# Time complexity – Big-O comparison of two algorithms \$ sqrt {n} \$ and \$ 2 ^ { sqrt { log _ {2} n}} \$

I have given 2 algorithms with their temporal complexity $$t_a (n) = sqrt {n}$$ and $$t_b (n) = 2 ^ { sqrt { log _ {2} n}}$$ and I have to show $$t_b (n) = O (t_a (n))$$,

I have created a program to check this statement and it seems that this is for everyone $$c> 0, forall n geq16$$ it applies, but I don't know how to prove it formally because I can't find a simplification for it $$t_b$$,

I know I have to prove $$exists c : forall n geq N: t_b (n) leq c * t_a (n)$$ with big-o notation.

A hint / solution would be really great.