Trying to solve the recurrence relation by comparing 3 cases of master theorem I am trying to understand how the master theorem is invoked on the following recurrence relation:

$$ T(n) = \sqrt{6006} T(n/2) + n^{\sqrt{6006}}. $$

So basically, I found the following source where they have Master theorem definition defined.

I found the similar solution (See the Link #1 in the comment below) where they are trying to solve the same problem (Problem #c). I 

In the solution, they have mentioned the following :

We have : 

n0.5logb(a)  =  nlogb(6006)1/2

=  n0.5log2(6006)1/2   

and so on and so forth. I am not able to understand how they compared the results with the given equation.
You determine $a$, $b$, and $f$ using pattern matching.

The Master theorem applies for recurrences of the form

$\qquad\displaystyle T(n) = a T(n/b) + f(n)$;

you have

$\qquad\displaystyle T(n) = \sqrt{6006} \cdot T(n/2) + n^{\sqrt{6006}}$.

Just by comparing the two formulae, you determine that

$\qquad\begin{align*}
  a &amp;= \sqrt{6006},\\
  b &amp;= 2, \quad\text{and} \\
  f &amp;= n \mapsto n^{\sqrt{6006}}.
\end{align*}$

Now you go through the three cases and check which applies (if any). See also here.