title: (log n)^k = O(n)? For k greater or equal to 1
body: <p><code>(log n)^k = O(n)? For k greater or equal to 1.</code></p>
  
  <p>My professor presented us with this statement in class, however I am not sure what it means for a function to a have a time complexity of O(n). Even stuff like <code>n^2 = O(n^2)</code>, how can a function f(x) have a run time complexity? </p>
  
  <p>As for the statement how does it equal O(n) rather than O((logn)^k)?</p>