Proving the Existence of a Large n for (2^n) > K

[iwonde]

Homework Statement

Show that for any natural number K, there is an n large enough so that (2^n) > K.

Homework Equations

The Attempt at a Solution

K is a natural number -> the smallest possible K would be 0 (lower bound?) and the smallest 2^n is 1 when n = 0, and the upper bound for both sides are infinite. So if I set n=0 and K =0 I get 2^n > K. I'm not sure if this is the right approach.

 

[Beam me down]

Take some number K, find the possible n's that satisfy 2^n>K. Ie. solve for n.