How to solve $5000 n \log(n) \leq 2^{n/2}$ I'm trying to solve the following problem:
What is the smallest value of n so that an algorithm with a runtime of $5000 n \log(n)$ runs faster than an algorithm with a runtime of $2^{2/n}$ on the same computer?
So I figured it was just a matter of solving the equation $5000 n \log(n) \leq 2^{n/2}$ but I have no idea how I should do it, I've tried for a while using algebra but that's leading me nowhere.

some modification you have: $ ln(5000)=8.52<n/2*(0.7)-ln(ln(n)*n)$
Then:
$ 8.52<0.35*n+ ln(n*ln(n))$
For n= 30 we have 8.52>7.some
For n = 40 we have 8.52<9.some 
We conclude that number is bwthween 30 and 40.
Metohd od half use 35: we have:  8.52>7.5
=> number is in [35, 40]
Use n=38 => 8.52>8.37 n is in [39, 40]
Us n=39 ->8.52<8.68 => n=39 is final soultuon(min n with that property).
Maybe it would be some mistakes, but this method you may use.
