I was reading this paper, about the complexity of decision problems for reversal bounded counter machines. I got to Theorem 1 on Page 6. The theorem shows that there's a log-space NTM which can determine if a non deterministic reversal-bounded counter machine is empty or not. (A log-space NTM can be converted into a polynomial time DTM).
The proof shows that, for input machine represented as a string of length $n$, with $m$ counters, that $O(m\log n)$ space is required.
Here's where I get lost. The paper says that, since $m$ is fixed, we can consider the machine to take $O(\log n)$ space.
Does this mean that the algorithm is only uses log-space if $m$ is fixed? Would the corresponding deterministic algorithm then be exponential in terms of $m$?