# Get arsinh from sinh

I need to establish the inverse function of the hyperbolic sine: I am trying to do this by setting $$y = \sinh(x)$$ and solving for $$x$$, however I got stuck at this: $$y=\frac{e^x -e^{-x}}{2}$$ $$2y=e^x - e^{-x}$$ I dont know how to solve for x at this point, though. Taking the logarithm seems nonsensical with a sum on the right side. The end goal is the arsinh given by $$y = \log(x + \sqrt{x^2 +1})$$

• Note that $e^{-x} = \frac{1}{e^x}$ you have a quadratic equation in $e^x$. – user10354138 May 16 '19 at 15:24

Hint: make replacement $$e^x = z$$ and use $$e^{-x} = 1 / z$$ to get quadratic equation on $$z$$.
• You mean $e^{-x}=1/z$, not $e^{-x}=z$. – user10354138 May 16 '19 at 15:25
Hint: Write $$2y=e^x-\frac{1}{e^x}$$, now substitute $$t=e^x$$ and solve the quadratic.