How do I apply the product log function (W(x)) to this equation?

Question

I have the following:

$$3(e^{hv\over kT}-1)v^2 = e^{hv\over kT}\frac {hv^3}{kT}$$

Which is the numerator of the derivative of Planck's energy distribution formula when the derivative is set to $0$.

According to WolframAlpha I can manipulate this equation to be:

$$v = \frac {kT(W({-3\over e^3}) + 3)}{h}$$

I understand how the $W(x)$ product log function works, I'm just confused about what happens in the middle here to produce that final equation. If you divide both sides by $v^2$ the equation looks incredibly close to being in product log form, however there is one final leap which I am missing.

Thanks in advance.

Answer 1

The basic identity satisfied by $W$ is $W(x) \exp(W(x)) = x$. Now if you let $t = h v/(kT)$, your equation says $$ 3 (e^t - 1) = e^t t$$ or $$ (3-t) e^t = 3$$ or $$ (t-3) e^{t-3} = -3 e^{-3}$$ So take $x = -3 e^{-3}$, and if $t - 3 = W(x) = W(-3 e^{-3})$, you do indeed get $$(t-3) e^{t-3} = W(x) e^{-W(x)} = x = -3 e^{-3}$$