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Thread: Log

  1. #1
    Senior Member
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    Log

    Prove that if $\displaystyle 5^{3t-2}=10^{t+1}$ then $\displaystyle t=\frac{log 250}{log 12.5}$

    I know how to solve it. just duno how to rearrange to get answer required.

    thanks
    Last edited by BabyMilo; Dec 2nd 2009 at 12:19 PM.
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  2. #2
    Senior Member
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    Quote Originally Posted by RRH View Post
    $\displaystyle
    5^{3t-2}=10^{t-1}
    $

    $\displaystyle
    log 5^{3t-2}=log 10^{t-1}
    $

    $\displaystyle
    (3t-2)log 5=(t-1)log 10
    $

    $\displaystyle
    3tlog 5-2log5=tlog 10-log 10
    $

    $\displaystyle
    3tlog 5-tlog 10=2log 5-log10
    $

    $\displaystyle
    t(3log 5-log 10)=2log 5-log 10
    $

    $\displaystyle
    t(log 5^3-log10)=log 5^2-log 10
    $

    $\displaystyle
    t(log125-log 10)=log 25-log 10
    $

    $\displaystyle
    t(\frac{log 125}{log 10})=\frac{log 25}{log 10}
    $

    $\displaystyle
    t=\frac{log 25}{log 10}*\frac{log 10}{log 125}
    $

    hmmm, I get a different answer than you do

    $\displaystyle
    t=\frac{log 250}{log 1250}
    $
    sry it's plus 1 not minus 1
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  3. #3
    RRH
    RRH is offline
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    Quote Originally Posted by BabyMilo View Post
    sry it's plus 1 not minus 1
    no problem I will rework it so you can see the steps

    $\displaystyle
    5^{3t-2}=10^{t+1}
    $

    $\displaystyle
    log 5^{3t-2}=log 10{t+1}
    $

    $\displaystyle
    (3t-2)log 5=(t+1)log 10
    $

    $\displaystyle
    3tlog 5-2log 5=tlog 10+log 10
    $

    $\displaystyle
    3tlog 5-tlog 10=log 10+2log 5
    $

    $\displaystyle
    t(3log 5-log 10)=log 10+2log5
    $

    $\displaystyle
    t(log5^3-log10)=log 10+log5^2
    $

    $\displaystyle
    t(log 125-log 10)=log 10+log 25
    $

    $\displaystyle
    tlog (\frac{125}{10})=log 250
    $

    $\displaystyle
    tlog 12.5=log 250
    $

    $\displaystyle
    t=\frac{log 250}{log 12.5}
    $

    much better
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