Lambert W function with rational polynomial During my research I ran into the following general type of equation: 
$$\exp(ax+b)=\frac{cx+d}{hx+f}. $$
Does anyone have an idea how to go about solving this equation?
Thanks in advance.
 A: Given:
$$\exp(ax+b)=\frac{cx+d}{hx+f}. $$
substituting $ax+b=z$
we obtain:
$$e^z=\frac{cz-bc+da}{hz-bh+fa} $$
setting:
$$A=c/h$$
$$t=(bc-da)/c$$
$$s=(bh-fa)/h$$
the equation becomes:
$$e^z=A\frac{z-t}{z-s} $$
This equation can be solved by means of a generalization of Lambert W function that has been described in:
"On the generalization of the Lambert W function with applications in theoretical physics", István Mező, Árpád Baricz , http://arxiv.org/abs/1408.3999
The solution (obtained by Lagrange inversion) can be written as a series involving Laguerre polynomials. In particular the authors found that above seen equation can be formally solved by Lagrange series (related to Lambert W function):
$$z(A,t,s)=t- (t-s) \sum_{n=1} \frac{L_n' (n(t-s))}{n} e^{-nt} A^n$$
where:
$L_n(x)$ is n-th Laguerre polynomial and $L_n'(x)$ its first derivative
Replacing $t$, $s$, $A$, $z$, the formal solution of $\exp(ax+b)=\frac{cx+d}{hx+f}$ has been obtained.
A: I think that, for the most general case, there is no closed form to the solution of the equation $$e^{a x+b}=\frac{c x+d}{h x+f}$$ For the particular case where $c=0$, you would have $$x=\frac{W\left(\frac{a d e^{\frac{a f}{h}-b}}{h}\right)}{a}-\frac{f}{h}$$ For the particular case where $h=0$, you would have $$x=-\frac{W\left(-\frac{a f e^{b-\frac{a d}{c}}}{c}\right)}{a}-\frac{d}{c}$$ where effectively appear Lambert function. Even the case $c=h$ does not seem to have a solution.
For your more general case $(c \neq 0,h\neq 0)$, I wonder if numerical method should not be the only practical way.
If there is a specific case you would like to be solved, post the values of the six parameters and I should try to show you.
