I have a simple question for which I am looking for a closed form expression (If there exits one). In other words, given:
$y=W(e^{ax+b})-W(e^{cx+d})+zx$
where $W$ is the Lambert $W$ function and $a,b,c,d,z$ are some constants, what is the function $f$, such that $x=f(y)$
Thanks alot in advance.
EDIT : If there exists no closed form solution, I will be happy to see nice arguments supporting this.
$\rightarrow$EDIT2 : As can be seen, we have a solution for the simplified version of this problem. If there exists a solution I have the following ideas to resolve the full version of the problem:
$1$- Is it possible to write $W(e^{f(x)})=W(e^{ax+b})-W(e^{cx+d})$
$2$- Having $y_1=W(e^{ax+b})+z_1x$ and $y_2=-W(e^{cx+d})+z_2x$
where $z=z_1+z_2$, $y=y_1+y_2$ and $f^{-1}(y_1)$ and $f^{-1}(y_2)$ are known functions as already found. Can we say that $f^{-1}(y)=f^{-1}(y_1)+f^{-1}(y_2)$? or can we modify this idea to get somethign useful?