Consider this equation:

$$x^{t-1}e^{-x} = a$$

I am aware that this is what you integrate from $0$ to $\infty$ in respect to $x$ to get the Gamma Function, but I do not want to worry about it here. I want to solve this for $x$, but you can see that it is both the base of an exponent and an exponent itself. I have a feeling I will need to use the Lambert W Function.

  • $\begingroup$ I suggest you keep in mind that any equation which can write $A+Bx+C\log(D+Ex)=0$ as solutions in terms of Lambert function(s). $\endgroup$ – Claude Leibovici Jul 19 '15 at 5:03

As surmised, the Lambert W is indeed the way to go. To that end, we have

$$x^{t-1}e^{-x}=a\implies xe^{-x/(t-1)}=a^{1/(t-1)}\implies \frac{-x}{t-1}e^{-x/(t-1)}=\frac{-a^{1/(t-1)}}{t-1}$$

Inasmuch as the $W$ function is defined by $z=W(z)e^{W(z)}$, we see immediately that

$$\bbox[5px,border:2px solid #C0A000]{x=-(t-1)W\left(\frac{-a^{1/(t-1)}}{t-1}\right)}$$

  • $\begingroup$ Thank you! But how did you get $xe^{-x/{t-1}}$. Could you clear that up for me? $\endgroup$ – Sam Jul 19 '15 at 3:12
  • $\begingroup$ @Sam You're welcome. My pleasure. In that first step, we raised both sides to the $1/(t-1)$ power. So, using $(e^x)^y=e^{xy}$, we have $(e^{-x})^{1/(t-1)}=e^{-x/(t-1)}$. Make sense? $\endgroup$ – Mark Viola Jul 19 '15 at 3:15
  • $\begingroup$ Oh, I apologize. I didn't see that you had taken $a$ (both sides) to the $1/{t-1}$ power. I thought you had simplified that down to that in some way. Thank you, I understand. $\endgroup$ – Sam Jul 19 '15 at 3:16
  • $\begingroup$ @Sam No need to apologize at all. And pleased to hear that you've got it now! Please let me know if I can improve further on the answer. I just want to give you the best answer I can. $\endgroup$ – Mark Viola Jul 19 '15 at 3:18
  • $\begingroup$ Your answer is already perfect! I asked what the solution is for x and you were able to come up with that. I completely understand how you did it. $\endgroup$ – Sam Jul 25 '15 at 2:12

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