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.
 A: To use Lambert W you may proceed this way :
\begin{align}
100\,n^2 &= 2^n\\
(10\,n)^2 &= e^{n\log 2}\\
10\,n &= \pm e^{n\log 2/2}\\
\left(-\,n \log 2/2\right) e^{-n\log 2/2}&= \mp (\log 2)/{20}\\
-\,n \log 2/2&=W\left(\mp (\log 2)/20\right)\\
\end{align}
The negative argument (see wikipedia) will be between $-\dfrac 1e$ and $0$ and return you two solutions :


*

*$-\frac{2}{\log 2}W_{-1}\left(-\frac{\log 2}{20}\right)\approx 14.324727837\quad$ and 

*$-\frac{2}{\log 2}W\left(-\frac{\log 2}{20}\right)\approx 0.1036578164$
while the positive argument will return simply 

*$-\frac{2}{\log 2}W\left(\frac{\log 2}{20}\right)\approx -0.0967040343267$ 

A: $$2^n=100n^2$$
$$2^n n^{-2}=100$$
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.
A: We obtain
$$n_1=-2\,{\frac {{\rm W} \left(-1/20\,\ln  \left( 2 \right) \right)}{\ln 
 \left( 2 \right) }}
$$
$$n_2=-2\,{\frac {{\rm W} \left(-1,-1/20\,\ln  \left( 2 \right) \right)}{
\ln  \left( 2 \right) }}
$$
$$n_3=-2\,{\frac {{\rm W} \left(1/20\,\ln  \left( 2 \right) \right)}{\ln 
 \left( 2 \right) }}
$$
