Solving a transcendental equation How do I go about solving the following equation?
$$x = A + B \log\left( \cosh\left(\frac{x}{C}\right)\right)$$
 A: If you let $u=e^{x/BC}$, you can rewrite the equation in the form
$$ku^C = u^B+u^{-B}$$
where $k=2e^{-A/B}$.  If $B$ and $C$ are commensurable, this can be transformed into a polynomial equation in a variable that converts back to $x$.  But that's about as far as it goes, I'd say.
(Yves Daoust and I posted almost simultaenously.)
A: Set $t=e^{x/C}$. The equation turns to
$$C\log t=A+B\log\left(\frac12\left(t+\frac1t\right)\right),$$
which can be rewritten as
$$at^b-t^2-1=0,$$
with $a=2e^{-A/B}$ and $b=(C+1)/B$.
There will be closed formulas (some terrible, solutions of quartics) at least for $b=-6,-4,-2,-1,-\frac23,0,\frac12,\frac23,1,\frac43,\frac32,2,\frac83,3,4,6,8$.
For instance, with $b=4$,
$$x=C\ln\sqrt{\pm\frac{1\pm\sqrt{1+4a}}{2a}},$$
and $b=-1$
$$x=C\ln\left(\sqrt[3]{\sqrt{\frac{a^2}4+\frac1{27}}+\frac a2}+\sqrt[3]{-\sqrt{\frac{a^2}4+\frac1{27}}+\frac a2}\right).$$

For a numerical solution (using Newton for instance), rewrite as
$$\log\cosh u=cu+d,$$
where $x=Cu$, and use
$$\color{blue}{\log\cosh u}\approx\color{green}{\frac{u^2}2}$$
for small $u$ and 
$$\color{blue}{\log\cosh u}\approx\color{magenta}{|u|-\log2}$$
for large $u$.
This should give you good starting points by intersecting with the straight line.

