{"id":3536,"date":"2016-08-05T05:30:12","date_gmt":"2016-08-05T05:30:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&p=3536"},"modified":"2019-07-24T21:34:03","modified_gmt":"2019-07-24T21:34:03","slug":"evaluate-logarithms","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/evaluate-logarithms\/","title":{"raw":"Expand and Condense Logarithms","rendered":"Expand and Condense Logarithms"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
    \r\n \t
  • Combine the product, power, and quotient rules to expand logarithmic expressions<\/li>\r\n \t
  • Combine the product, power, and quotient rules to condense logarithmic expressions<\/li>\r\n<\/ul>\r\n<\/div>\r\n

    Taken together, the product rule, quotient rule, and power rule are often called \"laws of logs.\" Sometimes we apply more than one rule in order to simplify an expression. For example:<\/p>\r\n\r\n

    [latex]\\begin{array}{c}{\\mathrm{log}}_{b}\\left(\\frac{6x}{y}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(6x\\right)-{\\mathrm{log}}_{b}y\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}6+{\\mathrm{log}}_{b}x-{\\mathrm{log}}_{b}y\\hfill \\end{array}[\/latex]<\/div>\r\n

    We can also use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal (fraction) has a negative power:<\/p>\r\n\r\n

    [latex]\\begin{array}{c}{\\mathrm{log}}_{b}\\left(\\frac{A}{C}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(A{C}^{-1}\\right)\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left(A\\right)+{\\mathrm{log}}_{b}\\left({C}^{-1}\\right)\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}A+\\left(-1\\right){\\mathrm{log}}_{b}C\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}A-{\\mathrm{log}}_{b}C\\hfill \\end{array}[\/latex]<\/div>\r\n

    We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.<\/p>\r\n\r\n

    \r\n\r\n\"traffic-sign-160659-300x265\"Remember that we can only apply laws of logs to products, quotients, powers, and roots\u2014never with addition or subtraction inside the argument of the logarithm. Consider the following example\r\n

    [latex]\\begin{array}{c}\\mathrm{log}\\left(10+100\\right)\\overset{?}{=}\\end{array}\\mathrm{log}\\left(10\\right)+\\mathrm{log}\\left(100\\right)\\\\\\mathrm{log}\\left(110\\right)\\overset{?}{=}1+2\\\\2.04\\ne3[\/latex]<\/p>\r\nBe careful to only apply the product rule when a logarithm has an argument that is a product or when you have a sum of logarithms.\r\n\r\n<\/div>\r\nIn our first example, we will show that a logarithmic expression can be expanded by combining\u00a0several of the rules of logarithms.\r\n

    \r\n

    Example<\/h3>\r\nRewrite [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)[\/latex] as a sum or difference of logs.\r\n[reveal-answer q=\"396207\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"396207\"]\r\n

    First, because we have a quotient of two expressions, we can use the quotient rule:<\/p>\r\n\r\n

    [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)=\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/div>\r\n

    Then seeing the product in the first term, we use the product rule:<\/p>\r\n\r\n

    [latex]\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)=\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/div>\r\n

    Finally, we use the power rule on the first term:<\/p>\r\n\r\n

    [latex]\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)=4\\mathrm{ln}\\left(x\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can also use the rules for logarithms to simplify the logarithm of a radical expression.\r\n
    \r\n

    Example<\/h3>\r\nExpand [latex]\\mathrm{log}\\left(\\sqrt{x}\\right)[\/latex].\r\n[reveal-answer q=\"485495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"485495\"][latex]\\begin{array}{c}\\mathrm{log}\\left(\\sqrt{x}\\right)\\hfill & =\\mathrm{log}{x}^{\\left(\\frac{1}{2}\\right)}\\hfill \\\\ \\hfill & =\\frac{1}{2}\\mathrm{log}x\\hfill \\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
    \r\n

    Think About it<\/h3>\r\nCan we expand [latex]\\mathrm{ln}\\left({x}^{2}+{y}^{2}\\right)[\/latex]? Use the space below to develop an argument one way or the other before you look at the solution.[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"213713\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"213713\"]\r\n\r\nNo. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm. Rewrite the expression as an equation and express it as an exponential to give yourself some proof.\r\n\r\n[latex]m=\\mathrm{ln}\\left({x}^{2}+{y}^{2}\\right)[\/latex]\r\n\r\nIf you rewrite this as an exponential, you get:\r\n\r\n[latex]e^m={x}^{2}+{y}^{2}[\/latex]\r\n\r\nFrom here, there is not much more you can do to make this expression more simple.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nLet us do one more example with an expression that contains several different mathematical operations.\r\n
    \r\n

    Example<\/h3>\r\nExpand [latex]{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)[\/latex].\r\n[reveal-answer q=\"89786\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"89786\"]\r\n

    We can expand by applying the Product and Quotient Rules.<\/p>\r\n\r\n

    [latex]\\begin{array}{c}{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)\\hfill & ={\\mathrm{log}}_{6}64+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill & \\text{Apply the Quotient and Product Rules}.\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{6}{2}^{6}+{\\mathrm{log}}_{6}{x}^{3}+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill & {\\text{Simplify by writing 64 as 2}}^{6}.\\hfill \\\\ \\hfill & =6{\\mathrm{log}}_{6}2+3{\\mathrm{log}}_{6}x+{\\mathrm{log}}_{6}\\left(4x+1\\right)-{\\mathrm{log}}_{6}\\left(2x - 1\\right)\\hfill & \\text{Apply the Power Rule}.\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we show another example of expanding logarithms.\r\n\r\n[embed]https:\/\/www.youtube.com\/watch?v=bzV3lbfNhn8&feature=youtu.be[\/embed]\r\n

    Condense Logarithms<\/h2>\r\n

    We can use the rules of logarithms we just learned to condense sums and differences with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.<\/p>\r\n\r\n

    \r\n

    Example<\/h3>\r\nWrite [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex] as a single logarithm.\r\n[reveal-answer q=\"171557\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"171557\"]\r\n

    Using the product and quotient rules:<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)={\\mathrm{log}}_{3}\\left(5\\cdot 8\\right)={\\mathrm{log}}_{3}\\left(40\\right)[\/latex]<\/div>\r\n

    This simplifies our original expression to:<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]<\/div>\r\n

    Then, using the quotient rule:<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)={\\mathrm{log}}_{3}\\left(\\frac{40}{2}\\right)={\\mathrm{log}}_{3}\\left(20\\right)[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example, we show how to simplify a more complex logarithm by condensing it.\r\n
    \r\n

    Example<\/h3>\r\nCondense [latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)[\/latex].\r\n[reveal-answer q=\"982712\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"982712\"]\r\n

    We apply the power rule first:<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+\\frac{1}{2}{\\mathrm{log}}_{2}\\left(x - 1\\right)-3{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{2}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/div>\r\n

    Next we apply the product rule to the sum:<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\right)+{\\mathrm{log}}_{2}\\left(\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)[\/latex]<\/div>\r\n

    Finally, we apply the quotient rule to the difference:<\/p>\r\n\r\n

    [latex]{\\mathrm{log}}_{2}\\left({x}^{2}\\sqrt{x - 1}\\right)-{\\mathrm{log}}_{2}\\left({\\left(x+3\\right)}^{6}\\right)={\\mathrm{log}}_{2}\\frac{{x}^{2}\\sqrt{x - 1}}{{\\left(x+3\\right)}^{6}}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nMore examples of condensing logarithms are in the following video.\r\n\r\n[embed]https:\/\/www.youtube.com\/watch?v=jR5UJ0AUz0c&feature=youtu.be[\/embed]\r\n

    Summary<\/h2>\r\n

    Given a sum or difference of logarithms with the same base, we can write an equivalent expression as a single logarithm.<\/p>\r\n\r\n

      \r\n \t
    1. Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.<\/li>\r\n \t
    2. Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.<\/li>\r\n \t
    3. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.<\/li>\r\n<\/ol>\r\n<\/section>","rendered":"
      \n
      \n

      Learning Outcomes<\/h3>\n
        \n
      • Combine the product, power, and quotient rules to expand logarithmic expressions<\/li>\n
      • Combine the product, power, and quotient rules to condense logarithmic expressions<\/li>\n<\/ul>\n<\/div>\n

        Taken together, the product rule, quotient rule, and power rule are often called “laws of logs.” Sometimes we apply more than one rule in order to simplify an expression. For example:<\/p>\n

        [latex]\\begin{array}{c}{\\mathrm{log}}_{b}\\left(\\frac{6x}{y}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(6x\\right)-{\\mathrm{log}}_{b}y\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}6+{\\mathrm{log}}_{b}x-{\\mathrm{log}}_{b}y\\hfill \\end{array}[\/latex]<\/div>\n

        We can also use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal (fraction) has a negative power:<\/p>\n

        [latex]\\begin{array}{c}{\\mathrm{log}}_{b}\\left(\\frac{A}{C}\\right)\\hfill & ={\\mathrm{log}}_{b}\\left(A{C}^{-1}\\right)\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}\\left(A\\right)+{\\mathrm{log}}_{b}\\left({C}^{-1}\\right)\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}A+\\left(-1\\right){\\mathrm{log}}_{b}C\\hfill \\\\ \\hfill & ={\\mathrm{log}}_{b}A-{\\mathrm{log}}_{b}C\\hfill \\end{array}[\/latex]<\/div>\n

        We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.<\/p>\n

        \n

        \"traffic-sign-160659-300x265\"Remember that we can only apply laws of logs to products, quotients, powers, and roots\u2014never with addition or subtraction inside the argument of the logarithm. Consider the following example<\/p>\n

        [latex]\\begin{array}{c}\\mathrm{log}\\left(10+100\\right)\\overset{?}{=}\\end{array}\\mathrm{log}\\left(10\\right)+\\mathrm{log}\\left(100\\right)\\\\\\mathrm{log}\\left(110\\right)\\overset{?}{=}1+2\\\\2.04\\ne3[\/latex]<\/p>\n

        Be careful to only apply the product rule when a logarithm has an argument that is a product or when you have a sum of logarithms.<\/p>\n<\/div>\n

        In our first example, we will show that a logarithmic expression can be expanded by combining\u00a0several of the rules of logarithms.<\/p>\n

        \n

        Example<\/h3>\n

        Rewrite [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)[\/latex] as a sum or difference of logs.<\/p>\n

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

        First, because we have a quotient of two expressions, we can use the quotient rule:<\/p>\n

        [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)=\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/div>\n

        Then seeing the product in the first term, we use the product rule:<\/p>\n

        [latex]\\mathrm{ln}\\left({x}^{4}y\\right)-\\mathrm{ln}\\left(7\\right)=\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/div>\n

        Finally, we use the power rule on the first term:<\/p>\n

        [latex]\\mathrm{ln}\\left({x}^{4}\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)=4\\mathrm{ln}\\left(x\\right)+\\mathrm{ln}\\left(y\\right)-\\mathrm{ln}\\left(7\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        We can also use the rules for logarithms to simplify the logarithm of a radical expression.<\/p>\n

        \n

        Example<\/h3>\n

        Expand [latex]\\mathrm{log}\\left(\\sqrt{x}\\right)[\/latex].<\/p>\n

        Show Solution<\/span><\/p>\n
        [latex]\\begin{array}{c}\\mathrm{log}\\left(\\sqrt{x}\\right)\\hfill & =\\mathrm{log}{x}^{\\left(\\frac{1}{2}\\right)}\\hfill \\\\ \\hfill & =\\frac{1}{2}\\mathrm{log}x\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n
        \n

        Think About it<\/h3>\n

        Can we expand [latex]\\mathrm{ln}\\left({x}^{2}+{y}^{2}\\right)[\/latex]? Use the space below to develop an argument one way or the other before you look at the solution.