I didn't claim that. The log(2) term is multiplied, and that's implied by my use of "factor", which in this context is always understood to mean multiplicative factor.
You're correct. I think this comes down to confusion on the operator precedence. I read it as O(n^(3 log 2)) because otherwise it's kind of very obviously not correct (no way it's n^3, that's already known), but should have stated that, because according to usual notation/precedence it's not that.
