Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
3  
$y = \sinh^{-1} (x)$ –  Ian Mateus Dec 18 '12 at 13:23
add comment

2 Answers

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, $\sinh^{-1} x$.

share|improve this answer
    
Oh seems so simple now. Thanks! –  Smithnson Dec 18 '12 at 13:25
add comment

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.

share|improve this answer
    
Indeed! :^) $\quad +1$ –  amWhy Apr 15 '13 at 0:22
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.