I was asked to find the tightest upper bound for (logn)^2
What I think should be the right way of doing this is as follows:
Since we know log X < X, we can let X be n^(1/4)
log n^(1/4) < n^(1/4)
log n < 4 * n^(1/4)
(logn)^2 < 4 * n^(1/2)
Thus, (logn)^2 = O(n^(1/2))
Plotting the graph out for it seems to suggest it as well. On a side note, I tried substituting n^(1/8) and realised perhaps an even tighter bound would be O(n^(1/8))