First, you can analyze the time complexity of binary search in whatever case you wish, say "best case" and "worst case". In the best case, you use $f(n)$ time, while in the worst case you use $g(n)$ time. These are two functions of $n$, describing two different scenarios you have defined. Now, you can make statements such as $f(n) = O(1)$ or $g(n) = \Theta(\log n)$. However, the Big Oh of a function has nothing to do with time complexity, it is merely something applied to a function that could just as well be representing the number of apples you get for $n$ euros.
In particular, thinking along the lines of "$O$ is always for best case, $\Omega$ is for the worst case" (or whatever), is wrong and dangerous.
To you specific question: if a function $g(n) = \Omega(\log n)$ and $g(n) = O(\log n)$, then $g(n) = \Theta(\log n)$. The function $g(n)$ might be representing the time complexity of binary search or the number of apples, it doesn't matter. And as always, look at the definitions and you can stop guessing.