How come the time complexity of Binary Search is log n I am watching this professor's video on Binary Search but when he reached here, I am a bit lost. How come he came up the time coomplexity is log in just by breaking off binary tree and knowing height is log n
https://youtu.be/C2apEw9pgtw?t=969 . and then the time complexity become log 16/2 = 4... how that is $\log n$ time complexity?


First, it is important to note that the running time of an algorithm is usually represented as function of the input size. Then, we 'measure' the complexity by fitting this function into a class of functions. For instance, if $T(n)$ is the function describing your algorithm's running time and $g\colon\mathbb{N}\to\mathbb{R}$ is another function then
 $$T\in O(g) \iff \text { there exist } c,n_0\in\mathbb{R}_{++} \text{ such that } T(n)\leq c g(n) \text{ for each } n\geq n_0. $$
Similarly, we say that
$$T\in \Omega(g) \iff \text { there exist } c,n_0\in\mathbb{R}_{++} \text{ such that } T(n)\geq c g(n) \text{ for each } n\geq n_0. $$
If $T$ belongs to both $O(g)$ and $\Omega(g)$ then we say that $T\in\Theta(g)$. Let's conclude that for the binary search algorithm we have a running time of $\Theta(\log(n))$. Note that we always solve a subproblem in constant time  and then we are given a subproblem of size $\frac{n}{2}$. Thus, the running time of binary search is described by the recursive function $$T(n)=T\Big(\frac{n}{2}\Big)+\alpha.$$
Solving the equation above gives us that $T(n)=\alpha\log_2(n)$. Choosing constants $c=\alpha$ and $n_0=1$, you can easily conclude that the running time of binary search is $\Theta(\log(n))$.
