title: Big O if 2^n vs. 4^n
body: <p>I'm trying to figure out these two Big O's. Obviously the big O of 2^n is O(2^n), but I'm not sure if you can reduce 4^n. If so, I would do 4^n = (2^2)^n. Then we could distribute to make this 2^(2n), which I would reduce to O(2^n) since the constant in front of the n wouldn't matter. </p>
  
  <p>Is this correct? Thank you.</p>