{"id":1144,"date":"2025-07-23T16:27:00","date_gmt":"2025-07-23T16:27:00","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1144"},"modified":"2026-03-18T17:52:16","modified_gmt":"2026-03-18T17:52:16","slug":"logarithmic-properties-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/logarithmic-properties-learn-it-4\/","title":{"raw":"Logarithmic Properties: Learn It 4","rendered":"Logarithmic Properties: Learn It 4"},"content":{"raw":"

Expanding Logarithms<\/h2>\r\nTaken together, the product rule, quotient rule, and power rule are often called \"properties of logs.\" Sometimes we apply more than one rule in order to expand an expression.\r\n\r\n
For example:\r\n

[latex]\\begin{array}{l}{\\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]<\/p>\r\n\r\n<\/section>We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.\r\n\r\nWith practice, we can look at a logarithmic expression and expand it mentally and then just writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots\u2014never with addition or subtraction inside the argument of the logarithm.\r\n\r\n

Rewrite [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)[\/latex] as a sum or difference of logs with exponent of [latex]1[\/latex].[reveal-answer q=\"526416\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"526416\"]First, because we have a quotient of two expressions, we can use the quotient rule:\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]<\/center>\r\nThen seeing the product in the first term, we use the product rule:\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]<\/center>\r\nFinally, we use the power rule on the first term:\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]<\/center>[\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]321496[\/ohm_question]<\/section>In the next example we will recall that we can write roots as exponents, and use this quality to simplify logarithmic expressions.\r\n\r\n
Expand [latex]\\mathrm{log}\\left(\\sqrt{x}\\right)[\/latex].[reveal-answer q=\"914877\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"914877\"]
[latex]\\begin{array}{l}\\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]<\/center>[\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]321502[\/ohm_question]<\/section>Now we will provide\u00a0some examples that will require careful attention.\r\n\r\n
Expand [latex]{\\mathrm{log}}_{6}\\left(\\frac{64{x}^{3}\\left(4x+1\\right)}{\\left(2x - 1\\right)}\\right)[\/latex].[reveal-answer q=\"324399\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"324399\"]We can expand by applying the product and quotient rules.[latex]\\begin{array}{lllllll}{\\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 product and quotient rule}.\\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][\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]321507[\/ohm_question]<\/section>
[ohm_question hide_question_numbers=1]321508[\/ohm_question]<\/section>\r\n

Condensing Logarithms<\/h2>\r\nWe can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined.\r\n\r\n
How To: Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm<\/strong>\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. From left to right, apply the product and quotient properties. Rewrite sums of logarithms as the logarithm of a product and differences of logarithms as the logarithm of a quotient.<\/li>\r\n<\/ol>\r\n<\/section>
    Use the power rule for logs to rewrite [latex]4\\mathrm{ln}\\left(x\\right)[\/latex] as a single logarithm with a leading coefficient of [latex]1[\/latex].[reveal-answer q=\"959478\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"959478\"]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, [latex]4[\/latex], as the exponent and the argument, [latex]x[\/latex], as the base and rewrite the product as a logarithm of a power:
    [latex]4\\mathrm{ln}\\left(x\\right)=\\mathrm{ln}\\left({x}^{4}\\right)[\/latex]<\/center>[\/hidden-answer]<\/section>In our next few examples we will use a combination of logarithm rules to condense logarithms.\r\n\r\n
    Write [latex]{\\mathrm{log}}_{3}\\left(5\\right)+{\\mathrm{log}}_{3}\\left(8\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex] as a single logarithm.[reveal-answer q=\"484876\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"484876\"]From left to right, since we have the addition of two logs, we first use the product rule:\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]<\/center>\r\nThis simplifies our original expression to:\r\n
    [latex]{\\mathrm{log}}_{3}\\left(40\\right)-{\\mathrm{log}}_{3}\\left(2\\right)[\/latex]<\/center>\r\nUsing the quotient rule:\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]<\/center>[\/hidden-answer]<\/section>
    [ohm_question hide_question_numbers=1]321511[\/ohm_question]<\/section>
    Condense [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].[reveal-answer q=\"841660\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"841660\"]We apply the power rule first:\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]<\/center>\r\nFrom left to right, since we have the addition of two logs, we apply the product rule to the sum:\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]<\/center>\r\nFinally we apply the quotient rule to the difference:\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]<\/center>[\/hidden-answer]<\/section>
    Rewrite [latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)[\/latex] as a single logarithm.[reveal-answer q=\"598036\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"598036\"]We apply the power rule first:\r\n
    [latex]2\\mathrm{log}x - 4\\mathrm{log}\\left(x+5\\right)+\\frac{1}{x}\\mathrm{log}\\left(3x+5\\right)=\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/center>\r\nFrom left to right, since we have the difference of two logs, we apply the quotient rule to the difference:\r\n
    [latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/center>\r\nFinally we apply the product rule to the sum:\r\n
    [latex]\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}}\\right)+\\mathrm{log}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{{x}^{2}}{{\\left(3x+5\\right)}^{{x}^{-1}}}}{{\\left(x+5\\right)}^{4}}\\right)[\/latex]<\/center>[\/hidden-answer]<\/section>
    [ohm_question hide_question_numbers=1]321513[\/ohm_question]<\/section>","rendered":"

    Expanding Logarithms<\/h2>\n

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

    For example:<\/p>\n

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

    We can use the power rule to expand logarithmic expressions involving negative and fractional exponents. We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.<\/p>\n

    With practice, we can look at a logarithmic expression and expand it mentally and then just writing the final answer. Remember, however, that we can only do this with products, quotients, powers, and roots\u2014never with addition or subtraction inside the argument of the logarithm.<\/p>\n

    Rewrite [latex]\\mathrm{ln}\\left(\\frac{{x}^{4}y}{7}\\right)[\/latex] as a sum or difference of logs with exponent of [latex]1[\/latex].<\/p>\n