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.

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 :

1. $-\frac{2}{\log 2}W_{-1}\left(-\frac{\log 2}{20}\right)\approx 14.324727837\quad$ and
2. $-\frac{2}{\log 2}W\left(-\frac{\log 2}{20}\right)\approx 0.1036578164$
while the positive argument will return simply
3. $-\frac{2}{\log 2}W\left(\frac{\log 2}{20}\right)\approx -0.0967040343267$

$$2^n=100n^2$$

$$2^n n^{-2}=100$$

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.

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

• Apologies, there was indeed a typo in the original problem. Would you be able to provide some additional working prior to these final solutions to aid understanding? Thank you. – Joshua Aug 20 '15 at 17:43