Given information:
4n
Proof:
Let b and B be positive real numbers. f(x) is not O(g(x)) if and only if for all positive real numbers b and B, there exists some number x that is greater than b and that has the property |f(x)|>B|g(x)|.
Let n= max (b,B,log2B).
We then note that n > b and n > log2B.
|4n|=4n                                    4n is always positive=( 2 2)n                                                4=22=22n   Â