Are logarithmic Big-O complexities defined with constant base equal to those defined with variable base? Example: Deleting from a B-Tree (not to be confused with binary tree) has Big-O complexity of $ O(\log_t n) $ (where $t \in \mathbb{N}$ is the order of the tree).
There was one true/false question on exam which asks if the Big-O complexity of the operation mentioned in the example above is $ O(\log_2 n) $.
I am beginner in this topic but I understand that both $ O(\log_2 n) $ and $ O(\log_t n) $ belong to the same Big-O complexity category of $ O(\log n) $. The only thing which confuses me is whether it matters if the base of a logarithm is given as a constant or as a variable.
Additionally: Would the answer change if we swap the bases from the example and the question?
Edit (if relevant): The mentioned complexity is related to the number of disk accesses and it is not the time complexity.
Let's look at the root of question: we have  $\log_2 n=\frac{\ln n}{\ln 2}$ and $\log_t n=\frac{\ln n}{\ln t}$, so $\log_2 n = \frac{\ln t}{\ln 2} \log_t n$ and when $t$ is constant, then logarithms are in same complexity class.
Bit if $t=t(n)$, then we can ruin this relation and create any class $O(f(n))$ of complexity taking $t(n)=e^{\frac{\ln n}{f(n)}}$ we have $\log_t n=\frac{\ln n}{\ln t}=f(n)$.