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Thread: Logarithms help

  1. July 8th 2009, 12:09 PM #1
    greghunter
    greghunter is offline
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    Logarithms help

    Solve 2^{x+1}=3^x.
    I've done quite a bit of logs but am not sure how to work this one out.
    Thanks for any help.
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  2. July 8th 2009, 12:36 PM #2
    AlephZero
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    Just take the bit-log of each side:

    2^{x+1}=3^x
    \implies \log_2(2^{x+1})=\log_2 3^x
    \implies x+1=x\log_2 3
    \implies x(1-\log_2 3)=-1
    \implies x=\frac{-1}{1-\log_2 3}

    which I believe is equivalent to x=\frac{-\ln 2}{\ln 2 - \ln 3}, if you prefer.
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