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Is O(loglog n) equal to O(n^2) in polynomial time?

I know O(log n) is eqaul to O(n) in polynomial time because this can change as follows: log n > 2^(log n) > n^(log 2) > n.

So, my question is to determine whether O(loglog n) = O(n^2) or not.

Thank you for your time.

Kelin Donut
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  • $O(\log n)$ cannot equal to $O(n)$, because $n$ itself is in $O(n)$, but not in $O(\log n)$. – zkutch Apr 10 '21 at 09:44

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