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Math Help - One more exponential equation

  1. #1
    Member
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    One more exponential equation

    2^2x -2 x 2^x -8 = 0

    Please help solve this equation.

    Thank you very much
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  2. #2
    MHF Contributor
    Joined
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    2^2x -2 x 2^x -8 = 0

    I'm sure that is
    2^(2x) -2(2^x) -8 = 0
    In another form,
    (2^x)^2 -2(2^x) -8 = 0 -----------(i)

    (i) is a quadratic erquation in 2^x.
    So if y = 2^x,
    then (i) becomes
    y^2 -2y -8 = 0
    So,
    (y -4)(y+2) = 0
    y = 4 or -2

    When y = 4,
    2^x = 4
    Take the natural logs of both sides,
    x*ln(2) = ln(4)
    x = ln(4) / ln(2)
    x = 2ln(2) / ln(2)
    x = 2 ----------------answer.

    When y = -2,
    2^x = -2
    x*ln(2) = ln(-2)
    There are no real values for logs of negative numbers, so reject this.
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  3. #3
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    Oct 2007
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    Ticbol,

    thank you. You explained it so well. It's great that we have people like you here ready to explain and help.
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