The recurrence for binary search is $T(n)=T(n/2) + O(1)$. The general form for the Master Theorem is $T(n)=aT(n/b) + f(n)$. We take $a=1$, $b=2$ and $f(n)=c$, where $c$ is a constant. The key quantity is $\log_b a$, which in this case is $\log_2 1=0$.
If you look at the Wikipedia entry (through the link you posted) you will see that there are 3 main cases for the Master Theorem. Here we are in Case 2 since by taking $k=0$ we find that $n^{\log_b a}(\log n)^k=(n^0)(\log n)^0=1$; therefore $f(n)=c=\Theta(n^{\log_b a}\log^k n)$.
From Case 2 of the Master Theorem we know that $T(n)=\Theta(n^{\log_b a}(\log n)^{k+1}$ which in this case yields $T(n)=\Theta(n^0(\log n)^1)=\Theta(\log n)$.