Height of AVL after entries
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.