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Math Help - Need help with log question?

  1. #1
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    Need help with log question?

    I was doing practice exams and I got stuck on this question?

    Solve for x in the equation: 2a^x = b^(x+1)

    I tried it over and over and still can't get x by itself, how do you do this question? It's on the no calculator section.

    The answer's x = (log2 - log b) / (log b - log a)

    If it helps.
    Last edited by imperfections; January 27th 2009 at 08:19 PM.
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  2. #2
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    Are you sure the answer is correct? Anyway, I will show you my method and you compare it.
    2a^x=b^(x+1)
    then, (x)\lg{2a} = (x+1)(\lg{b})
    After which, x(\lg{2}+\lg{a}) = (x)\lg{b} + \lg{b}
    Finally,  x(\lg{2} + \lg{a} - \lg{b}) = \lg{b}
    You arrive at the answer, x = \frac{\lg{b}}{\lg{2} + \lg{a} - \lg{b}}
    This may be wrong.
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  3. #3
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    Hello, imperfections!

    Solve for x\!:\;\;2a^x \:=\: b^{x+1}

    Answer: . x \:= \:\frac{\log2 - \log b}{\log b - \log a}

    Take logs of both sides: . \log\left(2a^x\right) \:=\:\log\left(b^{x+1}\right)

    . . . . . . . \log2 + \log(a^x) \:=\:\log(b^{x+1})

    . . . . . . . \log2 + x\!\cdot\log a \:=\:(x+1)\log b

    . . . . . . . \log2 + x\!\cdot\log a \:=\:x\!\cdot\!\log b + \log b

    . . . . . . x\!\cdot\!\log a - x\!\cdot\!\log b \:=\:\log b - \log 2


    Factor: . x\!\cdot\!(\log a - \log b) \:=\:\log b - \log 2


    Therefore: . \boxed{x \;=\;\frac{\log b - \log 2}{\log a - \log b}}



    To get their answer, multiply by \tfrac{\text{-}1}{\text{-}1}

    . . x \;=\;\frac{\text{-}1}{\text{-}1}\cdot\frac{\log b - \log 2}{\log a - \log b} \;=\;\frac{\log 2 - \log b}{\log b - \log a}

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