# Is it possible to find a solution for any $ae^{bx}+cx+d=0$?

So I enjoy solving problems of the form$$ae^{bx}+cx+d=0$$$$a,b,c\ne0$$however what I don't particularly enjoy is having to solve that equation every single time I come across it. So what I want to know is: Is there a way to solve every non-algebraically solvable equation of that form?

Here is my attempt:$$ae^{bx}+cx+d=0\implies a+(cx+d)e^{-bx}=0$$$$\implies(cx+d)e^{-bx}=-a\quad\text{rearranging terms}$$$$\implies((-c/b)(-bx-bd/c))e^{-bx}=-a$$$$\implies(-bx-bd/c)e^{-bx}=ab/c$$$$\implies(-bx-bd/c)e^{-bx-bd/c}=(ab/c)e^{-bd/c}$$and now taking the Lambert W of both sides (note that the solution uses $W_k(z)$ since all branches of the product log function will yield a valid solution),$$-bx-bd/c=W_k((ab/c)e^{-bd/c})$$which leads to the solution$$x=(1/b)W_k((ab/c)e^{-bd/c})-d/c$$However, my question is: Is my solution valid, or how would one go about finding the solution for any $ae^{bx}+cx+d$ where $a,b,c\ne0$?

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