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Could anyone please show me how to solve for $x$ this equation:

$$1 = ax e^{-bx}$$

If solved in terms of the Lambert $W$-function, would that be considered a concrete answer?

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Emilio B.
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  • $$1=axe^{-bx}\implies 0=\log a+\log x-bx$$

    this is not what you wrote after "or" ...

     – DonAntonio Sep 21 '13 at 18:57
  • @DonAntonio, I suppose th OP only wanted to describe the structure of the equations. $a$ and $b$ in the first equation have nothing to do with $a$ and $b$ in the second one. Maybe it would be better to take $c$ and $d$ for the second equation if this was his intention. – miracle173 Sep 21 '13 at 19:28
  • $\large{1 \over a} > b{\rm e}^{-b^{2}}$: NO solution. $\large{1 \over a} = b{\rm e}^{-b^{2}}$: ONE solution: $\large x = b$. $\large{1 \over a} < b{\rm e}^{-b^{2}}$: TWO solutions: one $\large < b$ and the other one $\large > b$. – Felix Marin Sep 21 '13 at 19:44
  • @Emilio : you should read the comments and edit your question because your second equation doesn't appear to be the same as our first. Also please tell us what you tried. I did not downvote it, and I don't think it merits 3 downvotes. – Stefan Smith Sep 21 '13 at 22:38
  • You can find a solution in terms of Lambert W-function. – Mhenni Benghorbal Sep 21 '13 at 22:52

2 Answers2

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You will need to use the Lambert W-function.

The solution is:

$\displaystyle a\ne0,\quad b=0,\quad x=\dfrac1a\\\,\\a\ne0,\quad b\ne0,\quad x=-\dfrac{W_n\left( -\frac ba\right)}b,\quad n\in\mathbb Z$

Please don't hesitate to ask for further help or explanations.

  • Hey thanks for the solution. Just one question: how do I find the value of n? – Emilio B. Sep 22 '13 at 15:08
  • Have a look at this paper (you will find it online): it will provide you with all information you need about W function. Also, you can look up the voice "analytic continuation". Corless, R.M.; Gonnet, G.H.; Hare, D.E.G.; Jeffrey, D.J.; and Knuth, D.E. "On the Lambert W Function." Advances in Computational Mathematics, Vol. 5, (1996): 329-359. –  Sep 22 '13 at 16:18
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If $a,b\ne0$, $$ \begin{align} 1&=axe^{-bx}\\ -\frac ba&=-bxe^{-bx}\\ \mathrm{W}\left(-\frac ba\right)&=-bx\\ -\frac1b\mathrm{W}\left(-\frac ba\right)&=x \end{align} $$ There is an infinite sequence of branches for Lambert-W, just as there is for logarithm. Otherwise, yes, this would be considered a concrete solution. Unless $b=0$, there is no other closed-form solution.

robjohn
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