Comparing two Big O notations

Question

Please, could you help me with the question below. Demonstrate that O(log n^k) = O(log n):

Answer 1

Using properties of logs, log(n^k) can easily be transformed into klog(n). Then, because constant factors 'small out' as it were when using big O notation, O(klog(n))= O(log(n)).

Thus, O(log(n^k)) = O(klog(n)) = O(log(n)).

Answer 2

Hint. One may observe that $$ \log \left(n^k \right)=k\log \left(n \right). $$