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Math Help - problem with logarithms

  1. #1
    Member srirahulan's Avatar
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    Lightbulb problem with logarithms

    log(x-2)+log2=2logy;\  log(x-3y+3)=0 find out the value of (x) and (y).
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  2. #2
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    Re: problem with logarithms

    Note that \log(x-2) + \log 2 = \log(2x-4), and that 2 \log y = \log y^2. Therefore the first equation becomes

    \log(2x-4) = \log y^2 \Rightarrow 2x-4 = y^2

    The second equation, note that \log 1 = 0 (for any non-zero base), so x - 3y + 3 = 0.

    Solve the system of equations.
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  3. #3
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    Re: problem with logarithms

    Whoops, x-3y +3 = 1
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