{"id":11310,"date":"2015-07-14T19:40:06","date_gmt":"2015-07-14T19:40:06","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=11310"},"modified":"2021-12-29T19:16:49","modified_gmt":"2021-12-29T19:16:49","slug":"logarithmic-vs-exponential-form","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/logarithmic-vs-exponential-form\/","title":{"raw":"Logarithmic vs Exponential form of equation","rendered":"Logarithmic vs Exponential form of equation"},"content":{"raw":"
\r\n

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 x<\/em>\u00a0represents the difference in magnitudes on the Richter Scale<\/strong>. How would we solve for\u00a0x<\/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 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\"]\"Graph Figure 2<\/b>[\/caption]\r\n

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 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 x<\/em>\u00a0and y<\/em>\u00a0and solve for y<\/em>\u00a0to find the inverse function. To represent y<\/em>\u00a0as a function of x<\/em>, we use a logarithmic function of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. The base b<\/em>\u00a0logarithm<\/strong> of a number is the exponent by which we must raise b<\/em>\u00a0to get that number.<\/p>\r\n

We read a logarithmic expression as, \"The logarithm with base b<\/em>\u00a0of x<\/em>\u00a0is equal to y<\/em>,\" or, simplified, \"log base b<\/em>\u00a0of x<\/em>\u00a0is y<\/em>.\" We can also say, \"b<\/em>\u00a0raised to the power of y<\/em>\u00a0is 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

We can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\r\n\r\n

[latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b>0,b\\ne 1[\/latex]<\/div>\r\n

Note that the base b<\/em>\u00a0is always positive.\r\n\"Think<\/span><\/p>\r\n

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 x<\/em>.<\/p>\r\n

We can illustrate the notation of logarithms as follows:\r\n\"logb<\/span><\/p>\r\n

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

\r\n

A General Note: Definition of the Logarithmic Function<\/h3>\r\n

A logarithm<\/strong> base b<\/em>\u00a0of a positive number x<\/em>\u00a0satisfies the following definition.<\/p>\r\n

For [latex]x>0,b>0,b\\ne 1[\/latex],<\/p>\r\n\r\n

[latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equivalent to }{b}^{y}=x[\/latex]<\/div>\r\n

where,<\/p>\r\n\r\n

    \r\n \t
  • we read [latex]{\\mathrm{log}}_{b}\\left(x\\right)[\/latex] as, \"the logarithm with base b<\/em>\u00a0of x<\/em>\" or the \"log base b<\/em>\u00a0of x<\/em>.\"<\/li>\r\n \t
  • the logarithm y<\/em>\u00a0is the exponent to which b<\/em>\u00a0must be raised to get x<\/em>.<\/li>\r\n<\/ul>\r\n

    Also, since the logarithmic and exponential functions switch the x<\/em>\u00a0and y<\/em>\u00a0values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,<\/p>\r\n\r\n

      \r\n \t
    • the domain of the logarithm function with base [latex]b \\text{ is} \\left(0,\\infty \\right)[\/latex].<\/li>\r\n \t
    • the range of the logarithm function with base [latex]b \\text{ is} \\left(-\\infty ,\\infty \\right)[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n
      \r\n

      Q & A<\/h3>\r\n

      Can we take the logarithm of a negative number?<\/strong><\/p>\r\n

      No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.<\/em><\/p>\r\n\r\n<\/div>\r\n

      Converting from logarithmic to exponential form<\/h3>\r\n
      \r\n

      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
        \r\n \t
      1. Examine the equation [latex]y={\\mathrm{log}}_{b}x[\/latex] and identify b<\/em>, y<\/em>, and x<\/em>.<\/li>\r\n \t
      2. Rewrite [latex]{\\mathrm{log}}_{b}x=y[\/latex] as [latex]{b}^{y}=x[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n
        \r\n

        Example<\/h3>\r\n

        Write the following logarithmic equations in exponential form.<\/p>\r\n\r\n

          \r\n \t
        1. [latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]<\/li>\r\n \t
        2. [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"31307\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"31307\"]\r\n

          First, identify the values of b<\/em>,\u00a0y<\/em>, and\u00a0x<\/em>. Then, write the equation in the form [latex]{b}^{y}=x[\/latex].<\/p>\r\n\r\n

            \r\n \t
          1. [latex]{\\mathrm{log}}_{6}\\left(\\sqrt{6}\\right)=\\frac{1}{2}[\/latex]\r\n

            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

          2. [latex]{\\mathrm{log}}_{3}\\left(9\\right)=2[\/latex]\r\n

            Here, b\u00a0<\/em>= 3, y\u00a0<\/em>= 2, and 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]\r\n\r\n<\/div>\r\n

            \r\n

            Try It<\/h3>\r\n

            Write the following logarithmic equations in exponential form.<\/p>\r\n

            a. [latex]{\\mathrm{log}}_{10}\\left(1,000,000\\right)=6[\/latex]<\/p>\r\n

            b. [latex]{\\mathrm{log}}_{5}\\left(25\\right)=2[\/latex]<\/p>\r\n[reveal-answer q=\"864685\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"864685\"]\r\n\r\na. [latex] {\\mathrm{log}}_{10}\\left(1,000,000\\right)=6 \\; \\text{is equivalent to} \\; {10}^{6}=1,000,000[\/latex]\r\n\r\nb. [latex] {\\mathrm{log}}_{5}\\left(25\\right)=2 \\; \\text{is equivalent to} \\; {5}^{2}=25 [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

            Converting from exponential to logarithmic form<\/h3>\r\n

            To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base b<\/em>, exponent x<\/em>, and output y<\/em>. Then we write [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\r\n\r\n

            \r\n

            Example<\/h3>\r\n

            Write the following exponential equations in logarithmic form.<\/p>\r\n\r\n

              \r\n \t
            1. [latex]{2}^{3}=8[\/latex]<\/li>\r\n \t
            2. [latex]{5}^{2}=25[\/latex]<\/li>\r\n \t
            3. [latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"918184\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"918184\"]\r\n

              First, identify the values of b<\/em>, y<\/em>, and x<\/em>. Then, write the equation in the form [latex]x={\\mathrm{log}}_{b}\\left(y\\right)[\/latex].<\/p>\r\n\r\n

                \r\n \t
              1. [latex]{2}^{3}=8[\/latex]\r\n

                Here, b\u00a0<\/em>= 2, x\u00a0<\/em>= 3, and 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

              2. [latex]{5}^{2}=25[\/latex]\r\n

                Here, b\u00a0<\/em>= 5, x\u00a0<\/em>= 2, and 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

              3. [latex]{10}^{-4}=\\frac{1}{10,000}[\/latex]\r\n

                Here, b\u00a0<\/em>= 10, 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]\r\n\r\n<\/div>\r\n

                \r\n

                Try It<\/h3>\r\n

                Write the following exponential equations in logarithmic form.<\/p>\r\n

                a. [latex]{3}^{2}=9[\/latex]<\/p>\r\n

                b. [latex]{5}^{3}=125[\/latex]<\/p>\r\n

                c. [latex]{2}^{-1}=\\frac{1}{2}[\/latex]<\/p>\r\n\r\n[reveal-answer q=\"871823\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"871823\"]\r\n\r\na. [latex]{3}^{2}=9<\/annotation><\/semantics><\/math><\/span><\/span> \\; \\text{is equivalent to} \\;{\\mathrm{log}}_{3}\\left(9\\right)=2<\/annotation><\/semantics><\/math><\/span><\/span>[\/latex]\r\n\r\nb. [latex]{5}^{3}=125<\/annotation><\/semantics><\/math><\/span><\/span> \\; \\text{is equivalent to} \\;{\\mathrm{log}}_{5}\\left(125\\right)=3<\/annotation><\/semantics><\/math><\/span><\/span> [\/latex]\r\n\r\nc.\u00a0[latex]{2}^{-1}=\\frac{1}{2}\u00a0<\/annotation><\/semantics><\/math><\/span><\/span>\\; \\text{is equivalent to} \\;{\\mathrm{log}}_{2}\\left(\\frac{1}{2}\\right)=-1<\/annotation><\/semantics><\/math><\/annotation><\/semantics><\/math><\/span><\/span>\u00a0[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>","rendered":"
                \n

                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 x<\/em>\u00a0represents the difference in magnitudes on the Richter Scale<\/strong>. How would we solve for\u00a0x<\/em>?<\/p>\n

                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 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

                \"Graph<\/p>\n

                Figure 2<\/b><\/p>\n<\/div>\n

                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 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 x<\/em>\u00a0and y<\/em>\u00a0and solve for y<\/em>\u00a0to find the inverse function. To represent y<\/em>\u00a0as a function of x<\/em>, we use a logarithmic function of the form [latex]y={\\mathrm{log}}_{b}\\left(x\\right)[\/latex]. The base b<\/em>\u00a0logarithm<\/strong> of a number is the exponent by which we must raise b<\/em>\u00a0to get that number.<\/p>\n

                We read a logarithmic expression as, “The logarithm with base b<\/em>\u00a0of x<\/em>\u00a0is equal to y<\/em>,” or, simplified, “log base b<\/em>\u00a0of x<\/em>\u00a0is y<\/em>.” We can also say, “b<\/em>\u00a0raised to the power of y<\/em>\u00a0is 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>\n

                We can express the relationship between logarithmic form and its corresponding exponential form as follows:<\/p>\n

                [latex]{\\mathrm{log}}_{b}\\left(x\\right)=y\\Leftrightarrow {b}^{y}=x,\\text{}b>0,b\\ne 1[\/latex]<\/div>\n

                Note that the base b<\/em>\u00a0is always positive.
                \n\"Think<\/span><\/p>\n

                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 x<\/em>.<\/p>\n

                We can illustrate the notation of logarithms as follows:
                \n\"logb<\/span><\/p>\n

                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

                \n

                A General Note: Definition of the Logarithmic Function<\/h3>\n

                A logarithm<\/strong> base b<\/em>\u00a0of a positive number x<\/em>\u00a0satisfies the following definition.<\/p>\n

                For [latex]x>0,b>0,b\\ne 1[\/latex],<\/p>\n

                [latex]y={\\mathrm{log}}_{b}\\left(x\\right)\\text{ is equivalent to }{b}^{y}=x[\/latex]<\/div>\n

                where,<\/p>\n