If

`f(n)=O(g(n))`

, then shouldn't `f(n)∗log2(f(n)^c)=O(g(n)∗log2(g(n)))`

depend on the value of C? Here C is a positive constant. According to me if C is large then the statement would become false and if c is small it'd be true. Hence the outcome is dependent on c.

`log(x^c) = c * log(x) `

So,

`log2(f(n)^c) == c * log2(f(n)) `

Therefore,

`f(n)∗log2(f(n)^c) = c * f(n) * log2(f(n)) = O(g(n)∗log2(g(n)))`

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Posted By Blogger to HDGEM at 3/06/2017 09:11:00 PM