# Number of Binary Digits in Power of 10

## Theorem

Let $n$ be a positive integer.

Expressed in binary notation, the number of digits in the $n$th power of $10$:

$10^n$

is equal to:

$\ceiling {n \log_2 10}$

where $\ceiling x$ denotes the ceiling of $x$.

## Examples

### $1000$

When expressed in binary notation, the number of digits in $1000$ is $10$.

## Proof

Let $10^n$ have $m$ digits when expressed in binary notation.

By the Basis Representation Theorem and its implications, a positive integer $x$ has $m$ digits if and only if:

$2^{m - 1} \le x < 2^m$

Thus:

 $\displaystyle 2^{m - 1}$ $\le$ $\, \displaystyle 10^n \,$ $\, \displaystyle <\,$ $\displaystyle 2^m$ $\displaystyle \leadsto \ \$ $\displaystyle m - 1$ $\le$ $\, \displaystyle \map {\log_2} {10^n} \,$ $\, \displaystyle <\,$ $\displaystyle m$ $\displaystyle \leadsto \ \$ $\displaystyle m - 1$ $\le$ $\, \displaystyle n \log_2 10 \,$ $\, \displaystyle <\,$ $\displaystyle m$

Because a power of $10$ cannot equal a power of $2$, it will always be the case that:

$m - 1 < n \log_2 10 < m$

and so:

$m - 1 < n \log_2 10 \le m$
$m = \ceiling {n \log_2 10}$

$\blacksquare$