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Math Help - Logarithm

  1. #1
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    Logarithm

    Hi everyone... need some logarithms questions.... Thanks a lot for your help in advance. I appreciate it greatly.

    Question 1
    Prove that \log_{y} x = \frac{1}{log_{x} y}
    Given that x and y are real positive numbers.

    Find the possible values of x if 2(\log_ {9} x + \log_{x} 9) = 5

    Question 2
    If  a^2 + b^2 = 7ab
    show that 2\log_{10} \frac{a + b}{3} = \log_{10} a + log_{10} b
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  2. #2
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    Quote Originally Posted by ecogreen View Post
    Hi everyone... need some logarithms questions.... Thanks a lot for your help in advance. I appreciate it greatly.

    Question 1
    Prove that \log_{y} x = \frac{1}{log_{x} y}
    Given that x and y are real positive numbers.

    Find the possible values of x if 2(\log_ {9} x + \log_{x} 9) = 5

    Question 2
    If  a^2 + b^2 = 7ab
    show that 2\log_{10} \frac{a + b}{3} = \log_{10} a + log_{10} b

    1. Using the change of base rule...

    \log_y{x} = \frac{\ln{x}}{\ln{y}}

     = \frac{1}{\frac{\ln{y}}{\ln{x}}}

     =\log_x{y}.


    Using this information:

    2(\log_ {9} x + \log_{x} 9) = 5

    \log_9{x} + \log_x{9} = \frac{5}{2}

    \log_9{x} + \frac{1}{\log_9{x}} = \frac{5}{2}

    X + \frac{1}{X} = \frac{5}{2}, where X = \log_9{x}

    X^2 + 1 = \frac{5}{2}X

    X^2 - \frac{5}{2}X + 1 = 0

    2X^2 - 5X + 2 = 0

    2X^2 - 4X - X + 2 = 0

    2X(X - 2) - 1(X - 2) = 0

    (X - 2)(2X - 1) = 0

    X - 2 = 0 or 2X - 1 = 0

    X = -2 or X = \frac{1}{2}.


    So that means \log_9{x} = -2 or \log_9{x} = \frac{1}{2}.

    x = 9^{-2} or x = 9^{\frac{1}{2}}

    x = \frac{1}{81} or x = 3.
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