2
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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