Results 1 to 3 of 3

Thread: Need help with log question?

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    1

    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; Jan 27th 2009 at 07:19 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jan 2009
    Posts
    8
    Are you sure the answer is correct? Anyway, I will show you my method and you compare it.
    $\displaystyle 2a^x=b^(x+1) $
    then, $\displaystyle (x)\lg{2a} = (x+1)(\lg{b}) $
    After which, $\displaystyle x(\lg{2}+\lg{a}) = (x)\lg{b} + \lg{b} $
    Finally, $\displaystyle x(\lg{2} + \lg{a} - \lg{b}) = \lg{b}$
    You arrive at the answer, $\displaystyle x = \frac{\lg{b}}{\lg{2} + \lg{a} - \lg{b}}$
    This may be wrong.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849
    Hello, imperfections!

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

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

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

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

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

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

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


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


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



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

    . . $\displaystyle 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} $

    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum