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I'm trying to find and inverse function and I reached the equation $x=(e^{y}-e^{-y})/2$

How do I solve it for $y$?


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$y = \sinh^{-1} (x)$ – Ian Mateus Dec 18 '12 at 13:23
up vote 2 down vote accepted

Here you multiply through by $2 e^{y}$ on both sides to get

$$e^{2 y} - 2 x e^{y} - 1 = 0$$

Solve for $e^{y}$:

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

and get

$$y = \log{\left ( x \pm \sqrt{x^2+1} \right ) }$$

Which sign to use? If $y$ is real, then of course use the positive sign. This is, of course, $\sinh^{-1} x$.

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Oh seems so simple now. Thanks! – Smithnson Dec 18 '12 at 13:25

Besides to @rlgordonma's answer note that we can write $x=\frac{e^{y}-e^{-y}}{2}$ as $x=\sinh(y)$. So $y=\text{arcsinh}(x)$ on a proper interval.

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Indeed! :^) $\quad +1$ – amWhy Apr 15 '13 at 0:22

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