Using Lambert's W Function to solve this equation I'm attempting to solve the following equation (eventually with Lambert's W Function having checked the solution on Wolfram Alpha):
$$100n^2 = 2^n$$
I got as far as follows but I am unsure how to progress:
$$\ln100 + 2\ln(n) = n\ln2$$
Any suggestions would be much appreciated.

$$2^n=100n^2$$
$$2^n n^{-2}=100$$
Raising to the power of $-1/2$:
$$2^{-n/2} n= \pm\frac{1}{10}$$
$$e^{- 1/2 \log (2) \cdot n } n= \pm\frac{1}{10}$$
$$e^{- 1/2 \log (2) \cdot n } (- 1/2 \log (2) n)=\pm\frac{1/2 \log (2) }{10}$$
$$e^{- 1/2 \log (2) \cdot n } (- 1/2 \log (2) n)=\pm\frac{\log (2) }{20}$$
$$n = \frac{W\left(\pm\frac{\log (2) }{20}\right)}{-1/2 \log(2)}$$
$$n = -\frac{2W\left(\pm\frac{\log (2) }{20}\right)}{\log(2)}$$
Now we must consider the $\pm$ sign and the different branches of W to obtain the answers in Dr. Sonnhard Graubner's answer.
