I'm trying to find the solution for $x$ from the following equation, which I believe involves some sort of of Lambert $W$ function or product logarithm.
$$\frac{a}{x}\left[\frac{c}{a(b-1)} + \frac{e}{(b-1)d} \right]=\ln{\left(\frac{a}{x}\right)}$$
under the following the parameter ranges: $0<c<x<a,\quad b>\frac{d+e}{d}, \quad d>0, \quad e>0 $.
Can anyone help please?
Solve $$ \frac{a}{x}\;B = \log\frac{a}{x} $$ Try to get it in the form $v=ue^u$. Then we can switch it to $W(v) = u$. $$ \frac{a}{x}\;B = \log\frac{a}{x} \\ e^{Ba/x} = \frac{a}{x} \\ 1 = \frac{a}{x}e^{-Ba/x}\; \\ -B = \frac{-Ba}{x}e^{-Ba/x}\; \\ W(-B) = -\frac{Ba}{x} \\ x = \frac{-Ba}{W(-B)} $$