Are there infinitely many $n$ such that $2^n$ and $3^n$ both have first digit $7$? (exercise in ergodic theory)

Question

I am in a reading group that is reading Invitation to Ergodic Theory by C.E. Silva. Section 3.2, exercise 10 asks whether there are infinitely many positive integers $n$ such that both $2^n$ and $3^n$ have leading digit $7.$ We think the answer is almost certainly yes, but we have not been able to solve the problem ourselves, or to find a solution online.

The section discusses the related problem of leading digits of powers of $2$ by first introducing a transformation $R_\alpha:[0,1)\to [0,1),$ given by $R_\alpha(x)=x+\alpha \bmod 1$ for some fixed $\alpha.$ Then, a theorem of Kronecker (3.2.3 in the text) states that, for any irrational $\alpha$ and any $x\in[0, 1),$ the set $\{R^n_\alpha(x)\}_{n\ge0}$ is dense in $[0,1).$

Using this theorem, one can show that there are infinitely many $2^n$ with leading digit $7:$ $2^n$ has leading digit $7$ exactly when there exists integer $k$ s.t. $$7\cdot 10^k \le 2^n < 8\cdot 10^k$$ so, after applying the base $10$ logarithm to all of these terms, $$\log_{10} 7 \le n\log_{10} 2 - k < \log_{10} 8$$ or, equivalently, $$R^n_{\log_{10}2}(0)\in [\log_{10} 7, \log_{10} 8).$$ Since $\log_{10}2$ is irrational, there are infinitely many $n$ that satisfy this property. Similar reasoning shows that there are infinitely many powers of $3$ with any leading digit, or, indeed, any leading string of digits.

However, we are stuck on showing that there are infinitely many $n$ where both $R^n_{\log_{10}2}(0)$ and $R^n_{\log_{10}3}(0)$ are in the desired interval for the same value of $n.$ Intuitively it seems like it must be true - there are infinitely many $m$ for which $R^m_{\log_{10}2}(0)\in [\log_{10} 7, \log_{10} 8),$ and it seems like the values of $R^m_{\log_{10}3}(0)$ for those $m$ should be dense in $[0,1),$ but we have been unable to show this so far.

Are there infinitely many $n$ such that both $2^n$ and $3^n$ have leading digit $7?$ Any answer is appreciated, but one using the methods of the section, as discussed above, would be especially preferred. Thanks in advance.

Answer 1

Let $r,s\in[0,1)$ be irrational, and let $m\in\Bbb Z^+$. By the pigeonhole principle there must be distinct $i,j\in\{1,\ldots,m^2+1\}$ and $k,\ell\in\{0,\ldots,m-1\}$ such that

$$\langle ir\bmod 1,is\bmod 1\rangle,\langle jr\bmod 1,js\bmod 1\rangle\in\left[\frac{k}m,\frac{k+1}m\right)\times\left[\frac{\ell}m,\frac{\ell+1}m\right)\;.$$

Then $x=|(j-i)r|\bmod 1<\frac1m$ and $y=|(j-i)s|\bmod 1<\frac1m$, and

$$\langle x,y\rangle\in\left[0,\frac1m\right)\times\left[0,\frac1m\right)\;.$$

Now assume further that $\frac{s}r$ is irrational. Then the points $\langle nx,ny\rangle$ for $n\in\Bbb Z$ are spaced less than $\frac{\sqrt2}m$ apart on a line of irrational slope, and it’s well known that the image of that line in $\Bbb R^2/\Bbb Z^2$ is dense in the unit square; see here for instance. Since $\frac{\log_{10}3}{\log_{10}2}$ is irrational, and we can choose arbitrarily large $m$, the desired result follows.