Use the result of exercise 47 above to prove the following: For every integer n ≥ 1 , if x is any real number with x > ( 2 n ) 2 n , then x n < 2 x .

To prove:

For every integer n≥1, if x is any real number with x>(2n)2n,xn<2x

Given information:

For every integer n≥1, if x is any real number with x>(2n)2n,log2x<x1n

Definition used:

f is of order g:

f(x) is Θ(g(x)) if there exists a positive real numbers, B and a nonnegative real number k such that |f(x)|≤B|g(x)| whenever x>k

Proof:

Let n be a positive integer and let x be a positive real number such that x>(2n)2n.

If x>(2n)2n, then xn>[( 2n)2n]n=(2n)2n2>(2n)2n.

Let us then use the property log2u<u1n for the real number u=xn