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Math Help - Need some Help with an Exponential Equation

  1. #1
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    Need some Help with an Exponential Equation

    Thanks in advance for any help!

    I am having trouble with this equation:

    2^(3*x)-3*2^(2*x)-6*2^x+8=0


    My main problem is how to deal with the exponents that are multiplied by a constant, here:

    2^(3*x)-3*2^(2*x)-6*2^x+8=0

    I cant just multiply these can I ? and end up with 12^x and so on, can I?

    Thank you!
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  2. #2
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    Re: Need some Help with an Exponential Equation

    Let 2^{x}=X and you will find that you have a cubic equation in X.
    One solution of the equation is easy to spot and you can then factorise.
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  3. #3
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    Re: Need some Help with an Exponential Equation

    I'm sorry but I'm not sure I understand your answer? I should replace the power, 2x with X? So can I just multiply the constants in front of the powers by the constants, then?

    Thanks again!
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  4. #4
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    Re: Need some Help with an Exponential Equation

    The second term, for example, 3*2^{2x} can be rewritten as 3*(2^{x})^{2}=3X^{2}.
    Substitute in the first and third terms in the same way and you have a cubic equation in X.
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  5. #5
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    Re: Need some Help with an Exponential Equation

    Hello, Rhys101!

    2^{3x} - 3\cdot2^{2x} - 6\cdot2^x+8\:=\:0

    We have: . (2^x)^3 - 3(2^x)^2 - 6(2^x) + 8 \:=\:0

    Let u = 2^x\!:\;\;u^3 - 3u^2 - 6u + 8 \:=\:0

    Factor: . (u-1)(u+2)(u-4) \:=\:0

    We have three roots: . u \;=\;1,\,\text{-}2,\,4

    Back-substitute: . \begin{Bmatrix}2^x \,=\,1 & \Rightarrow & x \,=\,0 \\ 2^x \,=\,\text{-}2 & \Rightarrow & \text{no real root} \\ 2^x \,=\,4 & \Rightarrow & x \,=\,2 \end{array}


    Therefore: . x\;=\;0,\,2

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  6. #6
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    Re: Need some Help with an Exponential Equation

    Thanks so much for all your help, both of you! I really appreciated it!
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