Solve the following equation for $x$: $$x=e^{a\frac{\ln(1+b/x)}{\ln(1+b/x)+c}}$$ where $a,b,c>0$
We can show that $1+b/x=1+b e^{-a\frac{\ln(1+b/x)}{\ln(1+b/x)+c}}$. Thus, $$\ln(1+b/x)=\ln\left(1+b e^{-a\frac{\ln(1+b/x)}{\ln(1+b/x)+c}}\right)$$
Now, let $y=\ln(1+b/x)$. Then, $y=\ln\left(1+b e^{-a\frac{y}{y+c}}\right)$. Therefore,
$$e^y=1+b e^{-a\frac{y}{y+c}}$$
I don't know how to continue from here. Any idea what is the answer?
Even if I find the answer in series, that would be nice
This equation is a trascendent equation. There is no formula to solve it.
The best you can do, for given $a=1;\;b=1;\;c=1$ for instance is to look for a graphic solution and/or a numerical solution
To solve $x=e^{\frac{\ln(1+1/x)}{\ln(1+1/x)+1}}$ you plot in the same graph the line $y=x$ and the curve $y = e^{\frac{\ln(1+1/x)}{\ln(1+1/x)+1}}$