# Is log(n) equivalent to (log(n))^x for big-O analysis?

My professor noted that we could treat any logarithmic function with an exponent as equivalent to log(n) for the purposes of big-O analysis.

ie. $$(n log(n) + 1)^2 + (log(n) + 1)(n^2 + 1)$$

From the left I would get $$(n^2)(log(n))^2$$ and from the right I would get $$(n^2)log(n)$$. According to my professor I can just say this function is $$O(n^2log(n))$$. However, I don't see how these functions grow in the same order. Looking at their graphs it would seem they are quite different.

$$f(n)$$ is not $$O(n^2\log n)$$
If we denote $$f(n) = (n\log n+1)^2+(log n+1)(n^2+1)$$ then: $$f(n) = n^2 \log^2n + 2n\log n + 1 + n^2\log n + n^2 + \log n + 1 \geq n^2log^2n$$
since all the other terms are obviously greater than 0. Therefore by definition $$f(n) = \Omega (n^2log^2n)$$. Which strictly means it cannot be bounded from above by anything asymptotically greater.
Note that it also holds that $$f(n) \leq 10n^2log^2n$$, so $$f(n) = O(n^2log^2n)$$ and also $$\Theta$$ of the same.