Let $r_1, r_2 > 1$ be the two bases, then the smallest number requiring $n+1$ digits in base $r_1$ is $r_1^n$ which has representation $10\ldots 0$ (where we have $n$ zeros), therefore an arbitrary number $m$ needs $n+1$ digits if and only if
$$
r_1^n \le m < r_1^{n+1}
$$
and this relation holds in any base. Also for the same $m$ we have
$$
r_2^{n \cdot \log_{r_2} r_1} \le m < r_2^{(n+1) \cdot \log_{r_2} r_1}.
$$
So the number of digits needed to represent $m$ in base $r_2$ is contained between $\lfloor n\log_{r_2} r_1 \rfloor + 1, \ldots, \lceil (n+1)\log_{r_2} r_1 \rceil + 1$. So your formula is indeed a good approximation and some times it is right. For example consider $r_1 = 2, r_2 = 3$, and:
$$
\begin{array}{l|l|l|l|c|c}
m & r_1 & r_2 & d_2 & d_3 & \lfloor (d_2-1)\cdot \log_3 2 \rfloor + 1 & \lceil d_2\cdot \log_3 2\rceil + 1 \\ \hline
0 & 0 & 0 & 1 & 1 & 1 & 2 \\
1 & 1 & 1 & 1 & 1 & 1 & 2 \\
2 & 10 & 2 & 2 & 1 & 1 & 3 \\
3 & 11 & 10 & 2 & 2 & 1 & 3 \\
4 & 100 & 11 & 3 & 2 & 2 & 3 \\
5 & 101 & 12 & 3 & 2 & 2 & 3 \\
6 & 110 & 100 & 3 & 3 & 2 & 3 \\
7 & 111 & 101 & 3 & 3 & 2 & 3 \\
\end{array}
$$
As you can see in the table, for $m = 3$ the number of digits needed in base $3$ is strictly between the bounds, for $m \in \{0,1,2,4,5\}$ the lower bound is assumed, and for $m \in \{ 6,7 \}$ the upper bound is assumed.
