Running time correct with Omega? I have the following statement. I would say it's correct as it's either equal or higher than $\Omega(\log^{10}(n))$.

Because:
I know $\log(2^n) = n$.
By that I would guess the same goes for $\log(n^{10})$ unimportant if the number is higher or not. By that it should be n. With this knowledge we already know it's higher than our $\Omega(\log^{10}(n))$.

We also know $\log(n)^2 = \log(n)$ again it will be higher or equal to our omega.

Is this correct?

$$(3 \log^2 n + 55 \log(n^{10}) + 8 \log n) \cdot \log n = \Omega(\log^{10} n)$$
Don't guess, but look at the definition which tells you that in order for $f(n) = \Omega(g(n))$ to hold, it must be the case that $\lim_{n \to \infty} f(n)/g(n) &gt; 0$.

If I plug in $f(n)$ and $g(n)$ according to your description, I think I will get $\lim_{n \to \infty} f(n)/g(n) = 0$. You should check whether I made a mistake and draw the appropriate conclusions.