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Math Help - How would you solve this log?

  1. #1
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    How would you solve this log?

    2^{2x} +2^{x+2} -12=0 They want you to solve the eqation.

    since it's a subtraction you divide. I multiplied 2x(x+2) and got 2x^2frac/-12

     2x^2/12=0

    I don't think I'm even close, but am not sure how to go from here.


    Thanks

    Jason
    Last edited by Darkhrse99; December 2nd 2007 at 07:44 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    2^{2x} +2^{x+2} -12=0 They want you to solve the eqation.
    Let's rewrite this as
    2^{2x} + 2^2 \cdot 2^x - 12 = 0

    2^{2x} + 4 \cdot 2^x - 12 = 0

    This is a quadratic in 2^x as Jhevon said.

    For further clarity, let's define y = 2^x. Then your equation becomes:
    y^2 + 4y - 12 = 0

    -Dan
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