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Math Help - Solve for t using logarithms?

  1. #1
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    Solve for t using logarithms?

    Wf^t=Qb^t
    I solved a similar problem that involved natural logs, but this one confused me more.
    Can someone help me?
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  2. #2
    Senior Member
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    Hi iheartthemusic29.

    Quote Originally Posted by iheartthemusic29 View Post
    Wf^t=Qb^t
    I solved a similar problem that involved natural logs, but this one confused me more.
    Can someone help me?
    It is a very good idea to use logarithms, lets see

    Wf^t=Qb^t


    \frac{f^t}{b^t} = \frac{Q}{W}


    \displaystyle (\frac{f}{b})^t = \frac{Q}{W}


    LN((\frac{f}{b})^t) = LN(\frac{Q}{W})


    t*LN(\frac{f}{b}) = LN(\frac{Q}{W})


    t= \frac{LN(\frac{Q}{W})}{LN(\frac{f}{b})}


    t= \frac{LN(\frac{Q}{W})}{LN(f)-LN(b)}


    t= \frac{LN(Q)-LN(W)}{LN(f)-LN(b)}

    Any questions?

    regards
    Rapha
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  3. #3
    MHF Contributor

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    Quote Originally Posted by iheartthemusic29 View Post
    Wf^t=Qb^t
    I solved a similar problem that involved natural logs, but this one confused me more.
    Can someone help me?
    I think I would be inclined to just take the logarithm directly:
    log(Wf^t)= log(Qb^t)
    log(W)+ log(f^t)= log(Q)+ log(b^t)

    log(W)+ t log(f)= log(Q)+ t log(b)

    log(W)+ t log(f)- t log)b)= log(Q)

    t(log(f)- log(b)= log(Q)- log(W)

    t log(f/b)= log(Q/W)

    t= log(Q/W)/log(f/b)
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