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Math Help - 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; January 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 3^{x + 2} = 3^x\cdot 3^2 = 9\left(3^x\right).

    So the equation becomes

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

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

    So the equation becomes

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

    This is a Quadratic Equation.

    Bring in a dummy variable X = 3^x.

    So the equation becomes

    X^2 - 9X + 8 = 0

    X^2 - 8X - X + 8 = 0

    X(X - 8) - 1(X - 8) = 0

    (X - 1)(X - 8) = 0.


    Therefore X - 1 = 0 or X - 8 = 0

    X = 1 or X = 8.


    So 3^x = 1 or 3^x = 8

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


    You could also work out x using natural logarithms.

    3^x = 8

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

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

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

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