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Thread: Solving an exponential equation of quadratic form

  1. #1
    Member smmmc's Avatar
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    3^(2x) - 3^(x+2) +8=0

    i tried 3^(x)=y but what about the 3^(x+2) ?
    Last edited by mr fantastic; Jan 8th 2010 at 10:07 PM. Reason: Moved from another thread
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  2. #2
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    Quote Originally Posted by smmmc View Post
    3^(2x) - 3^(x+2) +8=0

    i tried 3^(x)=y but what about the 3^(x+2) ?
    Notice that $\displaystyle 3^{x + 2} = 3^x\cdot 3^2 = 9\left(3^x\right)$.

    So the equation becomes

    $\displaystyle 3^{2x} - 9\left(3^x\right) + 8 = 0$.

    Also notice that $\displaystyle 3^{2x} = \left(3^x\right)^2$.

    So the equation becomes

    $\displaystyle \left(3^x\right)^2 - 9\left(3^x\right) + 8 = 0$.

    This is a Quadratic Equation.

    Bring in a dummy variable $\displaystyle X = 3^x$.

    So the equation becomes

    $\displaystyle X^2 - 9X + 8 = 0$

    $\displaystyle X^2 - 8X - X + 8 = 0$

    $\displaystyle X(X - 8) - 1(X - 8) = 0$

    $\displaystyle (X - 1)(X - 8) = 0$.


    Therefore $\displaystyle X - 1 = 0$ or $\displaystyle X - 8 = 0$

    $\displaystyle X = 1$ or $\displaystyle X = 8$.


    So $\displaystyle 3^x = 1$ or $\displaystyle 3^x = 8$

    $\displaystyle x = 0$ or $\displaystyle x = \log_3{8} = \log_3{2^3} = 3\log_3{2}$.


    You could also work out $\displaystyle x$ using natural logarithms.

    $\displaystyle 3^x = 8$

    $\displaystyle \ln{3^x} = \ln{8}$

    $\displaystyle x\ln{3} = \ln{2^3}$

    $\displaystyle x\ln{3} = 3\ln{2}$

    $\displaystyle x = \frac{3\ln{2}}{\ln{3}}$.
    Last edited by mr fantastic; Jan 8th 2010 at 10:07 PM.
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