Height of AVL after entries 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?
To show that this isn't always the case, imagine you added $e^n$ elements, then it would take

$$O(\log(n+e^n))=O(\log(e^n))=O(n)$$

But what does that really mean? It's not that your binary tree takes $O(n)$ time on $n$ elements to find something. Rather, it takes $O(\log(m))$ time for $m=n+e^n$ elements.

I can understand why you would be confused; if you had to search linearly through a list, and then you added $n^2$ elements, it would take $O(n^2$) time in terms of the original input, but not the new input, for which it would take $O(m)$ time on $m=n+n^2$ elements.

Chances are you didn't want to solve for the asymptotic solution; it generally doesn't make sense to. What you probably want is the approximation $\log(n+n^2)\approx\log(n^2)=2\log(n)$; that is, 


  For a self-balancing tree, squaring the input size doubles the search time.