Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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

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

share|cite|improve this answer

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 $W(z) e^{W(z)} = z$.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.