O(n^2) vs O (n(logn)^2)

Question

Is time complexity O(n^2) or O (n(logn)^2) better?

I know that when we simplify it, it becomes

O(n) vs O((logn)^2)

Answer 1

(Log n)^2 is better because if you do a variable change n by exp m, then m^2 is better than exp m