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 $y$?
Thanks!
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 $y$?
Thanks!
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, $\sinh^{-1} x$.
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.