# time complexity – I want to show, without taking a limit, that \$2^sqrt{2logn} in Ω(log^2n)\$ and \$2^sqrt{2logn} in O(sqrt{2}^{logn})\$

To prove $$2^{sqrt{2log n}} in O(sqrt{2}^{log n})$$, you should note that $$sqrt{2}^{log n} = left(sqrt{2}^{sqrt{log n}}right)^{sqrt{log n}}$$.

To prove $$2^{sqrt{2log n}} in Omega(log^2n)$$, you should note that $$log ^2n = 2^{2log log n}$$.

It will simplify the comparisons you have to do.