In the book A Comprehensive Course in Number Theory by Alan Baker. The author mentions that even though the average order of $\tau(n)$ is $\log n$, almost all numbers have about $(\log n)^{\log 2}$ divisors.
$\tau(n)$=number of divisors of n.
I was wondering how one would prove this result. If anyone has any ideas it would be great. Thanks.
We have $$ \sum_{n\le x}\tau(n)\sim \sum_{n\le x}\log(n), $$ so that the average value of $\tau(n)$ is indeed $\log(n)$, but from Hardy and Ramanujan we know, since $2^{\omega(n)}\le \tau(n)\le 2^{\Omega(n)}$, that for most numbers $n$, $$ \tau(n)=\log(n)^{\log(2)+o(1)}, $$ where $\log (2)$ is around $0.693$. The normal value of $\omega (n)$ resp. $\Omega(n)$ is $\log(\log(n))$, by Hardy and Ramanujan.