Solve $100n^2 = 2^n$

Question

I don't understand how to solve this... I try to get rid of the $2^n$ via logarithms and it leads me to $2log_210 + 2log_2n = n$ and I don't know how to proceed.

Thanks for the help.

Answer 1

Lambert W method: $y = xe^x$ if and only if $W(y) = x$.

For this problem: $$ 100 n^2 = 2^n $$ take square-root, $$ 10 n = 2^{n/2}\qquad\text{or}\qquad 10 n = -2^{n/2} $$ Lets do the first one $$ 10 n = 2^{n/2} \\ 10 n = \exp\left(\frac{n (\log 2)}{2}\right) \\ \frac{1}{10 n} = \exp\left(\frac{-n (\log 2)}{2}\right) \\ \frac{1}{10} = n\;\exp\left(\frac{-n (\log 2)}{2}\right) \\ -\frac{\log 2}{20} = \frac{-n \log 2}{2}\exp\left(\frac{-n (\log 2)}{2}\right) $$ All of that was to get it in the form $y = x e^x$. Then go to $W(y) = x$: $$ W\left(-\frac{\log 2}{20}\right) = \frac{-n (\log 2)}{2} \\ -\frac{2}{\log 2}W\left(-\frac{\log 2}{20}\right) = n $$

Note. Students of high-school algebra are not expected to know the Lambert W funcion. And they are not expected to be able to solve this problem explicitly.