What is the correct representation of Master Theorem? What I'm taught in my class -


  $T(n)=aT(\frac{n}{b})+\theta(n^k\log^pn)$
  where $a\geq1$, $b&gt;1$, $k\geq1$ and $p$ is a real number.
  
  
  if $a&gt;b^k$ then, $T(n)=\theta(n^{\log_ab})$
  if $a=b^k$ then,
  
  if $p&gt;-1$, then $T(n)=\theta(n^{\log_ab}\log^{p+1}n)$
  if $p = -1$, then $T(n)=\theta(n^{\log_ab}\log\log n)$
  if $p&lt;-1$, then $T(n)=\theta(n^{\log_ab})$
  
  if $a&lt;b^k$, then
  
  if $p\geq 0$, then $T(n)=\theta(n^k\log^pn)$
  if $p&lt;0$, then $T(n)=O(n^k)$
  
  


On certain websites, the above representation for Master Theorem is slightly different as follows - 


  $T(n)=aT(\frac{n}{b})+\theta(n^k(\log n)^p)$


This ambiguity creates a great confusion while solving problems such as : 


  What is the value of the recurrence : $T(n)=T(\sqrt{n})+\theta(\log\log n)$


Substituting $n=2^m$, we get a new expression :


  $S(m)=S(\frac{m}{2})+\theta(\log m)$
  Here  $a=1$, $b=2$, $k=0$ and $p=1$


If I apply Master Theorem as per the way I am taught in class, the result I get is : 


  $T(n)=\theta(\log\log\log n)$


and if I solve using the other formula, I get the result as : 


  $T(n)=\theta((\log\log n)^2)$


Which is the correct one? This confusion is making me question every problem that I've solved.
By convention, the notation $log^p(x)$ is defined to be $(log(x))^p$, not $p$ iterations of the $log$ function. This is similar to the trigonometric functions, which gives us identities like $sin^2(x)+cos^2(x)=1$.