Problem: Suppose $V$ is an AVL tree (a self-balancing binary search tree) of $n$ elements. After the insertion of $n^2$ elements, what would be its height?
My idea: the height of an AVL tree is originally $O(\log(n))$ where $n$ is the number of elements. After insertion of $n^2$ elements, its height will be:$$O(\log(n+n^2))=O(\log(n^2))=O(2\log(n))=O(\log(n))$$
My answer would be $O(\log(n))$ but I'm having doubts.
Why is the asymptotic complexity of the result the same despite the fact that there are more elements?