Solving for x in $Ae^{bx} +cx = d$ I ran into this problem, and after taking the logarithm of each side didn't make the solution apparent to me, i ran out of tools to solve this one. Any clues as to how i should proceed to get a solution for x?
$Ae^{bx} +cx = d$
 A: The traditional answer is: Use the Lambert W- function(s). 
But to discuss the equation it’s good to know something about the curve of $\,\displaystyle y=x^\frac{1}{x}\,$ .
The equation $\,\displaystyle Ae^{bx}+cx=d\,$ is equivalent to $\,\displaystyle y=z^{\frac{1}{z}}\,$ with $\,\displaystyle y=e^{-\frac{b}{c}Ae^{\frac{bd}{c}}}\,$ and $\,\displaystyle z=e^{-\frac{b}{c}(d-cx)}\,$ .  
If $\,\displaystyle y>e^\frac{1}{e}\,$ there are no solutions. If $\,0<y\leq 1\,$ or $\,\displaystyle y= e^\frac{1}{e}\,$ then we have one solution. 
And for $\,\displaystyle 1<y< e^\frac{1}{e}\,$ there are two solutions.
A: There is no algebraic exact solution for this. An advice is to use the Lambert W Function.
Another possibility is an asymptotic evaluation, but you need suppositions which may be incorrect, so treat this just like a sort of "case". The one in which $bx$ is very small, that is we are searching for a solution near the origin.
Set $bx = z$ hence $x = \frac{z}{b}$ and we assume $b>>0$. Then with Taylor
$$Ae^{bx} + cx = d$$
$$Ae^z + c\frac{z}{b} = d$$
$$A(1 + z) + \beta z = d$$
Where $\beta = \frac{c}{b}$
$$A + Az + \beta z - d = 0$$
$$z(A + \beta) = d - A$$
$$z = \frac{d - A}{A + \beta}$$
Then go back for $x$
$$bx = \frac{d - A}{A + \frac{c}{b}}$$
$$x = \frac{1}{b}\frac{d - A}{A + \frac{c}{b}}$$
I repeat it again: it's a special case
