TITLE: Solution to $a\cdot e^{bx} - cx = d$
QUESTION [6 upvotes]: Similar to the question asked here: Solving $e^x + x = 5$ for $x$ without using a numerical method?
How can I get a solution for $a\cdot e^{bx} - cx = d$, where a, b, c, d are constants? Is there a way I can get it in terms of the Lambert W function?
Thanks,
Mike

REPLY [5 votes]: Write $y=(b/c)(cx+d)$ to get $(b/c) \cdot a e^{bx}= y$. Note $bx=y-bd/c$, so this is $(ab/c)e^{y-bd/c}=y$, and then multiply by $-e^{-y}$ to get $-(ab/c) e^{-bd/c} = -ye^{-y}$. This means $-y=W(-\frac{ab}{c} e^{-bd/c})$, so
$$x=-\frac{d}{c}-\frac{1}{b}W\left(-\frac{ab}{c}e^{-bd/c}\right).$$