Solving for $x$ in logarithmic equation

Question

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?

Answer 1

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.

Answer 2

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.