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Math Help - Logarithmic and radical equations

  1. #1
    Junior Member
    Joined
    Aug 2008
    Posts
    65

    Logarithmic and radical equations

    Greetings,
    I have encountered inability to resolve the below shown problem.
    Attached Thumbnails Attached Thumbnails Logarithmic and radical equations-log.gif  
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
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    France
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    1,458
    Hi

    2\:\log_{4}(3y-4) - \log_{2}(x-4) = 1 \implies\: \:2\:\frac{\ln(3y-4)}{\ln 4} -\frac{\ln(x-4)}{\ln 2} = 1

    and since ln(4) = ln(2) = 2 ln(2)

    \ln(3y-4)-\ln(x-4) = \ln 2 \implies \ln \frac{3y-4}{x-4} = \ln 2 \implies \frac{3y-4}{x-4} = 2 \implies x = \frac32 \:y + 2

    Now you can substitute this expression into the first equation
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