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Math Help - Solve equation

  1. #1
    Junior Member
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    Solve equation

    Hello,

    How do I solve the following equation? 2^2^x-3*2^x-40 = 0

    I tried the following, but apparently this is not correct, i.e. score = 0%.

    2^2^x-3*2^x-40 = 0
    2^2^x-\sqrt{3}*2^2^x = 40
    (1-\sqrt{3})*2^2^x = 40
    2^2^x = \frac{40}{1-\sqrt{3}}
    2x = \log_{2}\frac{40}{1-\sqrt{3}}
    x = \frac{\log_{2}\frac{40}{1-\sqrt{3}}}{2}
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  2. #2
    Senior Member
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    Re: Solve equation

    You changed 3 to root3 Why?
    Let y=2^x Then 2^2x= (2^x)^2=y^2
    So equation becomes y^2-3y-40=0
    (y-8)(y+5)=0 y=8 0r-5
    So 2^x=8 giving x=3 ( and 2^x=-5 has no solution)
    Thanks from Lotte1990
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