I am trying to find an analytic solution for $\beta$ in the following expression. However, Mathematica was not able to manage to get the solution.
Solve[2 Cosh[$\beta \nu$] == Exp[$\beta$], $\beta$]
On the other hand, solution for $\nu$ is as follows:
$\nu=\dfrac{-\cosh ^{-1}\left(\frac{e^{\beta }}{2}\right)+2 i \pi}{\beta }$
However, I can not Inverse this function. $\beta$ and $\nu$ are both positive real numbers.
I would be grateful if you can show me the way to do this.