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Math Help - Solve the following exponential equation

  1. #1
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    Solve the following exponential equation

    Hi, i need some help solving the following equation, for some reason i can't get my head around it:

    4(2^2x)+31(2^x)-8=0

    it would really help me if you can post an explantion on how you solved it from the begining to the end.
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  2. #2
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    Re: Solve the following exponential equation

    Hello, wolflessalex!

    4\!\cdot\!2^{2x}+31\!\cdot\!2^x-8\:=\:0

    We have a quadratic: . 4(2^x)^2 + 31(2^x) - 8 \;=\;0

    Let u = 2^x

    Substitute: . 4u^2 + 31u - 8 \;=\;0

    Factor: . (u+8)(4u-1) \:=\:0

    And we have: . \begin{Bmatrix}u+8\:=\:0 & \Rightarrow & u \:=\:\text{-}8 \\4u-1\:=\:0 & \Rightarrow & u \:=\:\frac{1}{4} \end{Bmatrix}


    Back-substitute:

    . . 2^x \:=\:\text{-}8 \quad \text{No real root}

    . . 2^x \:=\:\tfrac{1}{4} \quad\Rightarrow\quad & 2^x \:=\:2^{-2} \quad\Rightarrow\quad x \:=\:\text{-}2
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  3. #3
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    Re: Solve the following exponential equation

    Thank you very much!!! Your explanation was perfect!!
    Finally i understood this annoying equation!!
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