How to find the inverse function of a logarithm?

Question

I was wondering if someone had any idea how to solve this directive?

Find the inverse function of $$y=\frac{1}{2}(2^x - 2^{-x}).$$

Thank you! Any help is appreciated.

Answer 1

$$f(x)=y=\frac{2^x-2^{-x}}{2} \implies 2^x= (y\pm\sqrt{1+y^2})$$ $$\implies x=\log_2 [y + \sqrt{1+y^2}]$$ $+$ sign has been chosen. So $$f^{-1}(x)=\log_2[x+\sqrt{1+x^2}]$$

Answer 2

Hint:

Rewrite $y$ as $$y=\frac{\mathrm e^{x\ln2}-\mathrm e^{-x\ln2}}{2}=\sinh(x\ln 2).$$