How to solve $\displaystyle \frac{n^2}{2^n} < \frac{1}{1000}$ ??

Results 1 to 5 of 5

- Nov 28th 2011, 11:34 AM #1

- Joined
- Feb 2008
- Posts
- 383

- Nov 28th 2011, 11:50 AM #2

- Nov 28th 2011, 11:55 AM #3

- Joined
- Feb 2008
- Posts
- 383

- Nov 28th 2011, 12:10 PM #4
## Re: Log problem.

To do so you will need the Lambert W function Lambert W function - Wikipedia, the free encyclopedia

Numerically will be your best bet. Here is a start, by algebra we have that

$\displaystyle 1000n^2 < 2^n$

$\displaystyle \begin{array}{c|c|c} n & 1000n^2 & 2^n \\ \hline 10 & 100000 & 1024 \\ 15 & 225000 & 32768 \\ 20 & 400000 & 1078576 \\ \end{array}$

So we can see the answer is between 15 and 20, just keep going

- Nov 28th 2011, 12:13 PM #5

- Joined
- Feb 2008
- Posts
- 383