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Math Help - Values that satisfy exponential equation

  1. #1
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    Values that satisfy exponential equation

    Determine all values of x that satisfy the equation.

    i. x^24^{\frac{x}{2}} - 2x2^{x + 1} - 32^x = 0

    x^22^x - 2x\cdot2^x - 32^x

    2^x\left(x^2 - 4x - 3\right)

    2^x\left(x + 2\sqrt{7}\right)\left(x - 2\sqrt{7}\right)

    Values are x = \pm 2\sqrt{7}


    Also, why can't 4^{\frac{x}{2}} be considered as \left(\sqrt{4}\right)^x = 2^x?
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  2. #2
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    Re: Values that satisfy exponential equation

    How have yo got 32^x = 3*2^x. In fact 32^x= (2^5)^x = 2^5x
    secondly what you are thinking about ( sqrt4)^x = 2^x is absolutely correct.
    please check the question once again if it has been correctly written, in present form we get
    2^x[ x^2 - 4x - 2^(4x) ]=0
    Thanks from Unreal
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  3. #3
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    Re: Values that satisfy exponential equation

    Quote Originally Posted by ibdutt View Post
    How have yo got 32^x = 3*2^x. In fact 32^x= (2^5)^x = 2^5x
    secondly what you are thinking about ( sqrt4)^x = 2^x is absolutely correct.
    please check the question once again if it has been correctly written, in present form we get
    2^x[ x^2 - 4x - 2^(4x) ]=0
    I noticed that too, but the book gives works with 32^x as 3\cdot2^x.
    4^{\frac{x}{2}} is evaluated as \left(2^2\right)^\frac{x}{2} = 2^x

    I'm trying to figure out the problem along these lines, whether the book is correct or not, I don't know :-)
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    Re: Values that satisfy exponential equation

    x^2 - 4x - 3 can be written as x^2 - 4x+ 4 -7. Now factorise.
    Thanks from Unreal
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  5. #5
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    Re: Values that satisfy exponential equation

    Quote Originally Posted by sa-ri-ga-ma View Post
    x^2 - 4x - 3 can be written as x^2 - 4x+ 4 -7. Now factorise.
    \left(x^2 - 4x - 3\right)

    x = \pm\sqrt{7} + 2
    Last edited by Unreal; April 28th 2013 at 11:20 PM.
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  6. #6
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    Re: Values that satisfy exponential equation

    I would use the quadratic formula instead...you should find:

    x=2\pm\sqrt{7}
    Thanks from Unreal
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  7. #7
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    Re: Values that satisfy exponential equation

    Quote Originally Posted by MarkFL View Post
    I would use the quadratic formula instead...you should find:

    x=2\pm\sqrt{7}
    And, are those the values?
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  8. #8
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    Re: Values that satisfy exponential equation

    Yes, as 2^x=0 has no solution.
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  9. #9
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    Re: Values that satisfy exponential equation

    Quote Originally Posted by Unreal View Post
    \left(x^2 - 4x - 3\right)

    \left(x + 2\sqrt{7}\right)\left(x - 2\sqrt{7}\right)
    I did understand how did you get the above factors. In my hint I expected you to proceed by rewriting the expression as
    (X -2)2 - (71/2)2, then factories. Or use quadratic formula to find the factors.
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