Big Oh Notation O ((log n) ^ k) = O (log n)?

Question

In the notation with a large O, there is O((log n)^k) = O(log n), where kis some constant (for example, the number of logarithmic for cycles), true?

My professor told me that this statement is true, but he said that it will be proved later during the course. I was wondering if any of you would be able to demonstrate your credibility or have a link where I could confirm whether this is true.

Answer 1

(1) It is true that O (log (n ^ k)) = O (log n).

(2) It is not true that O (log ^ k (n)) (also written O ((log n) ^ k)) = O (log n).

Observation: (1) has been proven by nmjohn.

: (2). (: f (n) = log ^ 2 n O (log ^ 2 n). O (log n)? c , n, n0, c log n > log ^ 2 n?)

EDIT:

, / , "Computer Science" StackExchange. . , !

http://area51.stackexchange.com/proposals/35636/computer-science-non-programming?referrer=rpnXA1_2BNYzXN85c5ibxQ2

Answer 2

, O (log n ^ k), O (k * log n) = k * O (log n) = O (log n).

Answer 3

O (log n) is a class of functions. You cannot perform calculations such as ^ k on it. Thus, the term O (log n) ^ k does not even seem reasonable to me.