I have to find the solution of the following equation $$e^{\frac{a}{ax+1}}=(ax+1)^{\frac{1}{x}}$$ where $a>0$. Any help in this regard will be much appreciated. Thanks in advance.
Here is my attempt:
taking $(ax+1)$ power on both sides we get $$e^{a}=(ax+1)^{a+\frac{1}{x}}$$ Now taking $x$ power on both sides we get $$e^{ax}=(ax+1)^{ax+1}$$ by using $ax+1=z$ we can write $$e^{z-1}=z^z$$ which I think is pretty much solve able but I am not sure.