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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?

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1 Answer 1

up vote 6 down vote accepted

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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Useful functions don't just solve a single equation, but a class of equations... –  Fabian Apr 20 '11 at 7:06

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