Exercise 4.5.14* (The probability that a power of $2$ has leading digit $1$ is $\log 2 / \log 10$ )

Proof.

(a)

Consider the two subsets of ${\mathbb{N}}^{\text{∗}}=\{1,2,3,\dots \}$ given by $$\begin{array}{llll}\hfill A& =\left\{n\in {\mathbb{N}}^{\text{∗}}\mid \exists <!--\; FUNCTION\; APPLICATION\; -->k\in {\mathbb{N}}^{\text{∗}},n=\lfloor k\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2}\rfloor \right\},& \hfill & \\ \hfill & =\{3,6,9,13,16,19,23,26,29,33,36,39,43,46,49,53,56,59,63,\dots \}& \hfill & \\ \hfill B& =\left\{n\in {\mathbb{N}}^{\text{∗}}\mid \exists <!--\; FUNCTION\; APPLICATION\; -->k\in {\mathbb{N}}^{\text{∗}},n=\lfloor k\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->5}\rfloor \right\}& \hfill & \\ \hfill & =\{1,2,4,5,7,8,10,11,12,14,15,17,18,20,21,22,24,25,27,\dots \}.& \hfill & \end{array}$$

In others terms,

$$\begin{array}{llll}\hfill A& =\left\{\lfloor \mathit{k\alpha }\rfloor \mid j\in {\mathbb{N}}^{\text{∗}}\right\},& \hfill & \\ \hfill B& =\left\{\lfloor \mathit{k\beta }\rfloor \mid k\in {\mathbb{N}}^{\text{∗}}\right\},& \hfill & \end{array}$$

where

$$\alpha =\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2},\beta =\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->5}.$$

Then $\alpha ,\beta$ are positive irrational numbers: if $\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2}=\frac{p}{q}$, where $p,q$ are positive integers, then $2^p=10^q=2^q{5}^q$. The unique factorization theorem shows that $q=0$. This is a contradiction, so $\alpha =\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2}$ is irrational, and similarly $\beta$ is irrational.

Since

$$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}+\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->5}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}=1,$$

Problem 4.1.23 shows that $A$ and $B$ are complementary subsets of ${\mathbb{N}}^{\text{∗}}$, that is

$$\begin{array}{lll}\hfill A\cup B={\mathbb{N}}^{\text{∗}},A\cap B=\varnothing .& & \hfill \text{(1)}\end{array}$$

Moreover, by Problem 4.5.13 (a), for every positive integer $n$, $2^{n+1}$ has leading digit $1$ if and only if there is some $k\ge 2$ such that $n+1=\lfloor (k-1)\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2}\rfloor +1$, that is if and only if $n\in A$.

By part (b) of the same problem, $5^n$ has not leading digit $1$ if and only if there is some positive integer $k\in {\mathbb{N}}^{\text{∗}}$ such that $n=\lfloor k\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->5}\rfloor$, that is if and only if $n\in B$. Therefore $5^n$ has leading digit $1$ if and only if $n\in {\text{∁}}_{{\mathbb{N}}^{\text{∗}}}B=A$.

This shows that for every $n\in {\mathbb{N}}^{\text{∗}}$, $5^n$ has leading digit $1$ if and only if $2^{n+1}$ has leading digit $1$.

(b)

If $n\in [[1,N]]$, then $2^n$ has leading coefficient $1$ if and only if $n-1\in A$, that if if and only if $n=1+\lfloor k\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2}\rfloor$ for some positive integer $k$. Then $1\le \lfloor k\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2}\rfloor \le N-1$. Moreover, for every positive integer $k$, $$\begin{array}{llll}\hfill 1\le \lfloor k\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2}\rfloor \le N-1& ⟺1\le k\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2}<N& \hfill & \\ \hfill & ⟺1\le k<N\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}& \hfill & \\ \hfill & ⟺1\le k\le \lfloor N\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}\rfloor .& \hfill & \end{array}$$

( $\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}$ is an irrational number, thus $N\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}$ is never an integer.)

If $K=\lfloor N\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}\rfloor$, the favorable case among $2,2^2,\dots ,2^N$ are

$$2^{1+\lfloor \frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2}\rfloor },2^{1+\lfloor 2\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2}\rfloor },\dots ,2^{1+\lfloor K\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->10}{\ln <!--\; FUNCTION\; APPLICATION\; -->2}\rfloor },$$

so the number of favorables cases in $2,2^2,\dots ,2^N$ is $K$, and the probability that an arbitrarily selected integer from $2,2^2,2^3,\dots ,2^N$ has leading digit $1$ is

$$p_N=\frac{K}{N}=\frac{1}{N}\lfloor N\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}\rfloor .$$

Since $\lfloor N\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}\rfloor \le N\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}<\lfloor N\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}\rfloor +1$,

$$\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}-\frac{1}{N}\le p_N\le \frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10},$$

thus

$$\underset{N\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{p}_N=\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}.$$

There are as many powers $5^n$ in $5,5^2,\dots ,5^N$ such that $5^n$ has leading coefficient $1$ than powers $2^{n+1}$ in $2^2,\dots ,2^{N+1}$ such that $2^{n+1}$ has leading coefficient $1$, thus the “probability” that a power of $5$ has leading digit $1$ is also $\frac{\ln <!--\; FUNCTION\; APPLICATION\; -->2}{\ln <!--\; FUNCTION\; APPLICATION\; -->10}$.