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Thread: Logarithmic Equation

  1. #1
    Senior Member Stroodle's Avatar
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    Logarithmic Equation

    Hi. I'm having trouble trying to solve the following equation for $\displaystyle x$

    $\displaystyle 2^{2x}-12\times2^x+32=0$

    Any help will be appreciated. THanks
    Last edited by Stroodle; Sep 27th 2009 at 01:01 AM. Reason: clarity
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  2. #2
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    Quote Originally Posted by Stroodle View Post
    Hi. I'm having trouble trying to solve the following equation for $\displaystyle x$

    $\displaystyle 2^{2x}-12\times2^x+32=0$

    Any help will be appreciated. THanks
    Let $\displaystyle X = 2^x$, so that the equation becomes

    $\displaystyle X^2 - 12X + 32 = 0$

    $\displaystyle (X - 8)(X - 4) = 0$

    $\displaystyle X = 8$ or $\displaystyle X = 4$.


    Thus $\displaystyle 2^x = 8$ or $\displaystyle 2^x = 4$.


    It should be pretty obvious that $\displaystyle x = 3$ or $\displaystyle x = 2$.
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  3. #3
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    Quote Originally Posted by Stroodle View Post
    Hi. I'm having trouble trying to solve the following equation for $\displaystyle x$

    $\displaystyle 2^{2x}-12\times2^x+32=0$

    Any help will be appreciated. THanks
    for this question, you don't need to use logarithm.

    $\displaystyle let 2^x be y.$
    $\displaystyle y^2-12y+32=0$
    $\displaystyle (y-4)(y-8)=0$

    y=4 or y=8
    $\displaystyle 2^x=4$ or $\displaystyle 2^x=8$
    $\displaystyle 2^x=2^2$ or $\displaystyle 2^x=2^3$
    By comparing indices,
    $\displaystyle x=2$ or $\displaystyle 3$
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  4. #4
    Senior Member Stroodle's Avatar
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    Thanks heaps for your help guys. Knew it would be something simple that I missed.
    Was in the logarithm section, so I was trying to do it with logs...
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