Gear2d
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I was wondering how would I go about proving the Log of a Power Rule:
log(n^k) is O(log(n)) for any constant k > 0
log(n^k) is O(log(n)) for any constant k > 0
The discussion centers around proving the Log of a Power Rule, specifically that log(n^k) is O(log(n)) for any constant k > 0. The scope includes theoretical aspects of logarithmic properties and their implications in algorithm analysis.
Participants appear to have differing interpretations of the initial question, with some focusing on the big O relationship and others on the power rule itself. No consensus is reached on the specific approach to take for the proof.
There is an indication that assumptions about the logarithmic base and the nature of the proof may not be fully articulated, which could affect the clarity of the discussion.
This discussion may be useful for students or individuals interested in algorithm analysis, logarithmic properties, and mathematical proofs related to big O notation.
mistermath said:You should show what you've tried, but basically..
log (n^k) = k log (n), and use w/e technique you want to show the constant up front does not matter.
OR are you asking how to prove the power rule for logs? I'll give a hint..
log (x) = y <=> 10^y = x
(assuming log has base 10 and <=> means if and only if).