A solution to a transcendental equation

Question

I have a transcendental equation as follows:

$$ e^{ax} = b+ax $$

for some constants $a>0$ and $b>0$. Is there a known solution for $x$?

I tried on Wolfram and it gave a solution which depended on an inverse logarithm function. Was trying to at least get an upper bound for $x$ using Taylor series, but went in vain.

Thanks for any help regarding a solution or a suggestion or an upper bound to $x$.

Answer 1

Substitute $z=ax+b$. $$e^{z-b}=e^{-b}e^z=ke^z=z$$ $$-k=-ze^{-z}$$ The RHS is of the form where we can apply the Lambert W function, the inverse of $xe^x$: $$W(-k)=-z$$ Thus $$x=-\frac{W(-e^{-b})+b}a$$