Consider the following function $Z \colon \Bbb{Z}_{\ge2} \times [0,1) \to \Bbb{N} \cup \{\infty\}$. If $b \in \Bbb{Z}_{\ge2}$ is an integer greater than $1$ and $x \in [0,1)$, let $Z(b,x) \in \Bbb{N} \cup \{\infty\}$ be the first place after the radix point at which the base $b$ expansion of $x$ has a $0$. Let's say the first place after the radix is $1$, so $Z\left( 10,\frac{1}{10} \right) = 2$ and $Z\left( 10,\frac{1}{100} \right) = Z(10,0) = 1$. If the expansion has no zeroes, then $Z$ evaluates to $\infty$, so $Z\left( 10,\frac{1}{3} \right) = \infty$. In order for this function to be well-defined we need to choose a way of representing terminating expansions, because $0.1000... = 0.0999...$, so let us choose the representation ending in an infinite string of zeroes.
If we fix $x \in [0,1)$ we can consider the function $Z_x := Z(-,x) \colon \Bbb{Z}_{\ge2} \to \Bbb{N}$. There are numbers $x$ for which there is no $b$ such that $Z(b,x) = \infty$, for example the absolutely normal numbers. Among these numbers, are there $x$ for which $Z_x$ is a bounded function? If so, how dense are these numbers in $[0,1)$?
If we simply ignore the bases for which $Z(b,x) = \infty$, we can still ask if $Z_x$ restricted to $\Bbb{Z}_{\ge2} \setminus Z_x^{-1}(\infty)$ is bounded. For example, for any $\frac{p}{q} \in \Bbb{Q} \cap (0,1)$ (not even necessarily in minimal form) we have $$ \frac{p}{q} = 0.ppp... \; \text{(base $q + 1$)} $$ so $Z\left( q+1,\frac{p}{q} \right) = \infty$, but it still makes sense to ask if $Z_{p/q}$ is unbounded if we ignore the bases for which it is infinite. For what numbers $x$ can we always find bases $b$ such that $Z\left( b,x \right)$ is finite but arbitrarily large?
Note that $\sup_b Z(b,x) \ge 3$ for all $x \ne 0$, because of the following argument. If $x$ is rational we can find a $b$ such that $Z(b,x) = \infty$, so we can assume that $x$ has a non-terminating decimal expansion. Let $d_1$, $d_2$ be the first two non-zero digits in this expansion. If we let $-k$ be the negative power of $10$ that corresponds to the digit right before $d_2$, then the first two digits after the point of $x$ in base $10^k$ will be non-zero, so $Z(10^k,x) \ge 3$. For example, $$ 0.01001 \; \text{(base 10)} = 0.[100][1000] \; \text{(base 10000)},$$ where $[100]$ and $[1000]$ are the base $10000$ digits corresponding to those numbers. This argument does not generalize to higher than $3$, though, since increasing the base to make the third digit non-zero might mess up the second digit.