O(n^2 * 2^(log n)) == O(n^2) ?
Why I think that this could be the case:
In big O, you only take the parts of the term that are most relevant, right?
O(n^2 + 3n + 9) == O(n^2). The
n^2 has a lot more influence to the result of the term than the
2^(log n)-part. http://www.wolframalpha.com/input/?i=Plot%5B%7Bn%5E2,+2%5ELog%5Bn%5D%7D,+%7Bn,+0,+50%7D%5D