Answer
7 comparisons
Work Step by Step
Let $f (n)$ be the quantity of examinations required in a binary search of a rundown of $n $components.
We know that recurrence relation for binary search with $n$ elements is: $f(n) = f(n/2) +2$.
In this manner $f(128) = f(64) + 2$
$=f(64) = f(32 +2) + 2$
$= (f(16) + 4) +2 $
$= (f(8) + 6) +2$
$=( f(4) + 8) +2 $
$ = (f(2) + 10) +2$
$= (f(1) + 12) +2 $
$f(1)=2$ since two comparisons are required for one number
so $= 2 + 14 $
$= 16$
So 7 comparisons are needed for a binary search in a set of 128 elements
