1. Log problem.

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

2. Re: Log problem.

You can try solving this numerically, I get $n\approx 0.03187$

3. Re: Log problem.

is numerically the only way?
ie not do-able with algebra?

many thanks.

4. Re: Log problem.

Originally Posted by BabyMilo
is numerically the only way?
ie not do-able with algebra?

many thanks.
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

$1000n^2 < 2^n$

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

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

5. Re: Log problem.

ok thanks.
I was just wasnt sure whether i could do it algebraically.