Following Michael Burr's suggestion in the comments, if we let
$$
f(x) = \frac{\tan(\log x)}{x}
$$
then we can define a running measure of the $x$'s for which $|f(x)| > a$ for some fixed $a > 0$:
$$
m_a(x) = \mu\Bigl( \{ t \in \mathbb R : 1 < t < x \text{ and } |f(t)| > a \} \Bigr).
$$
Claim.
$$
m_a(x) \sim \frac{2}{\pi a} \log x
$$
as $x \to \infty$.
This gives us a notion of the density of the "bad" intervals (the intervals where $f(x)$ is not small) in the real line. For instance, if we had had
$$
\lim_{x \to \infty} \frac{m_a(x)}{x} = \frac{1}{2}
$$
we would interpret this to mean that the bad intervals take up roughly half of the real line. In our case $\lim_{x \to \infty} m_a(x)/x = 0$, and it tends to zero rather quickly too, so we can interpret this to mean that, proportionally, the bad intervals are pretty insignificant.
Proof sketch. To calculate the width of the intervals where $|f(x)| > a$ we can start by calculating the points at which $\tan(\log x) = \pm ax$. Away from its poles $f(x)$ will be very close to zero for large $x$, so we only need to investigate neighborhoods of the poles.
Following this idea, the method in this answer can be used to show that, for large $x$, the graph of $y = \tan(\log x)$
- intersects the graph of $y=ax$ at $x = e^{(2n+1)\pi/2} - \frac{1}{a} + o(1)$ and
- intersects the graph of $y=-ax$ at $x = e^{(2n+1)\pi/2} + \frac{1}{a} + o(1)$
for $n \in \mathbb N$ with $n \to \infty$. We observe that for large $x$ the length of the intervals for which $|f(x)| > a$ approaches $2/a$, and so for
$$
e^{(2n+1)\pi/2} + \frac{1}{a} + \epsilon < x < e^{(2n+3)\pi/2} - \frac{1}{a} - \epsilon \tag{1}
$$
and $n$ large we have
$$
m_a(x) \approx \sum_{k=0}^{n} \frac{2}{a} \approx \frac{2n}{a}. \tag{2}
$$
For $x$ in the range in $(1)$ we have $n = \frac{\log x}{\pi} + O(1)$, and thus $(2)$ becomes
$$
m_a(x) \approx \frac{2}{\pi a} \log x.
$$
The claim follows from this estimate.
