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Thread: Log equation is this correct?

  1. December 16th 2008, 10:59 PM #1
    needhelp75
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    Log equation is this correct?

    10^(1-x) = 6 ^(x)


    I got:

    log10 / log6 + log10 = x
    or x = .562
    Last edited by needhelp75; December 16th 2008 at 11:14 PM.
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  2. December 16th 2008, 11:20 PM #2
    nzmathman
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    10^{1-x} = 6^x

    \Rightarrow \frac{10^1}{10^x} = 6^x

    \Rightarrow 10 = 6^x \times 10^x

    \Rightarrow 10 = 60^x

    \Rightarrow x\ln60 = \ln 10

    \Rightarrow x = \frac{\ln 10}{\ln60} = 0.5624
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