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I need to solve something like this
$\exp(-a x) = \frac1{x^b}$ where $a$ and $b$ are positive real values. Do other results exist when $b > 1$ or do I have to rely on numerical inspection?

1 Answers 1

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Observe:

$x^b \exp(-ax)=1$

$x \exp\left(-\frac{a}{b}x\right)=1$

$-\frac{a}{b}x \exp\left(-\frac{a}{b}x\right)=-\frac{a}{b}$

$-\frac{a}{b}x=W\left(-\frac{a}{b}\right)$

$x=-\frac{b}{a}W\left(-\frac{a}{b}\right)$

Now, the truth of the matter is that the Lambert function has two real branches for $x\in\left[-\frac1{e},0\right)$: the principal branch $W_0(x)$ and the branch $W_{-1}(x)$; you will have to check with your application which of these two solutions makes sense if the argument of the Lambert function falls in that interval. For nonnegative $x$, on the other hand, $W_0(x)$ is the only branch that yields real results.

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