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?

asymptotic-complexity

Solution

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)))

Problem

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. I am taking a class on algorithms and this is one of the questions I was asked. According to me this should be dependent on constant c but the answer was wrong.

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