Values of k that make this true, $\log^k n \in \Omega(\sqrt n)$

I have seen some algorithms with complexities like $ \log^3 n$ and $ \sqrt n$ . In view of getting a better idea on how to compare these I wanted to know for which values of $ k$ does $ \log^k n \in \Omega(\sqrt n)$ hold true?

I suspect $ k$ would need to be a function of $ n$ . Because if it were a constant $ C_1$ , we could always find a larger constant $ C_2$ that makes the previous statement false.

This is what I have tried.

$ \log^k n \in \Omega(\sqrt n)$

$ \log^{2k} n \in \Omega(n)$

$ \log(\log^{2k}n) \in \Omega(\log n)$

$ 2k \log\log n \in \Omega(\log n)$

From here, I can see that if $ k$ is $ \log n$ , then $ 2\log n \log\log n \in \Omega(\log n)$ is true.

However, I doubt this is a tight bound.