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Math Help - log equation

  1. #1
    Junior Member JoanF's Avatar
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    log equation

    Hi

    How do I solve this analytically

    log3 (x^2 -5x+6) - log2 (2-x)=2 ???

    I had already tried but I didn't get x =S

    please help...


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  2. #2
    Super Member Anonymous1's Avatar
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    Big Red, NY
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    Try swinging the base.

    \log_{3} ((x-3)(x-2)) = 2 + log_{2} (2-x)
    (x-3)(x-2) = 3^{2 + log_{2} (2-x)}
    (x-3)(x-2) = 3^{2}\times3^{log_{2} (2-x)}
    \log_{2}(\frac{(x-3)(x-2)}{9}) = log_{2}(3^{log_{2} (2-x)})
    \frac{(x-3)(x-2)}{9 \times \log_{2}(3)} = (2-x)
    ...etc
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  3. #3
    Junior Member JoanF's Avatar
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    Quote Originally Posted by Anonymous1 View Post
    Try swinging the base.

    \log_{3} ((x-3)(x-2)) = 2 + log_{2} (2-x)
    (x-3)(x-2) = 3^{2 + log_{2} (2-x)}
    (x-3)(x-2) = 3^{2}\times3^{log_{2} (2-x)}
    \log_{2}(\frac{(x-3)(x-2)}{9}) = log_{2}(3^{log_{2} (2-x)})
    \frac{(x-3)(x-2)}{9 \times \log_{2}(3)} = (2-x)
    ...etc
    I think your resolution is wrong... =S

    The solution is: x = 1,97 ...

    I solved
    \frac{(x-3)(x-2)}{9 \times \log_{2}(3)} = (2-x)
    ...etc[/quote] and I got: x = -11,26...
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