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Thread: Inverse of equation involving e^(-x)

  1. #1
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    Inverse of equation involving e^(-x)

    Hi,

    My problem is to find the inverse of $\displaystyle y = f(x) = \frac{1}{2}(e^x - \frac{1}{e^x}) $.

    I can get up to using $\displaystyle v$ for $\displaystyle e^x$, and that $\displaystyle v > 0$

    $\displaystyle 2yv = v^2 -1$

    $\displaystyle 0 = v^2 - 2yv - 1$

    The using the quadratic formula I get $\displaystyle v = \frac{2y \pm \sqrt{(-2y)^2 + 4}}{2}.$

    Here I am stuck


    Sorry for bad format.
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  2. #2
    Super Member Deadstar's Avatar
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    Following on from where you left off...

    $\displaystyle v = \frac{2y}{2} \pm \frac{\sqrt{4(y^2 + 1)}}{2}$ (since $\displaystyle (-2y)^2 + 4 = 4y^2 + 4 = 4(y^2 + 1)$)

    Note that $\displaystyle 2 = \sqrt{4}$ so we get

    $\displaystyle v = y \pm \sqrt{(y^2 + 1)}$

    Now since $\displaystyle v = e^x$, take natural logs of both sides to get...

    $\displaystyle x = \ln(v) =$ ..? Can you finish?
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  3. #3
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    Yes.

    I kept going around in circles about what to do with the square root in the quadratic formula.

    All is good now. Thanks
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  4. #4
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    Quote Originally Posted by Deadstar View Post
    Following on from where you left off...

    $\displaystyle v = \frac{2y}{2} \pm \frac{\sqrt{4(y^2 + 1)}}{2}$ (since $\displaystyle (-2y)^2 + 4 = 4y^2 + 4 = 4(y^2 + 1)$)

    Note that $\displaystyle 2 = \sqrt{4}$ so we get

    $\displaystyle v = y \pm \sqrt{(y^2 + 1)}$

    Now since $\displaystyle v = e^x$, take natural logs of both sides to get...

    $\displaystyle x = \ln(v) =$ ..? Can you finish?
    I'll add a small note: One of the solutions (either the positive or the negative root) must be discarded ....
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