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Thread: Values that satisfy exponential equation

  1. #1
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    Values that satisfy exponential equation

    Determine all values of $\displaystyle x$ that satisfy the equation.

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

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

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

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

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


    Also, why can't $\displaystyle 4^{\frac{x}{2}}$ be considered as $\displaystyle \left(\sqrt{4}\right)^x = 2^x$?
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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
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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 $\displaystyle 32^x$ as $\displaystyle 3\cdot2^x$.
    $\displaystyle 4^{\frac{x}{2}}$ is evaluated as $\displaystyle \left(2^2\right)^\frac{x}{2} = 2^x$

    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.
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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.
    $\displaystyle \left(x^2 - 4x - 3\right)$

    $\displaystyle x = \pm\sqrt{7} + 2$
    Last edited by Unreal; Apr 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:

    $\displaystyle x=2\pm\sqrt{7}$
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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:

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

    Yes, as $\displaystyle 2^x=0$ has no solution.
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    Re: Values that satisfy exponential equation

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

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