of this tabular function,[latex]f\\left(t\\right)[\/latex], \u00a0is all the output values[latex]f\\left(t\\right)[\/latex] in miles: {20, 40, 60, 70}<\/p>\n<p><span style=\"text-decoration: underline;\">Domain and Range of the Inverse Function<\/span><\/p>\n<p id=\"fs-id1165137640334\">The domain for the inverse will be the outputs from the original, so the domain of \u00a0[latex]{f}^{-1}(x)[\/latex] is the output values from\u00a0[latex]f\\left(t\\right)[\/latex]:\u00a0{20, 40, 60, 70}<\/p>\n<p>The range for the inverse will be the inputs from the original:\u00a0{30, 50, 70, 90}<\/p>\n<p>This translates to putting in a number of miles and getting out how long it took to drive that far in minutes.<\/p>\n<p>2.\u00a0So in the expression [latex]{f}^{-1}\\left(70\\right)[\/latex], 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function [latex]f[\/latex], 90 minutes, so [latex]{f}^{-1}\\left(70\\right)=90[\/latex]. The interpretation of this is that, to drive 70 miles, it took 90 minutes.<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Define Logarithmic Functions<\/h2>\n<p id=\"fs-id1165135192781\">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 were 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. 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<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\/121\/2016\/08\/05051924\/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\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137662989\">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 data-effect=\"bold\">logarithm<\/strong> of a number is the exponent by which we must raise <em>b<\/em>\u00a0to get that number.<\/p>\n<p id=\"fs-id1165137404844\">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 id=\"fs-id1165137597501\">We can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\n<div id=\"eip-604\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b>0,b\\ne 1[\/latex]<\/div>\n<p id=\"fs-id1165137678993\">Note that the base <em>b<\/em>\u00a0is always positive.<span id=\"fs-id1165137696233\" data-type=\"media\" data-alt=\"\" data-display=\"block\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051926\/CNX_Precalc_Figure_04_03_0042.jpg\" alt=\"Think b to the y equals x.\" width=\"487\" height=\"83\" data-media-type=\"image\/jpg\" \/><\/span><\/p>\n<p id=\"fs-id1165137400957\">Because 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 id=\"fs-id1165137827516\">We can illustrate the notation of logarithms as follows:<span id=\"fs-id1165137771679\" data-type=\"media\" data-alt=\"\" data-display=\"block\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051928\/CNX_Precalc_Figure_04_03_0032.jpg\" alt=\"logb (c) = a means b to the A power equals C.\" width=\"487\" height=\"101\" data-media-type=\"image\/jpg\" \/><\/span><\/p>\n<p id=\"fs-id1165137575165\">Notice that, 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 id=\"fs-id1165137472937\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\n<h3 class=\"title\" data-type=\"title\">Definition of the Logarithmic Function<\/h3>\n<p id=\"fs-id1165137704597\">A <strong>logarithm<\/strong> base <em>b<\/em>\u00a0of a positive number <em>x<\/em>\u00a0satisfies the following definition.<\/p>\n<p id=\"fs-id1165137584967\">For [latex]x>0,b>0,b\\ne 1[\/latex],<\/p>\n<div id=\"fs-id1165137433829\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equivalent to }{b}^{y}=x[\/latex]<\/div>\n<p id=\"fs-id1165137893373\">where,<\/p>\n<ul id=\"fs-id1165135530561\">\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<\/ul>\n<p id=\"fs-id1165137547773\">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 id=\"fs-id1165137643167\">\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<p>In our first example we will convert logarithmic equations into exponential equations.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p id=\"fs-id1165137580570\">Write the following logarithmic equations in exponential form.<\/p>\n<ol id=\"fs-id1165137705346\" data-number-style=\"lower-alpha\">\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]\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q161275\">Show Answer<\/span><\/p>\n<div id=\"q161275\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137408172\">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 id=\"fs-id1165137705659\" data-number-style=\"lower-alpha\">\n<li>[latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]\n<p id=\"fs-id1165137602796\">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 equivalent to [latex]{6}^{\\frac{1}{2}}=\\sqrt{6}[\/latex].<\/p>\n<\/li>\n<li>[latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]\n<p id=\"fs-id1165137698078\">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 equivalent to [latex]{3}^{2}=9[\/latex].<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<p>In the following video we present more examples of rewriting logarithmic equations as exponential equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex:  Write Logarithmic Equations as Exponential Equations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/q9_s0wqhIXU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div id=\"fs-id1165137874700\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\n<h3 id=\"fs-id1165137806301\">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 id=\"fs-id1165137641669\" data-number-style=\"arabic\">\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 key-takeaways\">\n<h3>Think About It<\/h3>\n<p id=\"eip-id1549475\">Can we take the logarithm of a negative number? Re-read the definition of a logarithm and formulate an answer. \u00a0Think about the behavior of exponents. \u00a0You can use the textbox below to formulate your ideas before you look at an answer.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q162494\">Show Answer<\/span><\/p>\n<div id=\"q162494\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137653864\">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.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Convert from exponential to logarithmic form<\/h2>\n<p>To convert from exponents to logarithms, 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<\/h3>\n<p id=\"fs-id1165137804412\">Write the following exponential equations in logarithmic form.<\/p>\n<ol id=\"fs-id1165135192287\" data-number-style=\"lower-alpha\">\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]\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q516026\">Show Answer<\/span><\/p>\n<div id=\"q516026\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137474116\">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 id=\"fs-id1165137573458\" data-number-style=\"lower-alpha\">\n<li>[latex]{2}^{3}=8[\/latex]\n<p id=\"fs-id1165137466396\">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 equivalent to [latex]{\\mathrm{log}}_{2}\\left(8\\right)=3[\/latex].<\/p>\n<\/li>\n<li>[latex]{5}^{2}=25[\/latex]\n<p id=\"fs-id1165135193035\">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 equivalent to [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex].<\/p>\n<\/li>\n<li>[latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]\n<p id=\"fs-id1165135187822\">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 equivalent to [latex]{\\text{log}}_{10}\\left(\\frac{1}{10,000}\\right)=-4[\/latex].<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<p>In our last video we show more examples of writing logarithmic equations as exponential equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Ex:  Write Exponential Equations as Logarithmic Equations\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/9_GPPUWEJQQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Evaluate Logarithms<\/h2>\n<section id=\"fs-id1165137405741\" data-depth=\"1\">\n<p id=\"fs-id1165137422589\">Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider [latex]{\\mathrm{log}}_{2}8[\/latex]. We ask, &#8220;To what exponent must 2\u00a0be raised in order to get 8?&#8221; Because we already know [latex]{2}^{3}=8[\/latex], it follows that [latex]{\\mathrm{log}}_{2}8=3[\/latex].<\/p>\n<p id=\"fs-id1165137733822\">Now consider solving [latex]{\\mathrm{log}}_{7}49[\/latex] and [latex]{\\mathrm{log}}_{3}27[\/latex] mentally.<\/p>\n<ul id=\"fs-id1165137937690\">\n<li>We ask, &#8220;To what exponent must 7 be raised in order to get 49?&#8221; We know [latex]{7}^{2}=49[\/latex]. Therefore, [latex]{\\mathrm{log}}_{7}49=2[\/latex]<\/li>\n<li>We ask, &#8220;To what exponent must 3 be raised in order to get 27?&#8221; We know [latex]{3}^{3}=27[\/latex]. Therefore, [latex]{\\mathrm{log}}_{3}27=3[\/latex]<\/li>\n<\/ul>\n<p id=\"fs-id1165137456358\">Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let\u2019s evaluate [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\frac{4}{9}[\/latex] mentally.<\/p>\n<ul id=\"fs-id1165137584208\">\n<li>We ask, &#8220;To what exponent must [latex]\\frac{2}{3}[\/latex] be raised in order to get [latex]\\frac{4}{9}[\/latex]? &#8221; We know [latex]{2}^{2}=4[\/latex] and [latex]{3}^{2}=9[\/latex], so [latex]{\\left(\\frac{2}{3}\\right)}^{2}=\\frac{4}{9}[\/latex]. Therefore, [latex]{\\mathrm{log}}_{\\frac{2}{3}}\\left(\\frac{4}{9}\\right)=2[\/latex].<\/li>\n<\/ul>\n<p>In our first example we will evaluate logarithms mentally.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]y={\\mathrm{log}}_{4}\\left(64\\right)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q161686\">Show Answer<\/span><\/p>\n<div id=\"q161686\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137611276\">First we rewrite the logarithm in exponential form: [latex]{4}^{y}=64[\/latex]. Next, we ask, &#8220;To what exponent must 4 be raised in order to get 64?&#8221;<\/p>\n<p>We know<\/p>\n<div id=\"eip-id1165134583995\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{4}^{3}=64[\/latex]<\/div>\n<p id=\"fs-id1165137619013\">Therefore,<\/p>\n<div id=\"eip-id1165135606935\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\mathrm{log}{}_{4}\\left(64\\right)=3[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In our first video we will show more examples of evaluating logarithms mentally, this helps you get familiar with what a logarithm represents.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Ex 1:  Evaluate Logarithms Without a Calculator - Whole Numbers\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/dxj5J9OpWGA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In our next example we will evaluate the logarithm of a reciprocal.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Evaluate [latex]y={\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q534439\">Show Answer<\/span><\/p>\n<div id=\"q534439\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137638179\">First we rewrite the logarithm in exponential form: [latex]{3}^{y}=\\frac{1}{27}[\/latex]. Next, we ask, &#8220;To what exponent must 3 be raised in order to get [latex]\\frac{1}{27}[\/latex]&#8220;?<\/p>\n<p id=\"fs-id1165137552085\">We know [latex]{3}^{3}=27[\/latex], but what must we do to get the reciprocal, [latex]\\frac{1}{27}[\/latex]? Recall from working with exponents that [latex]{b}^{-a}=\\frac{1}{{b}^{a}}[\/latex]. We use this information to write<\/p>\n<div id=\"eip-id1165137550550\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{3}^{-3}=\\frac{1}{{3}^{3}}\\hfill \\\\ =\\frac{1}{27}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137585807\">Therefore, [latex]{\\mathrm{log}}_{3}\\left(\\frac{1}{27}\\right)=-3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3 id=\"fs-id1165137453770\">How To: Given a logarithm of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], evaluate it mentally.<\/h3>\n<ol id=\"fs-id1165134079724\" data-number-style=\"arabic\">\n<li>Rewrite the argument <em>x<\/em>\u00a0as a power of <em>b<\/em>: [latex]{b}^{y}=x[\/latex].<\/li>\n<li>Use previous knowledge of powers of <em>b<\/em>\u00a0identify <em>y<\/em>\u00a0by asking, &#8220;To what exponent should <em>b<\/em>\u00a0be raised in order to get <em>x<\/em>?&#8221;<\/li>\n<\/ol>\n<p id=\"fs-id1165137661970\">\n<\/div>\n<h2>\u00a0Natural logarithms<\/h2>\n<p><span style=\"line-height: 1.5;\">The most frequently used base for logarithms is <\/span><em style=\"line-height: 1.5;\">e<\/em><span style=\"line-height: 1.5;\">. Base <\/span><em style=\"line-height: 1.5;\">e<\/em><span style=\"line-height: 1.5;\">\u00a0logarithms are important in calculus and some scientific applications; they are called <\/span><strong style=\"line-height: 1.5;\" data-effect=\"bold\">natural logarithms<\/strong><span style=\"line-height: 1.5;\">. The base <\/span><em style=\"line-height: 1.5;\">e<\/em><span style=\"line-height: 1.5;\">\u00a0logarithm, [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex], has its own notation, [latex]\\mathrm{ln}\\left(x\\right)[\/latex].<\/span><\/p>\n<p id=\"fs-id1165137473872\">Most values of [latex]\\mathrm{ln}\\left(x\\right)[\/latex] can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, [latex]\\mathrm{ln}1=0[\/latex]. For other natural logarithms, we can use the [latex]\\mathrm{ln}[\/latex] key that can be found on most scientific calculators. We can also find the natural logarithm of any power of <em>e<\/em>\u00a0using the inverse property of logarithms.<\/p>\n<div id=\"fs-id1165137452317\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\n<h3 class=\"title\" data-type=\"title\">A General Note: Definition of the Natural Logarithm<\/h3>\n<p id=\"fs-id1165137579241\">A <strong>natural logarithm<\/strong> is a logarithm with base <em>e<\/em>. We write [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] simply as [latex]\\mathrm{ln}\\left(x\\right)[\/latex]. The natural logarithm of a positive number <em>x<\/em>\u00a0satisfies the following definition.<\/p>\n<p id=\"fs-id1165135613642\">For [latex]x>0[\/latex],<\/p>\n<div id=\"fs-id1165137580230\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]y=\\mathrm{ln}\\left(x\\right)\\text{ is equivalent to }{e}^{y}=x[\/latex]<\/div>\n<p id=\"fs-id1165137658264\">We read [latex]\\mathrm{ln}\\left(x\\right)[\/latex] as, &#8220;the logarithm with base <em>e<\/em>\u00a0of <em>x<\/em>&#8221; or &#8220;the natural logarithm of <em>x<\/em>.&#8221;<\/p>\n<p id=\"fs-id1165137566720\">The logarithm <em>y<\/em>\u00a0is the exponent to which <em>e<\/em>\u00a0must be raised to get <em>x<\/em>.<\/p>\n<p id=\"fs-id1165137705251\">Since the functions [latex]y=e{}^{x}[\/latex] and [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] are inverse functions, [latex]\\mathrm{ln}\\left({e}^{x}\\right)=x[\/latex] for all <em>x<\/em>\u00a0and [latex]e{}^{\\mathrm{ln}\\left(x\\right)}=x[\/latex] for <em>x\u00a0<\/em>&gt; 0.<\/p>\n<\/div>\n<p>In the next\u00a0example, we will evaluate a natural logarithm using a calculator.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Evaluate [latex]y=\\mathrm{ln}\\left(500\\right)[\/latex] to four decimal places using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q957920\">Show Answer<\/span><\/p>\n<div id=\"q957920\" class=\"hidden-answer\" style=\"display: none\">\n<ul id=\"fs-id1165137563770\">\n<li>Press <strong data-effect=\"bold\">[LN]<\/strong>.<\/li>\n<li>Enter 500, followed by <strong data-effect=\"bold\">[ ) ]<\/strong>.<\/li>\n<li>Press <strong data-effect=\"bold\">[ENTER]<\/strong>.<\/li>\n<\/ul>\n<p id=\"fs-id1165137645024\">Rounding to four decimal places, [latex]\\mathrm{ln}\\left(500\\right)\\approx 6.2146[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next video, we show more examples of how to evaluate natural logarithms using a calculator.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Ex:  Evaluate Natural Logarithms on the Calculator\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Rpounu3epSc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Common logarithms<\/h2>\n<p>Sometimes we may see a logarithm written without a base. In this case, we assume that the base is 10. In other words, the expression [latex]{\\mathrm{log}}_{}[\/latex] means [latex]{\\mathrm{log}}_{10}[\/latex] \u00a0We call a base-10 logarithm a <strong data-effect=\"bold\">common logarithm<\/strong>. Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars and the pH of acids and bases also use common logarithms.<\/p>\n<div class=\"textbox\">\n<h3>Definition of Common Logarithm: Log is an exponent<\/h3>\n<p>A common logarithm is a logarithm with base 10. \u00a0We write\u00a0[latex]{\\mathrm{log}}_{10}(x)[\/latex] \u00a0simpliy as\u00a0[latex]{\\mathrm{log}}_{}(x)[\/latex]. \u00a0The common logarithm of a positive number, x, satisfies the following definition:<\/p>\n<p>For [latex]x\\gt0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]y={\\mathrm{log}}_{}(x)[\/latex] is equivalent to [latex]10^y=x[\/latex]<\/p>\n<p style=\"text-align: left;\">We read [latex]{\\mathrm{log}}_{}(x)[\/latex] as &#8221; the logarithm with base 10 of x&#8221; or &#8220;log base 10 of x&#8221;.<\/p>\n<p style=\"text-align: left;\">The logarithm y is the exponent to which 10 must be raised to get x.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Evaluate [latex]{\\mathrm{log}}_{}(1000)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q80362\">Show Answer<\/span><\/p>\n<div id=\"q80362\" class=\"hidden-answer\" style=\"display: none\">We know 10^3=1000, therefore<\/p>\n<p>[latex]{\\mathrm{log}}_{}(1000)=3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Evaluate [latex]y={\\mathrm{log}}_{}(321)[\/latex] to four decimal places using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q782139\">Show Answer<\/span><\/p>\n<div id=\"q782139\" class=\"hidden-answer\" style=\"display: none\">\n<ul id=\"fs-id1165137786486\">\n<li>Press <strong data-effect=\"bold\">[LOG]<\/strong>.<\/li>\n<li>Enter 321<em data-effect=\"italics\">,<\/em> followed by <strong data-effect=\"bold\">[ ) ]<\/strong>.<\/li>\n<li>Press <strong data-effect=\"bold\">[ENTER]<\/strong>.<\/li>\n<\/ul>\n<p>Rounding to four decimal places,\u00a0[latex]{\\mathrm{log}}_{}(321)\\approx2.5065[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our last example we will use a logarithm to find the difference in magnitude of two different earthquakes.<\/p>\n<\/section>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. The equation [latex]10^x=500[\/latex] represents this situation, where x is 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=\"q735383\">Show Answer<\/span><\/p>\n<div id=\"q735383\" class=\"hidden-answer\" style=\"display: none\">We begin by rewriting the exponential equation in logarithmic form.<\/p>\n<p style=\"text-align: center;\">[latex]10^x=500[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]{\\mathrm{log}}_{}(500)=x[\/latex]<\/p>\n<p id=\"fs-id1165137419444\">Next we evaluate the logarithm using a calculator:<\/p>\n<ul id=\"fs-id1165137736356\">\n<li>Press <strong data-effect=\"bold\">[LOG]<\/strong>.<\/li>\n<li>Enter<span style=\"font-size: 14px; line-height: normal; white-space: nowrap;\">\u00a0500\u00a0<\/span>followed by <strong data-effect=\"bold\">[ ) ]<\/strong>.<\/li>\n<li>Press <strong data-effect=\"bold\">[ENTER]<\/strong>.<\/li>\n<li>To the nearest thousandth,\u00a0[latex]{\\mathrm{log}}_{}(500)\\approx2.699[\/latex]<span id=\"MathJax-Element-202-Frame\" class=\"MathJax\" style=\"box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: 0px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; color: #333333; font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-variant: normal; orphans: auto; widows: 1; -webkit-text-stroke-width: 0px; position: relative; background-color: #ededed;\" tabindex=\"0\" data-mathml=\"&lt;math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot; display=&quot;inline&quot;&gt;&lt;semantics&gt;&lt;mrow&gt;&lt;mrow&gt;&lt;mtext&gt;&amp;#x2009;&lt;\/mtext&gt;&lt;mi&gt;log&lt;\/mi&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mn&gt;500&lt;\/mn&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;mo&gt;&amp;#x2248;&lt;\/mo&gt;&lt;mn&gt;2.699.&lt;\/mn&gt;&lt;\/mrow&gt;&lt;\/mrow&gt;&lt;annotation-xml encoding=&quot;MathML-Content&quot;&gt;&lt;mrow&gt;&lt;mtext&gt;\u2009&lt;\/mtext&gt;&lt;mi&gt;log&lt;\/mi&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mn&gt;500&lt;\/mn&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;mo&gt;\u2248&lt;\/mo&gt;&lt;mn&gt;2.699.&lt;\/mn&gt;&lt;\/mrow&gt;&lt;\/annotation-xml&gt;&lt;\/semantics&gt;&lt;\/math&gt;\"><span id=\"MathJax-Span-2627\" class=\"math\"><\/span><\/span><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<h2>Graphs of Logarithmic Functions<\/h2>\n<p id=\"fs-id1165137748716\">Before working with graphs, we will take a look at the domain (the set of input values) for which the logarithmic function is defined.<\/p>\n<p id=\"fs-id1165137758495\">Recall that the exponential function is defined as [latex]y={b}^{x}[\/latex] for any real number <em>x<\/em>\u00a0and constant [latex]b>0[\/latex], [latex]b\\ne 1[\/latex], where<\/p>\n<ul id=\"fs-id1165137736024\">\n<li>The domain of <em>y<\/em>\u00a0is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<li>The range of <em>y<\/em>\u00a0is [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<p id=\"fs-id1165135641666\">In the last section we learned that the logarithmic function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the inverse of the exponential function [latex]y={b}^{x}[\/latex]. So, as inverse functions:<\/p>\n<ul id=\"fs-id1165137656096\">\n<li>The domain of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the range of [latex]y={b}^{x}[\/latex]:[latex]\\left(0,\\infty \\right)[\/latex].<\/li>\n<li>The range of [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] is the domain of [latex]y={b}^{x}[\/latex]: [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ul>\n<div id=\"fs-id1165137423048\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\n<h3 id=\"fs-id1165135173951\">How To: Given a logarithmic function, identify the domain.<strong><br \/>\n<\/strong><\/h3>\n<ol id=\"fs-id1165137823224\" data-number-style=\"arabic\">\n<li>Set up an inequality showing the argument greater than zero.<\/li>\n<li>Solve for <em>x<\/em>.<\/li>\n<li>Write the domain in interval notation.<\/li>\n<\/ol>\n<\/div>\n<p>In our first example we will show how to identify the domain of a logarithmic function.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>What is the domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q370398\">Show Answer<\/span><\/p>\n<div id=\"q370398\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137693442\">The logarithmic function is defined only when the input is positive, so this function is defined when [latex]x+3>0[\/latex]. Solving this inequality,<\/p>\n<div id=\"eip-id1165135381135\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}x+3>0\\hfill & \\text{The input must be positive}.\\hfill \\\\ x>-3\\hfill & \\text{Subtract 3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137638183\">The domain of [latex]f\\left(x\\right)={\\mathrm{log}}_{2}\\left(x+3\\right)[\/latex] is [latex]\\left(-3,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Here is another example of how to identify the domain of a logarithmic function.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>What is the domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q275313\">Show Answer<\/span><\/p>\n<div id=\"q275313\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137780875\">The logarithmic function is defined only when the input is positive, so this function is defined when [latex]5 - 2x>0[\/latex]. Solving this inequality,<\/p>\n<div id=\"eip-id1165135470032\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}5 - 2x>0\\hfill & \\text{The input must be positive}.\\hfill \\\\ -2x>-5\\hfill & \\text{Subtract }5.\\hfill \\\\ x<\\frac{5}{2}\\hfill & \\text{Divide by }-2\\text{ and switch the inequality}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137656879\">The domain of [latex]f\\left(x\\right)=\\mathrm{log}\\left(5 - 2x\\right)[\/latex] is [latex]\\left(-\\infty ,\\frac{5}{2}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>\u00a0Graph logarithmic functions<\/h2>\n<p id=\"fs-id1165135194555\">Creating a graphical representation of most functions\u00a0gives us another layer of insight for predicting future events. How do logarithmic graphs give us insight into situations? Because every logarithmic function is the inverse function of an exponential function, we can think of every output on a logarithmic graph as the input for the corresponding inverse exponential equation. In other words, logarithms give the <em data-effect=\"italics\">cause<\/em> for an <em data-effect=\"italics\">effect<\/em>.<\/p>\n<p id=\"fs-id1165137603580\">To illustrate, suppose we invest $2500 in an account that offers an annual interest rate of 5%, compounded continuously. We already know that the balance in our account for any year <em>t<\/em>\u00a0can be found with the equation [latex]A=2500{e}^{0.05t}[\/latex].<\/p>\n<p>But what if we wanted to know the year for any balance? We would need to create a corresponding new function by interchanging the input and the output; thus we would need to create a logarithmic model for this situation. By graphing the model, we can see the output (year) for any input (account balance). For instance, what if we wanted to know how many years it would take for our initial investment to double? Figure 1\u00a0shows this point on the logarithmic graph.<\/p>\n<div style=\"width: 910px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051930\/CNX_Precalc_Figure_04_04_0012.jpg\" alt=\"A graph titled,\" width=\"900\" height=\"459\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165134104063\">Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] along with all its transformations: shifts, stretches, compressions, and reflections.<\/p>\n<p id=\"fs-id1165137679088\">We begin with the parent function [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. Because every logarithmic function of this form is the inverse of an exponential function with the form [latex]y={b}^{x}[\/latex], their graphs will be reflections of each other across the line [latex]y=x[\/latex]. To illustrate this, we can observe the relationship between the input and output values of [latex]y={2}^{x}[\/latex] and its equivalent [latex]x={\\mathrm{log}}_{2}\\left(y\\right)[\/latex] in the table below.<\/p>\n<table style=\"width: 70%;\" summary=\"Three rows and eight columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>\u20133<\/td>\n<td>\u20132<\/td>\n<td>\u20131<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]{2}^{x}=y[\/latex]<\/strong><\/td>\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]{\\mathrm{log}}_{2}\\left(y\\right)=x[\/latex]<\/strong><\/td>\n<td>\u20133<\/td>\n<td>\u20132<\/td>\n<td>\u20131<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165135509175\">Using the inputs and outputs from the table above, we can build another table to observe the relationship between points on the graphs of the inverse functions [latex]f\\left(x\\right)={2}^{x}[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/p>\n<table style=\"width: 70%;\" summary=\"Two rows and eight columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\n<td>[latex]\\left(-3,\\frac{1}{8}\\right)[\/latex]<\/td>\n<td>[latex]\\left(-2,\\frac{1}{4}\\right)[\/latex]<\/td>\n<td>[latex]\\left(-1,\\frac{1}{2}\\right)[\/latex]<\/td>\n<td>[latex]\\left(0,1\\right)[\/latex]<\/td>\n<td>[latex]\\left(1,2\\right)[\/latex]<\/td>\n<td>[latex]\\left(2,4\\right)[\/latex]<\/td>\n<td>[latex]\\left(3,8\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex]<\/strong><\/td>\n<td>[latex]\\left(\\frac{1}{8},-3\\right)[\/latex]<\/td>\n<td>[latex]\\left(\\frac{1}{4},-2\\right)[\/latex]<\/td>\n<td>[latex]\\left(\\frac{1}{2},-1\\right)[\/latex]<\/td>\n<td>[latex]\\left(1,0\\right)[\/latex]<\/td>\n<td>[latex]\\left(2,1\\right)[\/latex]<\/td>\n<td>[latex]\\left(4,2\\right)[\/latex]<\/td>\n<td>[latex]\\left(8,3\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137761335\">As we\u2019d expect, the <em data-effect=\"italics\">x<\/em>&#8211; and <em data-effect=\"italics\">y<\/em>-coordinates are reversed for the inverse functions. The figure below\u00a0shows the graph of <em>f<\/em>\u00a0and <em>g<\/em>.<\/p>\n<figure class=\"small\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051932\/CNX_Precalc_Figure_04_04_0022.jpg\" alt=\"Graph of two functions, f(x)=2^x and g(x)=log_2(x), with the line y=x denoting the axis of symmetry.\" data-media-type=\"image\/jpg\" \/><\/figure>\n<p style=\"text-align: center;\"><strong>\u00a0<\/strong>Notice that the graphs of [latex]f\\left(x\\right)={2}^{x}[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex] are reflections about the line <em>y\u00a0<\/em>= <em>x<\/em>.<\/p>\n<p id=\"fs-id1165137406913\">Observe the following from the graph:<\/p>\n<ul id=\"fs-id1165137408405\">\n<li>[latex]f\\left(x\\right)={2}^{x}[\/latex] has a <em data-effect=\"italics\">y<\/em>-intercept at [latex]\\left(0,1\\right)[\/latex] and [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex] has an <em data-effect=\"italics\">x<\/em>-intercept at [latex]\\left(1,0\\right)[\/latex].<\/li>\n<li>The domain of [latex]f\\left(x\\right)={2}^{x}[\/latex], [latex]\\left(-\\infty ,\\infty \\right)[\/latex], is the same as the range of [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/li>\n<li>The range of [latex]f\\left(x\\right)={2}^{x}[\/latex], [latex]\\left(0,\\infty \\right)[\/latex], is the same as the domain of [latex]g\\left(x\\right)={\\mathrm{log}}_{2}\\left(x\\right)[\/latex].<\/li>\n<\/ul>\n<h3 class=\"title\" data-type=\"title\">A General Note: Characteristics of the Graph of the Parent Function, <em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = log<sub><em data-effect=\"italics\">b<\/em><\/sub>(<em data-effect=\"italics\">x<\/em>)<\/h3>\n<p id=\"fs-id1165135520250\">For any real number <em>x<\/em>\u00a0and constant <em>b\u00a0<\/em>&gt; 0, [latex]b\\ne 1[\/latex], we can see the following characteristics in the graph of [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]:<\/p>\n<ul id=\"fs-id1165137400150\">\n<li>one-to-one function<\/li>\n<li>vertical asymptote: <em>x\u00a0<\/em>= 0<\/li>\n<li>domain: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\n<li>range: [latex]\\left(-\\infty ,\\infty \\right)[\/latex]<\/li>\n<li><em data-effect=\"italics\">x-<\/em>intercept: [latex]\\left(1,0\\right)[\/latex] and key point [latex]\\left(b,1\\right)[\/latex]<\/li>\n<li><em data-effect=\"italics\">y<\/em>-intercept: none<\/li>\n<li>increasing if [latex]b>1[\/latex]<\/li>\n<li>decreasing if 0 &lt; <em>b\u00a0<\/em>&lt; 1<\/li>\n<\/ul>\n<figure id=\"CNX_Precalc_Figure_04_04_003\">\n<div style=\"width: 834px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051934\/CNX_Precalc_Figure_04_04_003G2.jpg\" alt=\"Two graphs of the function f(x)=log_b(x) with points (1,0) and (b, 1). The first graph shows the line when b&gt;1, and the second graph shows the line when 0&lt;b&lt;1.\" width=\"824\" height=\"367\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<\/figure>\n<p>Figure 3\u00a0shows how changing the base <em>b<\/em>\u00a0in [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex] can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. (<em data-effect=\"italics\">Note:<\/em> recall that the function [latex]\\mathrm{ln}\\left(x\\right)[\/latex] has base [latex]e\\approx \\text{2}.\\text{718.)}[\/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\/121\/2016\/08\/05051936\/CNX_Precalc_Figure_04_04_0042.jpg\" alt=\"Graph of three equations: y=log_2(x) in blue, y=ln(x) in orange, and y=log(x) in red. The y-axis is the asymptote.\" width=\"487\" height=\"363\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 4.\u00a0<\/strong>The graphs of three logarithmic functions with different bases, all greater than 1.<\/p>\n<\/div>\n<p>In our first example we will graph a logarithmic function of the form\u00a0[latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Graph [latex]f\\left(x\\right)={\\mathrm{log}}_{5}\\left(x\\right)[\/latex]. State the domain, range.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q486007\">Show Answer<\/span><\/p>\n<div id=\"q486007\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137501970\">Before graphing, identify the behavior and key points for the graph.<\/p>\n<ul id=\"fs-id1165135497154\">\n<li>Since <em>b\u00a0<\/em>= 5 is greater than one, we know the function is increasing. The left tail of the graph will approach the vertical line\u00a0<em>x\u00a0<\/em>= 0, and the right tail will increase slowly without bound.<\/li>\n<li>The <em data-effect=\"italics\">x<\/em>-intercept is [latex]\\left(1,0\\right)[\/latex].<\/li>\n<li>The key point [latex]\\left(5,1\\right)[\/latex] is on the graph.<\/li>\n<li>We plot and label the points, and draw a smooth curve through the points.<\/li>\n<\/ul>\n<figure id=\"CNX_Precalc_Figure_04_04_005\" class=\"small\"><span id=\"fs-id1165135508394\" data-type=\"media\" data-alt=\"Graph of f(x)=log_5(x) with labeled points at (1, 0) and (5, 1). The y-axis is the asymptote.\"><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051938\/CNX_Precalc_Figure_04_04_0052.jpg\" alt=\"Graph of f(x)=log_5(x) with labeled points at (1, 0) and (5, 1). The y-axis is the asymptote.\" data-media-type=\"image\/jpg\" \/><\/span><\/figure>\n<p id=\"fs-id1165135697920\" style=\"text-align: center;\"><strong>Figure 5.\u00a0<\/strong>The domain is [latex]\\left(0,\\infty \\right)[\/latex], the range is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3 id=\"fs-id1165137805513\">How To: Given a logarithmic function with the form [latex]f\\left(x\\right)={\\mathrm{log}}_{b}\\left(x\\right)[\/latex], graph the function.<\/h3>\n<ol id=\"fs-id1165135435529\" data-number-style=\"arabic\">\n<li>Plot the <em data-effect=\"italics\">x-<\/em>intercept, [latex]\\left(1,0\\right)[\/latex].<\/li>\n<li>Plot the key point [latex]\\left(b,1\\right)[\/latex].<\/li>\n<li>Draw a smooth curve through the points.<\/li>\n<li>State the domain, [latex]\\left(0,\\infty \\right)[\/latex], the range, [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\n<\/ol>\n<h2>Summary<\/h2>\n<p>The inverse of a function can be defined for one-to-one functions. \u00a0If a function is not one-to-one, it can be possible to restrict it&#8217;s domain to make it so. The domain of a function will become the range of it&#8217;s inverse. \u00a0The range of a function will become the domain of it&#8217;s inverse. \u00a0Inverses can be verified using tabular data as well as algebraically.<\/p>\n<p>The base <em>b<\/em>\u00a0<strong data-effect=\"bold\">logarithm<\/strong> of a number is the exponent by which we must raise <em>b<\/em>\u00a0to get that number. Logarithmic functions are the inverse of Exponential functions, and it is often easier to understand them through this lens.\u00a0We can never take the logarithm of a negative number, therefore\u00a0[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y[\/latex] is defined for [latex]b>0[\/latex].<\/p>\n<p>Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally because the logarithm is an exponent. \u00a0Logarithms most commonly sue base 10, and often use base\u00a0<em>e.\u00a0<\/em>Logarithms can also be evaluated with most kinds of calculator.<\/p>\n<p>To define the domain of a logarithmic function algebraically, set the argument greater than zero and solve. To plot a logarithmic function, it is easiest to find and plot the x-intercept, and the key point[latex]\\left(b,1\\right)[\/latex].<\/p>\n<h2><\/h2>\n<h2>Earthquake!<\/h2>\n<div id=\"attachment_3656\" style=\"width: 210px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3656\" class=\"size-medium wp-image-3656\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05230800\/AlaskaQuake-Tire-200x300.jpg\" alt=\"Image of A 2x6 plank driven through a ten-ply tire by the tsunami in Whittier.\" width=\"200\" height=\"300\" \/><\/p>\n<p id=\"caption-attachment-3656\" class=\"wp-caption-text\">A 2&#215;6 plank driven through a ten-ply tire by the tsunami in Whittier, AK.<\/p>\n<\/div>\n<p>Joan&#8217;s grandpa comes over for dinner every Sunday. \u00a0Grandpa loves to talk, but it&#8217;s really hard to understand him. Also, as he gets older his hearing has been getting worse. Usually Joan tries to find an excuse to be out after dinner on Sundays, but since she had food poisoning on Saturday night, this week she wasn&#8217;t feeling like leaving the house.<\/p>\n<p>Grandpa was feeling extra chatty this week and since Joan was a captive audience, she got the full force of his attention. His stories started with the neighbor&#8217;s dog who won&#8217;t stop barking, then on to his arthritis.<\/p>\n<p>&#8220;Joanie, do you know that I think I have arthritis in my ankles because of an earthquake injury?&#8221; says Grandpa.<\/p>\n<p>Grandpa went on to tell Joan the story of when he lived in Anchorage, Alaska in the 60&#8217;s and lived through the\u00a0Good Friday earthquake.<\/p>\n<p>The <b>1964 Alaskan earthquake<\/b>, also known as the <b>Great Alaskan earthquake<\/b>, occurred at 5:36 p.m. AST on Good Friday, March 27.\u00a0Across south-central Alaska, ground fissures, collapsing structures, and earthquake-created tsunamis caused about 139 deaths, some of them being friends of Joan&#8217;s grandpa.<\/p>\n<p>Lasting four minutes and thirty-eight seconds, the magnitude 8.5 (Richter scale) <a class=\"footnote\" title=\"&quot;Facts About the 1964 Alaska Earthquake.&quot; LiveScience. Accessed August 18, 2016. http:\/\/www.livescience.com\/44412-1964-alaska-earthquake-facts.html.\" id=\"return-footnote-3518-5\" href=\"#footnote-3518-5\" aria-label=\"Footnote 5\"><sup class=\"footnote\">[5]<\/sup><\/a>megathrust earthquake was the most powerful recorded in North American history, and the second most powerful recorded in world history. Soil liquefaction, fissures, landslides, and other ground failures caused major structural damage in several communities and much damage to property.\u00a0Post-quake tsunamis severely affected numerous\u00a0Alaskan communities, as well as people and property in British Columbia, Washington, Oregon, and California.<span style=\"font-size: 13.3333px; line-height: 20px;\">\u00a0<\/span>Tsunamis also caused damage in Hawai&#8217;i and Japan. Evidence of motion directly related to the earthquake was reported from all over the world.<\/p>\n<p>Surviving the quake inspired Joan&#8217;s grandpa to study Geology in college. \u00a0Joan had heard of the Richter scale before, but didn&#8217;t really understand what it meant, so she asked her grandpa.<\/p>\n<div id=\"attachment_3654\" style=\"width: 200px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3654\" class=\"size-medium wp-image-3654\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05230131\/CharlesRichter-190x300.jpg\" alt=\"Photo of Charles Richter\" width=\"190\" height=\"300\" \/><\/p>\n<p id=\"caption-attachment-3654\" class=\"wp-caption-text\">Charles Richter<\/p>\n<\/div>\n<p>He explained that the Richter scale magnitude is expressed in whole numbers and decimal fractions. For example, a magnitude 5.3 might be computed for a moderate earthquake, and a strong earthquake might be rated as magnitude 6.3. Because of the logarithmic basis of the scale, he said, each whole number increase in magnitude represents a tenfold increase in measured amplitude. For example, Grandpa explained, a magnitude 6.3 earthquake\u00a0corresponds to the release of about 31 times more energy than a magnitude 5.3.<a class=\"footnote\" title=\"&quot;Measuring the Size of an Earthquake.&quot; Measuring the Size of an Earthquake. Accessed August 05, 2016. http:\/\/earthquake.usgs.gov\/learn\/topics\/measure.php.\" id=\"return-footnote-3518-6\" href=\"#footnote-3518-6\" aria-label=\"Footnote 6\"><sup class=\"footnote\">[6]<\/sup><\/a><\/p>\n<p>Joan really enjoyed the conversation she had with her grandpa. She had never asked him about his work as a geologist, and she could tell it made him really happy that she did. At least something good emerged out of\u00a0Joan&#8217;s disastrous date and food poisoning fiasco from the night before.<\/p>\n<p>&#8220;1964 Quake: The Great Alaska Earthquake&#8221; is an eleven minute video created as part of USGS activities acknowledging the fifty year anniversary of the quake on March 27, 2014.\u00a0The video features USGS geologist George Plafker, who, in the 1960s, correctly interpreted the quake as a subduction zone event. This was a great leap forward in resolving key mechanisms of the developing theory of plate tectonics. The extreme loss of life and destruction from this earthquake and accompanying tsunamis inspired\u00a0the NOAA Tsunami Warning Centers and the USGS Earthquake Hazards Program.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-9\" title=\"1964 Quake: The Great Alaska Earthquake\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/lE2j10xyOgI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/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-3518\">\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>Ex 1: Determine if Two Functions Are Inverses. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/vObCvTOatfQ\">https:\/\/youtu.be\/vObCvTOatfQ<\/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>Ex 2: Determine if Two Functions Are Inverses. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/hzehBtNmw08\">https:\/\/youtu.be\/hzehBtNmw08<\/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>Ex: Write Exponential Equations as Logarithmic Equations. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/9_GPPUWEJQQ\">https:\/\/youtu.be\/9_GPPUWEJQQ<\/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>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><li>Ex 1: Evaluate Logarithms Without a Calculator - Whole Numbers. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/dxj5J9OpWGA\">https:\/\/youtu.be\/dxj5J9OpWGA<\/a>. <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>Ex 1: Composition of Function. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/r_LssVS4NHk\">https:\/\/youtu.be\/r_LssVS4NHk<\/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>Ex: Function and Inverse Function Values. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/IR_1L1mnpvw\">https:\/\/youtu.be\/IR_1L1mnpvw<\/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>Ex: Write Logarithmic Equations as Exponential Equations. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/q9_s0wqhIXU\">https:\/\/youtu.be\/q9_s0wqhIXU<\/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>Ex: Evaluate Natural Logarithms on the Calculator. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Rpounu3epSc\">https:\/\/youtu.be\/Rpounu3epSc<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/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><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-3518-1\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary\" target=\"_blank\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/#summary<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-3518-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-3518-2\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary\" target=\"_blank\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#summary<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-3518-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><li id=\"footnote-3518-3\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/\" target=\"_blank\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2010\/us2010rja6\/<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-3518-3\" class=\"return-footnote\" aria-label=\"Return to footnote 3\">&crarr;<\/a><\/li><li id=\"footnote-3518-4\"><a href=\"http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details\" target=\"_blank\">http:\/\/earthquake.usgs.gov\/earthquakes\/eqinthenews\/2011\/usc0001xgp\/#details<\/a>. Accessed 3\/4\/2013. <a href=\"#return-footnote-3518-4\" class=\"return-footnote\" aria-label=\"Return to footnote 4\">&crarr;<\/a><\/li><li id=\"footnote-3518-5\">\"Facts About the 1964 Alaska Earthquake.\" LiveScience. Accessed August 18, 2016. http:\/\/www.livescience.com\/44412-1964-alaska-earthquake-facts.html. <a href=\"#return-footnote-3518-5\" class=\"return-footnote\" aria-label=\"Return to footnote 5\">&crarr;<\/a><\/li><li id=\"footnote-3518-6\">\"Measuring the Size of an Earthquake.\" Measuring the Size of an Earthquake. Accessed August 05, 2016. http:\/\/earthquake.usgs.gov\/learn\/topics\/measure.php. <a href=\"#return-footnote-3518-6\" class=\"return-footnote\" aria-label=\"Return to footnote 6\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":21,"menu_order":3,"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\":\"cc\",\"description\":\"Ex 1: Composition of Function\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/r_LssVS4NHk\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Function and Inverse Function Values\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/IR_1L1mnpvw\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex 1: Determine if Two Functions Are Inverses\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/vObCvTOatfQ\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex 2: Determine if Two Functions Are Inverses\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/hzehBtNmw08\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Write Logarithmic Equations as Exponential Equations\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/q9_s0wqhIXU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex: Write Exponential Equations as Logarithmic Equations\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/9_GPPUWEJQQ\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Ex 1: Evaluate Logarithms Without a Calculator - 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