By Dirchlet's hyperbola method, one can prove that the average number of divisors of integers $1 \leq n \leq X$ is $\log X$. This question concerns the number of integers $n \leq X$ such that the number of divisors, $d(n)$, is substantially larger than average. Indeed, what is known about the size of the set
$$\displaystyle \{1 \leq n \leq X : d(n) > (\log X)^A \}$$
where $A > 1$ is considered to be a large (but fixed) positive number?
Theorem 1.11 and Theorem 1.22 of the paper by Norton, cited in the comment of Peter Humphries, show that for any fixed $A \ge \log 2$, $$ \frac{X (\log\log X)^{O(1)}}{(\log X)^{B(A)}} \ll_A |\{1\le n\le X:d(n) \ge (\log X)^A\}| \ll_A \frac{X}{(\log X)^{B(A)}}, $$ where $$ B(A):=1+\frac{A}{\log 2}\left(\log\left(\frac{A}{\log 2}\right) -1 \right). $$ Equation (1.37) of the same paper gives the correct order of magnitude: For every fixed $A>\log 2$, $$ |\{1\le n\le X:d(n) \ge (\log X)^A\}| \asymp_A \frac{X}{(\log X)^{B(A)} (\log\log X)^{1/2}}. $$
The normal order of $\log(d(n))$ is $\log(2)\log\log(n))$. So, for every $\epsilon>0$, $$ \log(d(n))<(1+\epsilon)\log(2)\log(\log(n)) $$ hold for almost all n: that is, if the proportion of $n\le x$ for which this does not hold tends to 0 as $x$ tends to infinity. Thus $$ d(n) < \log(n)^{(1+\epsilon)\log(2)} $$ holds for almost all $n$.
