{"id":11210,"date":"2015-07-14T18:51:59","date_gmt":"2015-07-14T18:51:59","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=11210"},"modified":"2021-11-15T02:32:10","modified_gmt":"2021-11-15T02:32:10","slug":"condense-logarithmic-expressions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/condense-logarithmic-expressions\/","title":{"raw":"Condense logarithmic expressions","rendered":"Condense logarithmic expressions"},"content":{"raw":"

We 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.<\/p>\r\n\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.<\/h3>\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<\/div>\r\n
    \r\n
    \r\n
    \r\n

    Example 9: Using the Product and Quotient Rules to Combine Logarithms<\/h3>\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.<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Solution<\/h3>\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 reduces 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<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n

    Try It 9<\/h3>\r\n

    Condense [latex]\\mathrm{log}3-\\mathrm{log}4+\\mathrm{log}5-\\mathrm{log}6[\/latex].<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

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

    Example 10: Condensing Complex Logarithmic Expressions<\/h3>\r\n

    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].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

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

    Example 11: Rewriting as a Single Logarithm<\/h3>\r\n

    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.<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Solution<\/h3>\r\n

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

    Next we apply the product rule to the sum:<\/p>\r\n\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({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/div>\r\n

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

    [latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n

    Try It 10<\/h3>\r\n

    Rewrite [latex]\\mathrm{log}\\left(5\\right)+0.5\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(7x - 1\\right)+3\\mathrm{log}\\left(x - 1\\right)[\/latex] as a single logarithm.<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n

    \r\n

    Try It 11<\/h3>\r\n

    Condense [latex]4\\left(3\\mathrm{log}\\left(x\\right)+\\mathrm{log}\\left(x+5\\right)-\\mathrm{log}\\left(2x+3\\right)\\right)[\/latex].<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\nThe following video gives more examples of combining logarithms.\r\n\r\n[embed]https:\/\/www.youtube.com\/watch?v=jR5UJ0AUz0c[\/embed]\r\n

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

    Example 12: Applying of the Laws of Logs<\/h3>\r\n

    Recall that, in chemistry, [latex]\\text{pH}=-\\mathrm{log}\\left[{H}^{+}\\right][\/latex]. If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Solution<\/h3>\r\n

    Suppose C<\/em>\u00a0is the original concentration of hydrogen ions, and P<\/em>\u00a0is the original pH of the liquid. Then [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex]. If the concentration is doubled, the new concentration is 2C<\/em>. Then the pH of the new liquid is<\/p>\r\n\r\n

    [latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)[\/latex]<\/div>\r\n

    Using the product rule of logs<\/p>\r\n\r\n

    [latex]\\text{pH}=-\\mathrm{log}\\left(2C\\right)=-\\left(\\mathrm{log}\\left(2\\right)+\\mathrm{log}\\left(C\\right)\\right)=-\\mathrm{log}\\left(2\\right)-\\mathrm{log}\\left(C\\right)[\/latex]<\/div>\r\n

    Since [latex]P=-\\mathrm{log}\\left(C\\right)[\/latex], the new pH is<\/p>\r\n\r\n

    [latex]\\text{pH}=P-\\mathrm{log}\\left(2\\right)\\approx P - 0.301[\/latex]<\/div>\r\n

    When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n

    Try It 12<\/h3>\r\n

    How does the pH change when the concentration of positive hydrogen ions is decreased by half?<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

    We 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.<\/p>\n

    \n

    How To: Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.<\/h3>\n
      \n
    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>\n
    2. Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.<\/li>\n
    3. Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.<\/li>\n<\/ol>\n<\/div>\n
      \n
      \n
      \n

      Example 9: Using the Product and Quotient Rules to Combine Logarithms<\/h3>\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.<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

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

      This reduces our original expression to<\/p>\n

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

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

      Try It 9<\/h3>\n

      Condense [latex]\\mathrm{log}3-\\mathrm{log}4+\\mathrm{log}5-\\mathrm{log}6[\/latex].<\/p>\n

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

      \n
      \n
      \n

      Example 10: Condensing Complex Logarithmic Expressions<\/h3>\n

      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].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

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

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

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

      Example 11: Rewriting as a Single Logarithm<\/h3>\n

      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.<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

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

      Next we apply the product rule to the sum:<\/p>\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({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)[\/latex]<\/div>\n

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

      [latex]\\mathrm{log}\\left({x}^{2}\\right)-\\mathrm{log}\\left({\\left(x+5\\right)}^{4}{\\left(3x+5\\right)}^{{x}^{-1}}\\right)=\\mathrm{log}\\left(\\frac{{x}^{2}}{{\\left(x+5\\right)}^{4}\\left({\\left(3x+5\\right)}^{{x}^{-1}}\\right)}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      Try It 10<\/h3>\n

      Rewrite [latex]\\mathrm{log}\\left(5\\right)+0.5\\mathrm{log}\\left(x\\right)-\\mathrm{log}\\left(7x - 1\\right)+3\\mathrm{log}\\left(x - 1\\right)[\/latex] as a single logarithm.<\/p>\n

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

      \n

      Try It 11<\/h3>\n

      Condense [latex]4\\left(3\\mathrm{log}\\left(x\\right)+\\mathrm{log}\\left(x+5\\right)-\\mathrm{log}\\left(2x+3\\right)\\right)[\/latex].<\/p>\n

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

      The following video gives more examples of combining logarithms.<\/p>\n