title: Big Oh with log (n) and exponents
body: <p>So I have a few given functions and need to fond Big Oh for them (which I did). </p>
  
  <ol>
  <li>n log(n) =  O(n log(n)) </li>
  <li>n^2 = O(n^2) </li>
  <li>n log(n^2) = O(n log(n)) </li>
  <li>n log(n)^2 = O(n^3) </li>
  <li>n = O(n)</li>
  </ol>
  
  <p>log is the natural logarithm.</p>
  
  <p>I am pretty sure that 1,2,5 are correct.
  For 3 I found a solution somewhere here: n log(n^2) = 2 n log (n) => O (n log n)
  But I am completely unsure about 4). n^3 is definitely bigger than n*log(n^2) but is it the Oh of it? My other guess would be O(n^2).</p>
  
  <p>A few other things:</p>
  
  <ul>
  <li>n^2 * log(n)</li>
  <li>n^2 * log(n)^2
  What would that be?</li>
  </ul>
  
  <p>Would be great if someone could <strong>explain</strong> it if it is wrong. Thank you!</p>