# Solution to $a\cdot e^{bx} - cx = d$

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

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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

$$x=-\frac{d}{c}-\frac{1}{b}W\left(-\frac{ab}{c}e^{-bd/c}\right).$$

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You don't need to assume $a = 1$. One solution is \begin{align} x = - \frac{bd + c \ W(-\frac{ab}{c}e^{-bd/c})}{bc}, \end{align} where $W$ is the Lambert $W$ function. This can be proved using the functional identity $W(z) e^{W(z)} = z$.

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You may assume $a=1$. Then Wolfram Alpha gives you the answer in terms of the Lambert $W$ function.

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