Prove that for any constant c > 0, (log n)^c = o(n).

Question

Answer 1

Let us assume n = 2^n here and put that in equations (log n)^c = o(n)

we will get (log (2^n))^c = o(2^n)

lets assume that base of the log is 2 hence we will get

n^c = o (2^n) {I don't know if that in the given question it is big Omega or small omega, but I assume it as small omega)

so the equation for asymptotic equation would be.
n^c < 2^n

now now putting both side under logs we will get.
c * log(n) <  n (log 2)

after neglecting constants we will get our output as log(n) < n

which is always True for any constants we take

Answer 2

The definition for Big O  notation is -For Constants N and c ,if for some n>N ,f(n)<=c*g(n) then we can say that g(n) bounds f(n)

OR

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

Here f(n) is (log n)^c and g(n) is n.

so for any constant c>0,it can be easily proved that (log n)^c<=n.

Hence (log n)^c=O(n)