Why to consider binary search running time complexity as log2N

Question

Can someone explain me when it comes to binary search we say the running time complexity is O(log n)? I searched it in Google and got the below,

"The number of times that you can halve the search space is the same as log₂ n".

I know we do halve until we find the search key in the data structure, but why we have to consider it as log₂ n? I understand that e^x is exponential growth and so the log₂ n is the binary decay. But I am unable to interpret the binary search in terms of my logarithm definition understanding.

Answer 1

Think of it like this:

If you can afford to half something m times, (i.e., you can afford to spend time proportional to m), then how large array can you afford to search?

Obviously arrays of size 2^m, right?

So if you can search an array of size n = 2^m, then the time it takes is proportional to m, and solving m for n look like this:

n = 2^m

log₂(n) = log₂(2^m)

log₂(n) = m


Put another way: Performing a binary search on an array of size n = 2^m takes time proportional to m, or equivalently, proportional to log₂(n).