{"id":3542,"date":"2016-08-05T05:30:11","date_gmt":"2016-08-05T05:30:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&p=3542"},"modified":"2019-07-24T21:33:50","modified_gmt":"2019-07-24T21:33:50","slug":"identify-the-domain-of-a-logarithmic-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/identify-the-domain-of-a-logarithmic-function\/","title":{"raw":"Quotient and Power Rules for Logarithms","rendered":"Quotient and Power Rules for Logarithms"},"content":{"raw":"
\r\n
\r\n

Learning Outcome<\/h3>\r\n
    \r\n \t
  • Define and use the quotient and power rules for logarithms<\/li>\r\n<\/ul>\r\n<\/div>\r\n

    For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents<\/em> to simplify division of like bases raised to powers by subtracting the exponents: [latex]\\frac{x^a}{x^b}={x}^{a-b}[\/latex]. The quotient rule for logarithms<\/strong> says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.<\/p>\r\n\r\n

    \r\n

    \u00a0The Quotient Rule for Logarithms<\/h3>\r\n

    The quotient rule for logarithms<\/strong> can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/div>\r\n<\/div>\r\n<\/section>\r\n
    <\/div>\r\n
    \r\n
    We can show\u00a0[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\r\n<\/div>\r\n
    <\/div>\r\n
    Given\u00a0positive real numbers M<\/em>, N<\/em>, and b<\/em>, where [latex]b>0[\/latex] we will show<\/div>\r\n
    \r\n
    [latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\r\n

    Let [latex]m={\\mathrm{log}}_{b}M[\/latex] and [latex]n={\\mathrm{log}}_{b}N[\/latex]. In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N[\/latex]. It follows that<\/p>\r\n\r\n

    [latex]\\begin{array}{c}{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(\\frac{{b}^{m}}{{b}^{n}}\\right)\\hfill & \\text{Substitute for }M\\text{ and }N.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left({b}^{m-n}\\right)\\hfill & \\text{Apply the quotient rule for exponents}.\\hfill \\\\ \\hfill & =m-n\\hfill & \\text{Apply the inverse property of logs}.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)\\hfill & \\text{Substitute for }m\\text{ and }n.\\hfill \\end{array}[\/latex]<\/div>\r\n
    \r\n
    \r\n

    Example<\/h3>\r\nExpand the following expression using the quotient rule for logarithms.\r\n\r\n[latex]\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right)[\/latex]\r\n[reveal-answer q=\"942932\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"942932\"]\r\n

    Factoring and canceling, we get:<\/p>\r\n\r\n

    [latex]\\begin{array}{l}\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right) & =\\mathrm{log}\\left(\\frac{2x\\left(x+3\\right)}{3\\left(x+3\\right)}\\right)\\hfill & \\text{Factor the numerator and denominator}.\\hfill \\\\ & =\\mathrm{log}\\left(\\frac{2x}{3}\\right)\\hfill & \\text{Cancel the common factors}.\\hfill \\end{array}[\/latex]<\/div>\r\n
    <\/div>\r\n
    \r\n

    Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.<\/p>\r\n\r\n

    [latex]\\begin{array}{l}\\mathrm{log}\\left(\\frac{2x}{3}\\right) & =\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\\\ & =\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the previous example, it was helpful to first factor the numerator and denominator and divide common terms. This gave us a simpler expression to use to write an equivalent expression. It is important to remember to subtract the\u00a0logarithm of the denominator from the logarithm of the numerator. Always check to see if you need to expand\u00a0further with the product rule.\r\n\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n

    Example<\/h3>\r\nExpand [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)[\/latex].\r\n[reveal-answer q=\"293418\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"293418\"]\r\n

    First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.<\/p>\r\n

    [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)={\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right)[\/latex]<\/p>\r\n

    Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor\u00a0[latex]15[\/latex] are\u00a0[latex]3[\/latex] and [latex]5[\/latex]<\/p>\r\n[latex]\\begin{array}{l}{\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right) & = \\left[{\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)\\right]-\\left[{\\mathrm{log}}_{2}\\left(3x+4\\right)+{\\mathrm{log}}_{2}\\left(2-x\\right)\\right]\\hfill \\\\ & ={\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)-{\\mathrm{log}}_{2}\\left(3x+4\\right)-{\\mathrm{log}}_{2}\\left(2-x\\right)\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    \u00a0Analysis of the Solution<\/h2>\r\n

    There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\\frac{4}{3}[\/latex] and\u00a0[latex]x=2[\/latex]. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that\u00a0[latex]x\\gt0[\/latex],\u00a0[latex]x\\gt1[\/latex], [latex]x\\gt-\\frac{4}{3}[\/latex], and\u00a0[latex]x\\lt2[\/latex]. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.<\/p>\r\nIn the following video, we show more examples of using the quotient rule for logarithms.\r\n\r\n[embed]https:\/\/www.youtube.com\/watch?v=89SGlxXO2rI&feature=youtu.be[\/embed]\r\n

    \r\n

    Using the Power Rule for Logarithms<\/h2>\r\n

    We have explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as [latex]{x}^{2}[\/latex]? One method is as follows:<\/p>\r\n\r\n

    [latex]\\begin{array}{c}{\\mathrm{log}}_{b}\\left({x}^{2}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(x\\cdot x\\right)\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}x\\hfill \\\\ \\hfill & =2{\\mathrm{log}}_{b}x\\hfill \\end{array}[\/latex]<\/div>\r\n

    Notice that we used the product rule for logarithms<\/strong> to simplify\u00a0the example above. By doing so, we have derived the power rule for logarithms\u00a0<\/strong>which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,<\/p>\r\n\r\n

    [latex]\\begin{array}{c}100={10}^{2}\\hfill & \\sqrt{3}={3}^{\\frac{1}{2}}\\hfill & \\frac{1}{e}={e}^{-1}\\hfill \\end{array}[\/latex]<\/div>\r\n
    \r\n

    \u00a0The Power Rule for Logarithms<\/h3>\r\n

    The power rule for logarithms<\/strong> can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{b}\\left({M}^{n}\\right)=n{\\mathrm{log}}_{b}M[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n

    Example<\/h3>\r\nExpand [latex]{\\mathrm{log}}_{2}{x}^{5}[\/latex].\r\n[reveal-answer q=\"342931\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"342931\"]\r\n

    The argument is already written as a power, so we identify the exponent,\u00a0[latex]5[\/latex], and the base, x<\/em>, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right)=5{\\mathrm{log}}_{2}x[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe power rule for logarithms is possible because we can use the product rule and combine like terms. In the next example, you will see that we can also rewrite an expression as a power in order to\u00a0use the power rule.\r\n\r\n<\/div>\r\n
    \r\n

    Example<\/h3>\r\nExpand [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.\r\n[reveal-answer q=\"621660\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"621660\"]\r\n

    Expressing the argument as a power, we get [latex]{\\mathrm{log}}_{3}\\left(25\\right)={\\mathrm{log}}_{3}\\left({5}^{2}\\right)[\/latex].<\/p>\r\n

    Next we identify the exponent,\u00a0[latex]2[\/latex], and the base,\u00a0[latex]5[\/latex], and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{3}\\left({5}^{2}\\right)=2{\\mathrm{log}}_{3}\\left(5\\right)[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNow let us use the power rule in reverse.\r\n
    \r\n

    Example<\/h3>\r\nRewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of\u00a0[latex]1[\/latex].\r\n[reveal-answer q=\"983893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"983893\"]\r\n

    Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\\mathrm{ln}\\left(x\\right)[\/latex], we identify the factor,\u00a0[latex]4[\/latex], as the exponent and the argument, x<\/em>, as the base, and rewrite the product as a logarithm of a power:<\/p>\r\n[latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    \u00a0Summary<\/h2>\r\n

    You can use the quotient rule of logarithms to write an equivalent difference of logarithms in the following way:<\/p>\r\n\r\n

      \r\n \t
    1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.<\/li>\r\n \t
    2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.<\/li>\r\n \t
    3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.<\/li>\r\n<\/ol>\r\n

      To use the power rule of logarithms to write an equivalent product of a factor and a logarithm, consider the following:<\/p>\r\n\r\n

        \r\n \t
      1. Express the argument as a power, if needed.<\/li>\r\n \t
      2. Write the equivalent expression by multiplying the exponent times the logarithm of the base.<\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n<\/section>","rendered":"
        \n
        \n

        Learning Outcome<\/h3>\n
          \n
        • Define and use the quotient and power rules for logarithms<\/li>\n<\/ul>\n<\/div>\n

          For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents<\/em> to simplify division of like bases raised to powers by subtracting the exponents: [latex]\\frac{x^a}{x^b}={x}^{a-b}[\/latex]. The quotient rule for logarithms<\/strong> says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.<\/p>\n

          \n

          \u00a0The Quotient Rule for Logarithms<\/h3>\n

          The quotient rule for logarithms<\/strong> can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.<\/p>\n

          [latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/div>\n<\/div>\n<\/section>\n
          <\/div>\n
          \n
          We can show\u00a0[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\n<\/div>\n
          <\/div>\n
          Given\u00a0positive real numbers M<\/em>, N<\/em>, and b<\/em>, where [latex]b>0[\/latex] we will show<\/div>\n
          \n
          [latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\text{=}{\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)[\/latex].<\/div>\n

          Let [latex]m={\\mathrm{log}}_{b}M[\/latex] and [latex]n={\\mathrm{log}}_{b}N[\/latex]. In exponential form, these equations are [latex]{b}^{m}=M[\/latex] and [latex]{b}^{n}=N[\/latex]. It follows that<\/p>\n

          [latex]\\begin{array}{c}{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(\\frac{{b}^{m}}{{b}^{n}}\\right)\\hfill & \\text{Substitute for }M\\text{ and }N.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left({b}^{m-n}\\right)\\hfill & \\text{Apply the quotient rule for exponents}.\\hfill \\\\ \\hfill & =m-n\\hfill & \\text{Apply the inverse property of logs}.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left(M\\right)-{\\mathrm{log}}_{b}\\left(N\\right)\\hfill & \\text{Substitute for }m\\text{ and }n.\\hfill \\end{array}[\/latex]<\/div>\n
          \n
          \n

          Example<\/h3>\n

          Expand the following expression using the quotient rule for logarithms.<\/p>\n

          [latex]\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right)[\/latex]<\/p>\n

          Show Solution<\/span><\/p>\n
          \n

          Factoring and canceling, we get:<\/p>\n

          [latex]\\begin{array}{l}\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right) & =\\mathrm{log}\\left(\\frac{2x\\left(x+3\\right)}{3\\left(x+3\\right)}\\right)\\hfill & \\text{Factor the numerator and denominator}.\\hfill \\\\ & =\\mathrm{log}\\left(\\frac{2x}{3}\\right)\\hfill & \\text{Cancel the common factors}.\\hfill \\end{array}[\/latex]<\/div>\n
          <\/div>\n
          \n

          Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.<\/p>\n

          [latex]\\begin{array}{l}\\mathrm{log}\\left(\\frac{2x}{3}\\right) & =\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\\\ & =\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

          In the previous example, it was helpful to first factor the numerator and denominator and divide common terms. This gave us a simpler expression to use to write an equivalent expression. It is important to remember to subtract the\u00a0logarithm of the denominator from the logarithm of the numerator. Always check to see if you need to expand\u00a0further with the product rule.<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          Example<\/h3>\n

          Expand [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)[\/latex].<\/p>\n

          Show Solution<\/span><\/p>\n
          \n

          First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.<\/p>\n

          [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)={\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right)[\/latex]<\/p>\n

          Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor\u00a0[latex]15[\/latex] are\u00a0[latex]3[\/latex] and [latex]5[\/latex]<\/p>\n

          [latex]\\begin{array}{l}{\\mathrm{log}}_{2}\\left(15x\\left(x - 1\\right)\\right)-{\\mathrm{log}}_{2}\\left(\\left(3x+4\\right)\\left(2-x\\right)\\right) & = \\left[{\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)\\right]-\\left[{\\mathrm{log}}_{2}\\left(3x+4\\right)+{\\mathrm{log}}_{2}\\left(2-x\\right)\\right]\\hfill \\\\ & ={\\mathrm{log}}_{2}\\left(3\\right)+{\\mathrm{log}}_{2}\\left(5\\right)+{\\mathrm{log}}_{2}\\left(x\\right)+{\\mathrm{log}}_{2}\\left(x - 1\\right)-{\\mathrm{log}}_{2}\\left(3x+4\\right)-{\\mathrm{log}}_{2}\\left(2-x\\right)\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          \u00a0Analysis of the Solution<\/h2>\n

          There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\\frac{4}{3}[\/latex] and\u00a0[latex]x=2[\/latex]. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that\u00a0[latex]x\\gt0[\/latex],\u00a0[latex]x\\gt1[\/latex], [latex]x\\gt-\\frac{4}{3}[\/latex], and\u00a0[latex]x\\lt2[\/latex]. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.<\/p>\n

          In the following video, we show more examples of using the quotient rule for logarithms.<\/p>\n