O (log (n)) * O (log (n)) in complexity theory

Question

Tackled with some kind of silly question on complexity.

I have a loop that runs time O(lg(n)). I have another loop inside that is also equal O(lg(n)), so all complexity O(lg(n)) * O(lg(n))or O (lg (n) ² ) . May I say that the final O is equal O(lg(n)), since since n is a power of 2, then

O (lg (n)) * O (lg (n)) = O (lg (n ² )) = O (2lg (n)) = O (lg (n)))

or cannot it be used in this way?

Answer 1

, ! :

O (lg (n)) * O (lg (n)) = O (lg (n ²))

. . O (lg (n) ²)