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Math Help - Solving simultaneously involving logs.

  1. #1
    Member smmmc's Avatar
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    Solving simultaneously involving logs.

    1) 52=x*10^m
    2) 80=x*10^3m

    my attempt

    from 1) rearrange to get x=52/10^m

    then sub into 2)
    so
    80=52/10^m * 10^3m

    so
    log 10(80) = log10(52) - log10(10^m) + log10(10^3m)
    thenn i get
    80=52 - m +3m


    Help, Thank You. im unsure abou the log 10(80) part i think it should be 10^80 ?
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by smmmc View Post
    1) 52=x*10^m
    2) 80=x*10^3m

    my attempt

    from 1) rearrange to get x=52/10^m [A]

    then sub into 2)
    so
    80=52/10^m * 10^3m

    so
    log 10(80) = log10(52) - log10(10^m) + log10(10^3m)
    ...
    You get:

    \log(80)-\log(52)=2m~\implies~m=\dfrac12(\log(80)-\log(52))

    which yields m \approx 0.09354...

    Re-substitute into [A] to get x.
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  3. #3
    MHF Contributor

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    Substitution always works but, with a problem like this, I think the simplest way to eliminate one unknown variable is to divide one equation by the other.

    \frac{80}{52}= \frac{x10^{3m}}{x10^m}= 10^{2m}
    and we have eliminated x. Now 2m= log\left(\frac{80}{52}\right)= log(80)- log(52)
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