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Thread: Exponential Equation.

  1. #1
    Newbie Jeet's Avatar
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    Smile Exponential Equation.

    I got this question and I am unable to solve it. Please help :-0

    For what value of k the equation e^(2x)=k(x^0.5) has only one solution?

    Help appreciated :-)

    Quote Originally Posted by Jeet View Post
    In this whole forum can't anyone help me !
    Its worth to mention the answer should be independent of x and in terms of e.
    Last edited by mr fantastic; September 20th 2010 at 03:41 PM. Reason: Merged posts, edited title.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Is this whole question in itself? Or there are other smaller parts preceding this one? Maybe there are hints given previously...
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  3. #3
    Newbie Jeet's Avatar
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    Thumbs down

    This is the whole question. No hints, nothing :-(
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Ok, I think I got the answer.

    If you sketch a graph of y= e^{2x} and that of y = k\sqrt{x}, you'll notice that if they meet at a point only, their gradients are the same.

    This means, their derivative are equal.

    e^{2x} = k\sqrt{x}

    2e^{2x} = \dfrac{k}{2\sqrt{x}}

    k = 4\sqrt{x}e^{2x}

    Plugging that in our equation:

    e^{2x} = (4\sqrt{x}e^{2x})\sqrt{x}

    1 = 4x

    x = \dfrac14

    Now, plugging this value of x into the first equation:

    e^{2\frac14} = k\sqrt{\frac14}

    e^{\frac12} = \frac{k}{2}

    k = 2\sqrt{e}

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