title: Comparing growth rate of exponential function?
body: <p>Suppose we have two functions f(n) = 2<sup>2<sup>n+1</sup></sup> and g(n)=2<sup>2<sup>n</sup></sup>. I want to compare their growth rates of these two functions using two different methods.</p>
  <h2>Method One: Take the Ratio</h2>
  <p>Let's define t(n) = f(n) / g(n). Then</p>
  <blockquote>
  <p>t(n) = f(n) / g(n)</p>
  <p>= 2<sup>2<sup>n+1</sup></sup> / 2<sup>2<sup>n</sup></sup></p>
  <p>= 2<sup>2<sup>n + 1</sup> - 2<sup>n</sup></sup></p>
  <p>= 2<sup>2<sup>n</sup></sup></p>
  </blockquote>
  <p>So we'll assume that f(n) grows much faster than g(n).</p>
  <h2>Method Two: Use Logarithms</h2>
  <p>As before, let t(n) = f(n) / g(n). Now, let's take log base two of both sides:</p>
  <blockquote>
  <p>lg t(n) = lg (f(n) / g(n))</p>
  <p>= lg (2<sup>2<sup>n+1</sup></sup> / 2<sup>2<sup>n</sup></sup>)</p>
  <p>= lg 2<sup>2<sup>n+1</sup></sup> - lg 2<sup>2<sup>n</sup></sup>)</p>
  <p>= 2<sup>n+1</sup> - 2<sup>n</sup></p>
  </blockquote>
  <p>Now, let's take the log base two of both sides:</p>
  <blockquote>
  <p>lg lg t(n) = (n + 1) lg 2 / n lg 2</p>
  <p>= (n + 1) / n</p>
  </blockquote>
  <p>Ignoring constant term, we get lg lg t(n) = 1, which is a constant, so f(n) and g(n) should have the same growth rate.</p>
  <p>Why am I getting the wrong answer using Method Two ?</p>