Proving n^2 = O(log n)

Question

This is how I would think it n^2 = O(log n)

f(n) <= c*g(n)

should c=1 ?

n0 = 1

n^2 <= log n

0 <= log n - n^2, for all n>=1

we take n=1 then 0 <= log(1) - 1^2

0 <= 0 - 1

0 <= -1

Therefore n^2 = O(logn) false?

Is this correct?