{"id":11204,"date":"2015-07-14T18:50:48","date_gmt":"2015-07-14T18:50:48","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=11204"},"modified":"2021-11-15T01:41:37","modified_gmt":"2021-11-15T01:41:37","slug":"use-the-product-rule-for-logarithms","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/use-the-product-rule-for-logarithms\/","title":{"raw":"Use the product rule for logarithms","rendered":"Use the product rule for logarithms"},"content":{"raw":"

Recall that the logarithmic and exponential functions \"undo\" each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.<\/p>\r\n\r\n

[latex]\\begin{cases}{\\mathrm{log}}_{b}1=0\\\\ {\\mathrm{log}}_{b}b=1\\end{cases}[\/latex]<\/div>\r\n

For example, [latex]{\\mathrm{log}}_{5}1=0[\/latex] since [latex]{5}^{0}=1[\/latex]. And [latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5[\/latex].<\/p>\r\n

Next, we have the inverse property.<\/p>\r\n\r\n

[latex]\\begin{cases}\\hfill \\\\ {\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x\\hfill \\\\ \\text{ }{b}^{{\\mathrm{log}}_{b}x}=x,x>0\\hfill \\end{cases}[\/latex]<\/div>\r\n

For example, to evaluate [latex]\\mathrm{log}\\left(100\\right)[\/latex], we can rewrite the logarithm as [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)[\/latex], and then apply the inverse property [latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex] to get [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)=2[\/latex].<\/p>\r\n

To evaluate [latex]{e}^{\\mathrm{ln}\\left(7\\right)}[\/latex], we can rewrite the logarithm as [latex]{e}^{{\\mathrm{log}}_{e}7}[\/latex], and then apply the inverse property [latex]{b}^{{\\mathrm{log}}_{b}x}=x[\/latex] to get [latex]{e}^{{\\mathrm{log}}_{e}7}=7[\/latex].<\/p>\r\n

Finally, we have the one-to-one<\/strong> property.<\/p>\r\n\r\n

[latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/div>\r\n

We can use the one-to-one property to solve the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)={\\mathrm{log}}_{3}\\left(2x+5\\right)[\/latex] for x<\/em>. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for x<\/em>:<\/p>\r\n\r\n

[latex]\\begin{cases}3x=2x+5\\hfill & \\text{Set the arguments equal}\\text{.}\\hfill \\\\ x=5\\hfill & \\text{Subtract 2}x\\text{.}\\hfill \\end{cases}[\/latex]<\/div>\r\n

But what about the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)+{\\mathrm{log}}_{3}\\left(2x+5\\right)=2[\/latex]? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.<\/p>\r\n

Recall that we use the product rule of exponents<\/em> to combine the product of exponents by adding: [latex]{x}^{a}{x}^{b}={x}^{a+b}[\/latex]. We have a similar property for logarithms, called the product rule for logarithms<\/strong>, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.<\/p>\r\n

Given any real number x<\/em>\u00a0and 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(MN\\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(MN\\right)\\hfill & ={\\mathrm{log}}_{b}\\left({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 product 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

Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider [latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)[\/latex]. Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:<\/p>\r\n\r\n

[latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)={\\mathrm{log}}_{b}w+{\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}y+{\\mathrm{log}}_{b}z[\/latex]<\/div>\r\n
\r\n

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

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

[latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\text{ for }b>0[\/latex]<\/div>\r\n<\/div>\r\n
\r\n

How To: Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.<\/h3>\r\n
    \r\n \t
  1. Factor the argument completely, expressing each whole number factor as a product of primes.<\/li>\r\n \t
  2. Write the equivalent expression by summing the logarithms of each factor.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
    \r\n
    \r\n
    \r\n

    Example 1: Using the Product Rule for Logarithms<\/h3>\r\n

    Expand [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Solution<\/h3>\r\n

    We begin by factoring the argument completely, expressing 30 as a product of primes.<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\cdot 3\\cdot 5\\cdot x\\cdot \\left(3x+4\\right)\\right)[\/latex]<\/div>\r\n

    Next we write the equivalent equation by summing the logarithms of each factor.<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\right)+{\\mathrm{log}}_{3}\\left(3\\right)+{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n

    Try It 1<\/h3>\r\n

    Expand [latex]{\\mathrm{log}}_{b}\\left(8k\\right)[\/latex].<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

    Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.<\/p>\n

    [latex]\\begin{cases}{\\mathrm{log}}_{b}1=0\\\\ {\\mathrm{log}}_{b}b=1\\end{cases}[\/latex]<\/div>\n

    For example, [latex]{\\mathrm{log}}_{5}1=0[\/latex] since [latex]{5}^{0}=1[\/latex]. And [latex]{\\mathrm{log}}_{5}5=1[\/latex] since [latex]{5}^{1}=5[\/latex].<\/p>\n

    Next, we have the inverse property.<\/p>\n

    [latex]\\begin{cases}\\hfill \\\\ {\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x\\hfill \\\\ \\text{ }{b}^{{\\mathrm{log}}_{b}x}=x,x>0\\hfill \\end{cases}[\/latex]<\/div>\n

    For example, to evaluate [latex]\\mathrm{log}\\left(100\\right)[\/latex], we can rewrite the logarithm as [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)[\/latex], and then apply the inverse property [latex]{\\mathrm{log}}_{b}\\left({b}^{x}\\right)=x[\/latex] to get [latex]{\\mathrm{log}}_{10}\\left({10}^{2}\\right)=2[\/latex].<\/p>\n

    To evaluate [latex]{e}^{\\mathrm{ln}\\left(7\\right)}[\/latex], we can rewrite the logarithm as [latex]{e}^{{\\mathrm{log}}_{e}7}[\/latex], and then apply the inverse property [latex]{b}^{{\\mathrm{log}}_{b}x}=x[\/latex] to get [latex]{e}^{{\\mathrm{log}}_{e}7}=7[\/latex].<\/p>\n

    Finally, we have the one-to-one<\/strong> property.<\/p>\n

    [latex]{\\mathrm{log}}_{b}M={\\mathrm{log}}_{b}N\\text{ if and only if}\\text{ }M=N[\/latex]<\/div>\n

    We can use the one-to-one property to solve the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)={\\mathrm{log}}_{3}\\left(2x+5\\right)[\/latex] for x<\/em>. Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for x<\/em>:<\/p>\n

    [latex]\\begin{cases}3x=2x+5\\hfill & \\text{Set the arguments equal}\\text{.}\\hfill \\\\ x=5\\hfill & \\text{Subtract 2}x\\text{.}\\hfill \\end{cases}[\/latex]<\/div>\n

    But what about the equation [latex]{\\mathrm{log}}_{3}\\left(3x\\right)+{\\mathrm{log}}_{3}\\left(2x+5\\right)=2[\/latex]? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining terms on the left side of the equation.<\/p>\n

    Recall that we use the product rule of exponents<\/em> to combine the product of exponents by adding: [latex]{x}^{a}{x}^{b}={x}^{a+b}[\/latex]. We have a similar property for logarithms, called the product rule for logarithms<\/strong>, which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.<\/p>\n

    Given any real number x<\/em>\u00a0and 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(MN\\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(MN\\right)\\hfill & ={\\mathrm{log}}_{b}\\left({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 product 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

    Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider [latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)[\/latex]. Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:<\/p>\n

    [latex]{\\mathrm{log}}_{b}\\left(wxyz\\right)={\\mathrm{log}}_{b}w+{\\mathrm{log}}_{b}x+{\\mathrm{log}}_{b}y+{\\mathrm{log}}_{b}z[\/latex]<\/div>\n
    \n

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

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

    [latex]{\\mathrm{log}}_{b}\\left(MN\\right)={\\mathrm{log}}_{b}\\left(M\\right)+{\\mathrm{log}}_{b}\\left(N\\right)\\text{ for }b>0[\/latex]<\/div>\n<\/div>\n
    \n

    How To: Given the logarithm of a product, use the product rule of logarithms to write an equivalent sum of logarithms.<\/h3>\n
      \n
    1. Factor the argument completely, expressing each whole number factor as a product of primes.<\/li>\n
    2. Write the equivalent expression by summing the logarithms of each factor.<\/li>\n<\/ol>\n<\/div>\n
      \n
      \n
      \n

      Example 1: Using the Product Rule for Logarithms<\/h3>\n

      Expand [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)[\/latex].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      We begin by factoring the argument completely, expressing 30 as a product of primes.<\/p>\n

      [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\cdot 3\\cdot 5\\cdot x\\cdot \\left(3x+4\\right)\\right)[\/latex]<\/div>\n

      Next we write the equivalent equation by summing the logarithms of each factor.<\/p>\n

      [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(2\\right)+{\\mathrm{log}}_{3}\\left(3\\right)+{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      Try It 1<\/h3>\n

      Expand [latex]{\\mathrm{log}}_{b}\\left(8k\\right)[\/latex].<\/p>\n

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

      \n\t\t\t

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