{"id":11206,"date":"2015-07-14T18:51:11","date_gmt":"2015-07-14T18:51:11","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=11206"},"modified":"2021-11-15T01:41:42","modified_gmt":"2021-11-15T01:41:42","slug":"use-the-quotient-rule-for-logarithms","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/use-the-quotient-rule-for-logarithms\/","title":{"raw":"Use the quotient and power rules for logarithms","rendered":"Use the quotient and power rules for logarithms"},"content":{"raw":"
\r\n

For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents<\/em> to combine the quotient of exponents by subtracting: [latex]{x}^{\\frac{a}{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

Given any real number x\u00a0<\/em>and positive real numbers M<\/em>, N<\/em>, and b<\/em>, where [latex]b\\ne 1[\/latex], we will show<\/p>\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{cases}{\\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{cases}[\/latex]<\/div>\r\n

For example, to expand [latex]\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right)[\/latex], we must first express the quotient in lowest terms. Factoring and canceling we get,<\/p>\r\n\r\n

[latex]\\begin{cases}\\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 \\\\ & \\text{ }=\\mathrm{log}\\left(\\frac{2x}{3}\\right)\\hfill & \\text{Cancel the common factors}.\\hfill \\end{cases}[\/latex]<\/div>\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{cases}\\mathrm{log}\\left(\\frac{2x}{3}\\right)=\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\\\ \\text{ }=\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\end{cases}[\/latex]<\/div>\r\n
\r\n

A General Note: The 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
\r\n

How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.<\/h3>\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<\/div>\r\n
    \r\n
    \r\n
    \r\n

    Example 2: Using the Quotient Rule for Logarithms<\/h3>\r\n

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

    \r\n

    Solution<\/h3>\r\n

    First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.<\/p>\r\n\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]<\/div>\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 15 are 3 and 5.<\/p>\r\n\r\n

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

    Analysis of the Solution<\/h3>\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 x\u00a0<\/em>= 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that x\u00a0<\/em>> 0, x\u00a0<\/em>> 1, [latex]x>-\\frac{4}{3}[\/latex], and x\u00a0<\/em>< 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n

    Try It 2<\/h3>\r\n

    Expand [latex]{\\mathrm{log}}_{3}\\left(\\frac{7{x}^{2}+21x}{7x\\left(x - 1\\right)\\left(x - 2\\right)}\\right)[\/latex].<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

    \r\n

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

    We\u2019ve 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{cases}{\\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{cases}[\/latex]<\/div>\r\n

    Notice that we used the product rule for logarithms<\/strong> to find a solution for the example above. By doing so, we have derived the power rule for logarithms<\/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{cases}100={10}^{2}\\hfill & \\sqrt{3}={3}^{\\frac{1}{2}}\\hfill & \\frac{1}{e}={e}^{-1}\\hfill \\end{cases}[\/latex]<\/div>\r\n
    \r\n

    A General Note: The 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

    How To: Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.<\/h3>\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<\/div>\r\n
      \r\n
      \r\n
      \r\n

      Example 3: Expanding a Logarithm with Powers<\/h3>\r\n

      Expand [latex]{\\mathrm{log}}_{2}{x}^{5}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      Solution<\/h3>\r\n

      The argument is already written as a power, so we identify the exponent, 5, 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<\/div>\r\n<\/div>\r\n<\/div>\r\n
      \r\n

      Try It 3<\/h3>\r\n

      Expand [latex]\\mathrm{ln}{x}^{2}[\/latex].<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

      \r\n
      \r\n
      \r\n

      Example 4: Rewriting an Expression as a Power before Using the Power Rule<\/h3>\r\n

      Expand [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      Solution<\/h3>\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, 2, and the base, 5, 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<\/div>\r\n<\/div>\r\n<\/div>\r\n
      \r\n

      Try It 4<\/h3>\r\n

      Expand [latex]\\mathrm{ln}\\left(\\frac{1}{{x}^{2}}\\right)[\/latex].<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

      \r\n
      \r\n
      \r\n

      Example 5: Using the Power Rule in Reverse<\/h3>\r\n

      Rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\r\n\r\n<\/div>\r\n

      \r\n

      Solution<\/h3>\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, 4, 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<\/div>\r\n<\/div>\r\n<\/div>\r\n

      \r\n

      Try It 5<\/h3>\r\n

      Rewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section>","rendered":"

      \n

      For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents<\/em> to combine the quotient of exponents by subtracting: [latex]{x}^{\\frac{a}{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

      Given any real number x\u00a0<\/em>and positive real numbers M<\/em>, N<\/em>, and b<\/em>, where [latex]b\\ne 1[\/latex], we will show<\/p>\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{cases}{\\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{cases}[\/latex]<\/div>\n

      For example, to expand [latex]\\mathrm{log}\\left(\\frac{2{x}^{2}+6x}{3x+9}\\right)[\/latex], we must first express the quotient in lowest terms. Factoring and canceling we get,<\/p>\n

      [latex]\\begin{cases}\\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 \\\\ & \\text{ }=\\mathrm{log}\\left(\\frac{2x}{3}\\right)\\hfill & \\text{Cancel the common factors}.\\hfill \\end{cases}[\/latex]<\/div>\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{cases}\\mathrm{log}\\left(\\frac{2x}{3}\\right)=\\mathrm{log}\\left(2x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\\\ \\text{ }=\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(3\\right)\\hfill \\end{cases}[\/latex]<\/div>\n
      \n

      A General Note: The 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
      \n

      How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms.<\/h3>\n
        \n
      1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.<\/li>\n
      2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.<\/li>\n
      3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.<\/li>\n<\/ol>\n<\/div>\n
        \n
        \n
        \n

        Example 2: Using the Quotient Rule for Logarithms<\/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<\/div>\n

        \n

        Solution<\/h3>\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]<\/div>\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 15 are 3 and 5.<\/p>\n

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

        Analysis of the Solution<\/h3>\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 x\u00a0<\/em>= 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that x\u00a0<\/em>> 0, x\u00a0<\/em>> 1, [latex]x>-\\frac{4}{3}[\/latex], and x\u00a0<\/em>< 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

        \n

        Try It 2<\/h3>\n

        Expand [latex]{\\mathrm{log}}_{3}\\left(\\frac{7{x}^{2}+21x}{7x\\left(x - 1\\right)\\left(x - 2\\right)}\\right)[\/latex].<\/p>\n

        Solution<\/a><\/p>\n<\/div>\n

        \n
        \n

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

        We\u2019ve 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>\n

        [latex]\\begin{cases}{\\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{cases}[\/latex]<\/div>\n

        Notice that we used the product rule for logarithms<\/strong> to find a solution for the example above. By doing so, we have derived the power rule for logarithms<\/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>\n

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

        A General Note: The Power Rule for Logarithms<\/h3>\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>\n

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

        How To: Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm.<\/h3>\n
          \n
        1. Express the argument as a power, if needed.<\/li>\n
        2. Write the equivalent expression by multiplying the exponent times the logarithm of the base.<\/li>\n<\/ol>\n<\/div>\n
          \n
          \n
          \n

          Example 3: Expanding a Logarithm with Powers<\/h3>\n

          Expand [latex]{\\mathrm{log}}_{2}{x}^{5}[\/latex].<\/p>\n<\/div>\n

          \n

          Solution<\/h3>\n

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

          [latex]{\\mathrm{log}}_{2}\\left({x}^{5}\\right)=5{\\mathrm{log}}_{2}x[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n

          Try It 3<\/h3>\n

          Expand [latex]\\mathrm{ln}{x}^{2}[\/latex].<\/p>\n

          Solution<\/a><\/p>\n<\/div>\n

          \n
          \n
          \n

          Example 4: Rewriting an Expression as a Power before Using the Power Rule<\/h3>\n

          Expand [latex]{\\mathrm{log}}_{3}\\left(25\\right)[\/latex] using the power rule for logs.<\/p>\n<\/div>\n

          \n

          Solution<\/h3>\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>\n

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

          [latex]{\\mathrm{log}}_{3}\\left({5}^{2}\\right)=2{\\mathrm{log}}_{3}\\left(5\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
          \n

          Try It 4<\/h3>\n

          Expand [latex]\\mathrm{ln}\\left(\\frac{1}{{x}^{2}}\\right)[\/latex].<\/p>\n

          Solution<\/a><\/p>\n<\/div>\n

          \n
          \n
          \n

          Example 5: Using the Power Rule in Reverse<\/h3>\n

          Rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\n<\/div>\n

          \n

          Solution<\/h3>\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, 4, as the exponent and the argument, x<\/em>, as the base, and rewrite the product as a logarithm of a power:<\/p>\n

          [latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          \n

          Try It 5<\/h3>\n

          Rewrite [latex]2{\\mathrm{log}}_{3}4[\/latex] using the power rule for logs to a single logarithm with a leading coefficient of 1.<\/p>\n

          Solution<\/a><\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n\n\t\t\t

          \n\t\t\t

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