I have an equation of the form

$$ x a + b = \exp(xc)$$

where I, in fact, know that $b =1$. Which implies that one solution to the equation is always at $x = 0$. I'm now searching for the other solution.

I learned from Wolfram Alpha about the existence of the Lambert W function, and tried to employ it here. I got

$$x_n = \frac{-a W_n(-c \frac{\exp(-bc/a)}{a}) - bc}{ac}$$

, where $W$ is the $n$ths branch of the Lambert W function.

However, I want to restrict my solution to the reals, and don't know how. Plugging in arbitrary numbers, I got that for $n=0$, I get $x=0$, and for $n=2$, I get another real solution, and anything else appears to be complex.

How do I determine $\{n \in \mathcal N: x_n \in \mathcal R\}$?

  • $\begingroup$ @JJacquelin I don't understand. with $x = 0$, the right hand side becomes $1$, which solves the equation if the left hand side is too one, which it is for $b=1$. $\endgroup$ – FooBar Sep 16 '15 at 8:45
  • $\begingroup$ Sorry for my mistake. $\endgroup$ – JJacquelin Sep 16 '15 at 8:50
  • $\begingroup$ On the real domain, there are only two branches. The main branch $W_0(x)$ and the branch $W_{-1}(x)$ :en.wikipedia.org/wiki/Lambert_W_function $\endgroup$ – JJacquelin Sep 16 '15 at 9:02

The only two branches of $W$ which can ever assume real values are the $n=0$ and $n=-1$ branches. For this to happen in your case, assuming that $a,b,c\in\mathbb{R}$, the argument to $W$ (from your solution, above) has to satisfy:

$$-\frac{1}{e}\le -\frac{c\cdot\exp\left(\frac{-cb}{a}\right)}{a},\,\,n=0$$ $$-\frac{1}{e}\le -\frac{c\cdot\exp\left(\frac{-cb}{a}\right)}{a}\lt 0,\,\,n=-1$$

If the first of the above conditions is satisfied you will get a real root for $n=0$. If the second is also satisfied you will get a second real root for $n=-1$. All other roots will be complex.


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.