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Math Help - Logarithm problem

  1. #1
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    Logarithm problem

    This equation has me at loggerheads...

    6(\log_{x}2 - \log_{4}x) + 7 = 0

    Even a hint would be appreciated!
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  2. #2
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    Hello,

    Quote Originally Posted by SVXX View Post
    This equation has me at loggerheads...

    6(\log_{x}2 - \log_{4}x) + 7 = 0

    Even a hint would be appreciated!
    Note that \log_n x=\frac{\ln x}{\ln n}

    So \log_x 2=\frac{\ln 2}{\ln x}

    And \log_4 x=\frac{\ln x}{\ln 4}

    Multiply by ln(x), and ln(4) if you want to, then susbtitute y=ln x
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  3. #3
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    We haven't done natural logarithms in class yet, LOL.
    Could I get some further tips for this? Or is there any other way to solve it?
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  4. #4
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    Ok...

    There is a formula :

    \log_b x=\frac{1}{\log_a b} \cdot \log_a x (took it from Wikipedia)

    Taking b=x and a=4, you get :

    \log_x 2=\frac{1}{\log_4 x} \cdot \log_4 2

    So now you're working on base 4

    This time, multiply by \log_4 x

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  5. #5
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    If I take \log_{4}x as y, it will turn out to be a quadratic equation. So I guess we'll have two answers...

    <br />
      \log_{x}2 - \log_{4}x = -7/6
    =>  \log_{4}2/\_log_{4}x - \log_{4}x = -7/6
    =>  1/2/y - y = -7/6
    =>  1/2y - y = -7/6
    =>  6y*y - 7y - 3 = 0


    The roots for y are 1.5 and 0.3333.
    Taking y = 1.5,
    <br />
\log_{4}x = 1.5<br />
    4 raised to 3/2 or 1.5 is x, or x = 8.
    Merci beaucoup, Madame :P
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  6. #6
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    Quote Originally Posted by SVXX View Post
    If I take \log_{4}x as y, it will turn out to be a quadratic equation. So I guess we'll have two answers...

    <br />
      \log_{x}2 - \log_{4}x = -7/6
    =>  \log_{4}2/\_log_{4}x - \log_{4}x = -7/6
    =>  1/2/y - y = -7/6
    =>  1/2y - y = -7/6
    =>  6y*y - 7y - 3 = 0


    The roots for y are 1.5 and 0.3333.
    Taking y = 1.5,
    <br />
\log_{4}x = 1.5<br />
    4 raised to 3/2 or 1.5 is x, or x = 8.
    Merci beaucoup, Madame :P
    De rien
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