Why does $\exp\left[W\left(b\left(\ln{n}\right)^2\right) - \ln{b} - \ln{\ln{n}}\right] = \frac{\ln{n}}{W(b\ln^2{n})}$?

Question

Why does $$\exp\left[W\left(b\left(\ln{n}\right)^2\right) \; - \; \ln{b} - \ln{\ln{n}}\right] = \frac{\ln{n}}{W(b\ln^2{n})}\;?$$

$W$ is the Lambert-W function and all variables are real and positive.

Answer 1

This is simple algebra; the LHS is: $\exp\left[W(b\ln^2 n)-\ln b-\ln\ln n\right] =$ $\dfrac{\exp\left[W(b\ln^2 n)\right]}{e^{\ln b}e^{\ln\ln n}} =$ $\dfrac{\exp\left[W(b\ln^2 n)\right]}{b\ln n}$. Now multiply both sides by $b\ln nW(b\ln^2 n)$.