title: When f(n) = n^.1 and g(n) = log(n)^10, does f(n) = Ω(g)?
body: <p>I was told that "any exponential trumps any logarithm".</p>
  
  <p>But when the exponential is between zero and one, doesn't the execution time of the logarithm grow much faster? So by that logic it would be f = O(g)</p>
  
  <p>I'm having trouble choosing whether to follow my intuition or what I've been told, but what I've been told may have been not totally accurate.</p>