Prove that the number of primes satisfying this is $\log(n)$

Question

Let $\textrm{ord}(n)$ denote the number of primes $p$ such that order of $10$ modulo $p$ is $n$. Prove that $\textrm{ord}(n) \sim \log(n)$.

Answer 1

If order of $10$ modulo $p$ is $n$, then,$$10^n\equiv 1(\textrm{mod}\ p)\\10^d\not\equiv1(\textrm{mod}\ p)\ \text{whenever $d|n$}.$$So, if $\omega(n)$ denotes number of distinct prime factors of $n$, then,$$\textrm{ord}(n)=\omega(10^n-1)-\#\bigcup_{n>d|n}\{p:p|10^d-1\}.$$If you wanted to mean that $\displaystyle\lim_{n\to\infty}\frac{\textrm{ord}(n)}{\log n}=1$ by $\textrm{ord}(n) \sim \log(n)$ then it may not be true. If there exist infinite repunit primes (which is still a conjecture) then $\textrm{ord}(n)<2$ infinitely many times, which means $\liminf\limits_{n\to\infty}\frac{\textrm{ord}(n)}{\log n}=0$, contrary to your result.

However, $\omega(10^n-1)=O(\log(10^n-1))=O(n)$ gives, $\textrm{ord}(n)<<n$.