determining the number of bits required to represent a number in binary In the example in the following slide, we follow the highlighted formula. With regard to the highlight, I'm confused why the number is greater or equal to $2^{n-1}$, while only need to be less than $2^n$ (not less than or equal to $2^n$)?

 A: Think of a number with $n$ bits. Each bit can be 0 or 1, so you have $2^n$ combinations. However one of the combinations is the number 0 (i.e. all $n$ bits are 0). So you can only count up to $2^n-1$ with $n$ bits and not all the way up to $2^n$. That's why you see $<2^n$ in your example and not $\leq 2^n$.
A: What the example is illustrating
is a general rule:
if you have a positive whole number $x$
that you want to write in binary,
and if
$$ 2^{n-1} \leq x < 2^n $$
where $n$ is a whole number,
you need exactly $n$ bits to write $x.$
I hope you will agree that since the number $48$
is not a power of two, it really does not matter
(for a number like that) whether we write
$\leq$ or $<.$
But what if we want to write $64$ in binary? If the rule were
$$ 2^{n-1} \leq x \leq 2^n $$
then $64$ (which is equal to $2^6$)
would fit the rule in two ways:
it would fit with $n=6,$
because $2^(6-1)\leq 64\leq 2^6,$
and it would fit with $n=7,$
because $2^{7-1}\leq 64\leq 2^7.$
But it cannot be true both that it takes exactly $6$ binary bits to write $64$
and that it takes exactly $7.$
So this is not a good rule.
In fact you need $7$ bits to write $64.$
Notice that this is the answer you get if you write $\leq$ for $2^{n-1}$ but
$<$ for $2^n.$
Another way to describe the rule with fewer symbols and more words is that you need exactly $n$ binary bits to write $x$
if $x$ is less than $2^n$ but not less than $2^{n-1}.$
