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Math Help - [SOLVED] Logarithm with different base

  1. #1
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    [SOLVED] Logarithm with different base

    Hi everyone

    I need help for this problem :

    If 2*\log_2 (x-2y)=\log_3 (xy) , find \frac{x}{y}


    2*\log_2 (x-2y)=\log_3 (xy)

    \log_2 (x-2y)^2=\log_3 (xy)

    How to deal with different base?

    Thx
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  2. #2
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    Quote Originally Posted by songoku View Post
    Hi everyone



    \log_2 (x-2y)^2=\log_3 (xy)

    How to deal with different base?

    Thx

    This may help

     <br />
\log_bx = \frac{\log_ax}{\log_ab} <br />
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  3. #3
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    Hi pickslides

    I've tried that and still....

    \frac{\log_2 (x-2y)^2}{\log_2 2}=\frac{\log_2 (xy)}{\log_2 3}

    \log_2 (x-2y)^2 * \log_2 3 = \log_2 (xy)

    Next hint?

    Thx
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  4. #4
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    Are you sure this is the way the problem is stated? I've tried to solve this several ways, and the best I can come up with is

     \frac{x}{y}\,=\, \frac{(xy)^{log_3 \sqrt{2}}}{y} +2

    Which doesn't look too satisfying.
    I can show my work, but I just wanted to post this first, and have you confirm this is really the problem. Also, has anyone else tried solving this?

    QM
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  5. #5
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    Hi QM deFuturo

    Yes, that's the whole problem and that's the way it is stated. I also have tried to solve it and always get stuck...

    Thx for trying
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  6. #6
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    I conclude that this problem can't be solved

    Thx
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