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Math Help - How do I solve for x?

  1. #1
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    How do I solve for x?

    If x=\frac{e^t-e^{-t}}{2}, solve for t.

    This is my work.

    2x=e^t-e^{-t}
    \ln 2x=\ln (e^t-e^{-t})

    I'm stuck trying to simplify \ln (e^t-e^{-t}). Help?

    Thanks.
    Last edited by chengbin; March 30th 2010 at 04:00 PM.
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  2. #2
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    Quote Originally Posted by chengbin View Post
    If x=\frac{e^t-e^{-t}}{2}, solve for t.

    This is my work.

    2x=e^t-e^{-t}
    \ln 2x=\ln (e^t-e^{-t})

    I'm stuck trying to simplify \ln (e^t-e^{-t}). Help?

    Thanks.
    x = \frac{e^t - e^{-t}}{2}<br />

    2x = e^t - e^{-t}<br />

    2xe^t = e^{2t} - 1

    0 = e^{2t} - 2xe^t - 1

    e^t = \frac{2x \pm \sqrt{4x^2 + 4}}{2}<br />

    e^t = \frac{2x \pm 2\sqrt{x^2 + 1}}{2}

    e^t = x \pm \sqrt{x^2+1}

    since \sqrt{x^2+1} > x ...

    e^t = x + \sqrt{x^2+1}<br />

    t = \ln\left(x + \sqrt{x^2+1}\right)
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  3. #3
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    Thanks for that solution.

    Is there a way to move on like the way I did it?
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  4. #4
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    No, your method was making things more complicated rather than simplifying. One thing that might make the solution easier to understand is to let [tex]y= e^t[tex] to start with. Since e^{-t}= \frac{1}{e^t}= \frac{1}{y}, your equation becomes x= \frac{y+ 1/y}{2}. Multiply both sides by 2y to get [tex]2xy= 2y^2+ 1[tex] or 2y^2- 2xy+ 1= 0. Use the quadratic formula to solve that for y and then take t= ln(y).
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