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Math Help - solving exponential equation

  1. #1
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    solving exponential equation

    4^x*\frac{1}{2}^{3-2x}=8*(2^x)^2

    This is as far as i got:
    \frac{1}{2}^{-2x}*\frac{1}{2}^{3-2x}=\frac{1}{2}^{-3}*\frac{1}{2}^{-x^2}

    2x*3-2x=3*-x^2

    6x-2x=-3x^2

    3x^2+4x=0
    any help would be much appriciated!
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  2. #2
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    Factor the x out front, and then solve each of the two factors.
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  3. #3
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    Quote Originally Posted by windir View Post
    4^x*\frac{1}{2}^{3-2x}=8*(2^x)^2

    This is as far as i got:
    \frac{1}{2}^{-2x}*\frac{1}{2}^{3-2x}=\frac{1}{2}^{-3}*\frac{1}{2}^{-x^2}

    2x*3-2x=3*-x^2

    6x-2x=-3x^2

    3x^2+4x=0
    any help would be much appriciated!
    4^x \cdot \left(\frac{1}{2}\right)^{3-2x} = 8 \cdot 2^{2x}

    2^{2x} \cdot 2^{2x-3} = 8 \cdot 2^{2x}

    2^{2x-3} = 8

    2^{2x-3} = 2^3

    2x-3 = 3

    x = 3
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  4. #4
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    oh wow. thank you so much! i cant believe i didnt see that before!
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