{"id":1146,"date":"2025-07-23T16:26:40","date_gmt":"2025-07-23T16:26:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1146"},"modified":"2026-03-18T15:15:56","modified_gmt":"2026-03-18T15:15:56","slug":"logarithmic-properties-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/logarithmic-properties-learn-it-2\/","title":{"raw":"Logarithmic Properties: Learn It 2","rendered":"Logarithmic Properties: Learn It 2"},"content":{"raw":"

Using the Product Rule for Logarithms<\/h2>\r\nRecall 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.\r\n\r\n
\r\n

the product rule for logarithms<\/h3>\r\nThe product rule for logarithms<\/strong> can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.\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]<\/p>\r\n\r\n<\/section>[reveal-answer q=\"173302\"]Where does the product rule come from?[\/reveal-answer]\r\n[hidden-answer a=\"173302\"]\r\n\r\nGiven any real number [latex]x[\/latex]\u00a0and positive real numbers [latex]M[\/latex], [latex]N[\/latex], and [latex]b[\/latex], where [latex]b\\ne 1[\/latex], we will show\r\n

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

[latex]\\begin{array}{lllllllll}{\\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{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\nNote that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors.\r\n\r\n

For example, consider [latex]\\mathrm{log}_{b}(wxyz)[\/latex].\u00a0Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:\r\n
[latex]\\mathrm{log}_{b}(wxyz)=\\mathrm{log}_{b}w+\\mathrm{log}_{b}x+\\mathrm{log}_{b}y+\\mathrm{log}_{b}z[\/latex]<\/div>\r\n<\/section>
Expand [latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)[\/latex].[reveal-answer q=\"603492\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"603492\"]We begin by writing\u00a0an equal\u00a0equation by summing the logarithms of each factor.\r\n
[latex]{\\mathrm{log}}_{3}\\left(30x\\left(3x+4\\right)\\right)={\\mathrm{log}}_{3}\\left(30x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)={\\mathrm{log}}_{3}\\left(30\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/center>\r\n

The final expansion looks like this. Note how the factor [latex]30x[\/latex] can be expanded into the sum of two logarithms:<\/p>\r\n\r\n

[latex]{\\mathrm{log}}_{3}\\left(30\\right)+{\\mathrm{log}}_{3}\\left(x\\right)+{\\mathrm{log}}_{3}\\left(3x+4\\right)[\/latex]<\/center>\r\n[\/hidden-answer]\r\n\r\n<\/section>
[ohm_question hide_question_numbers=1]321487[\/ohm_question]<\/section>
[ohm_question hide_question_numbers=1]321489[\/ohm_question]<\/section><\/section>\r\n

Using the Quotient Rule for Logarithms<\/h2>\r\nFor 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.\r\n\r\n
\r\n

the quotient rule for logarithms<\/h3>\r\nThe quotient rule for logarithms<\/strong> can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.\r\n

[latex]{\\mathrm{log}}_{b}\\left(\\frac{M}{N}\\right)={\\mathrm{log}}_{b}M-{\\mathrm{log}}_{b}N[\/latex]<\/p>\r\n\r\n<\/section>

[reveal-answer q=\"927754\"]Derivation of the quotient rule.[\/reveal-answer]\r\n[hidden-answer a=\"927754\"]Given any real number [latex]x[\/latex]and positive real numbers [latex]M[\/latex], [latex]N[\/latex], and [latex]b[\/latex], where [latex]b\\ne 1[\/latex], we will show\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].<\/p>\r\nLet [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\r\n

[latex]\\begin{array}{l}{\\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]<\/p>\r\n[\/hidden-answer]\r\n\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\r\n

[latex]\\begin{array}{lllll}\\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{array}[\/latex]<\/p>\r\nNext we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.\r\n

[latex]\\begin{array}{lll}\\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{array}[\/latex]<\/p>\r\n\r\n<\/section>

How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms<\/strong>\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<\/section><\/section>
    Expand [latex]{\\mathrm{log}}_{2}\\left(\\frac{15x\\left(x - 1\\right)}{\\left(3x+4\\right)\\left(2-x\\right)}\\right)[\/latex].[reveal-answer q=\"582435\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"582435\"]First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.\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]<\/center>\r\nNotice that the resulting terms are logarithms of products. To expand completely, we apply the product rule.\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) \\\\\\text{}= \\left[{\\mathrm{log}}_{2}\\left(15\\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(15\\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]<\/center>\r\nAnalysis of the Solution<\/strong>\r\n[latex]\\\\[\/latex]\r\nThere 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 [latex]x\u00a0= 2[\/latex]. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm that [latex]x\u00a0> 0[\/latex], [latex]x\u00a0> 1[\/latex], [latex]x>-\\frac{4}{3}[\/latex], and [latex]x\u00a0< 2[\/latex]. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.[\/hidden-answer]<\/section>
    [ohm_question hide_question_numbers=1]321493[\/ohm_question]<\/section><\/section>
    [ohm_question hide_question_numbers=1]321488[\/ohm_question]<\/section>","rendered":"

    Using the Product Rule for Logarithms<\/h2>\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.<\/p>\n

    \n

    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]<\/p>\n<\/section>\n