I don't know if he edited his comment, but as it is currently written the log is not in the exponent under standard order of operations.

Also, putting a constant factor like log(2) inside big-O notation is meaningless.

Constant factors in exponents are _not_ meaningless. One of those cases where the rules-of-thumb bite you.

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.

We have better than that. I think that what was meant is O(n^2*log(n))

That isn't just what's meant, that's also what's written.

The way that it was written didn't disambiguate whether or not the log is part of the exponent. Putting the * in there disambiguated that.

It doesn't. The multiplication is implied and it would have the same precedence as explicitly putting * there. So either both are ambiguous or neither is (and it isn't, the poster was just mistaken).

I'm not sure why you think that it is implied.

I know that I have a lot of math background, and my brain read that log(n) as part of the exponent. I figure that if I read it that way, you have to expect that others will as well.

No matter how you read a^b log(c), the multiplication is implied. The original confusion is whether it's to be read as (a^b)log(c) or a^(b log(c)). a^b*log(c) simply removes the implied multiplication, but it does not change whether that should be read as (a^b)*log(c) or a^(b*log(c)) and thus does not resolve the original confusion.