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Math Help - Equation that makes my life harder

  1. #1
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    Equation that makes my life harder

    Hello everybody.
    I have an equation that I am trying to solve for 7 hours. But nothing made me success, unfortunately. So I decided to ask for help there.
    Equation that makes my life harder-gif.latex.gif
    I will be very lucky if somebody helps me to solve my problem.

    P.S.: Sorry for my English, I am only beginner now.
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  2. #2
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    Re: Equation that makes my life harder

    Let x=2X
    (4+sqrt(15))^X + (4-sqrt(15))^X = 8^X
    An obvious solution is X=1 . So x=2.
    Thanks from Shakarri, topsquark and imclaren
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  3. #3
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    Re: Equation that makes my life harder

    Thank you for finding the solution but I have already found it.
    I must proof that this equation has only one decision. (I can't proof it using graph because it is enough difficult to construct y=(4+sqrt(15))^X + (4-sqrt(15))^X without, for example, WolframAlpha).
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  4. #4
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    Re: Equation that makes my life harder

    Let a = 4 + \sqrt{15} and b = 4 - \sqrt{15}

    so that

    a^{x/2} + b^{x/2} = 8^{x/2}

    Multiply this equation by a^{x/2}

    a^x + 1 = \left(\sqrt{8a}\right)^x since a b=1
    Thanks from imclaren
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  5. #5
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    Re: Equation that makes my life harder

    I'm probably idiot, but I don't really know, how to solve this equation. Sorry. Equation that makes my life harder-png.latex.png
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  6. #6
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    Re: Equation that makes my life harder

    Let ax=t2
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  7. #7
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    Re: Equation that makes my life harder

    I don't know how to solve this equation either and I don't see how letting a^x = t^2 will help.

    On the other hand if you graph the function

    f(x) = a^x+1 - \left(\sqrt{8a}\right)^x with x > 0

    you will notice that it is decreasing and that f(2) = 0

    This shows that x = 2 is the only solution

    To prove that it is the only solution you would have to show f ' (x) < 0
    Thanks from imclaren
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  8. #8
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    Re: Equation that makes my life harder

    Quote Originally Posted by Idea View Post
    I don't know how to solve this equation either and I don't see how letting a^x = t^2 will help.

    On the other hand if you graph the function

    f(x) = a^x+1 - \left(\sqrt{8a}\right)^x with x > 0

    you will notice that it is decreasing and that f(2) = 0

    This shows that x = 2 is the only solution

    To prove that it is the only solution you would have to show f ' (x) < 0
    My mistake ! You are right.
    Thanks from imclaren
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