how are n^log5 and 5^logn of the same order?

Knowing that n² < 5ⁿ i would assume that n^log5 < 5^logn

CodePudding user response：

This is because if you follow logarithmic laws you see that they are the same, following this:

CodePudding user response：

a^b = e^(b*log(a)) = (e^b)^(log(a)) = (exp(b))^(log(a)).

The first equality is a definition, the second one is a well-known rule on powers, the third one is just a notation.

With a=5 and b=log(n) this gives

5^(log(n)) = (exp(log(n))^(log(5)) = n^(log(5))

because exp(log(n))=n.