Why is b-tree search O(log n)?
You have introduced $n$ and $m$ as the order of B-tree, I will stick to $m$.

Their height will be in the best case $\lceil log_m(N + 1) \rceil$, and the worst case is height $\lceil log_{\frac{m}{2}}(N)\rceil$ but there is also a saturation factor $d$, that you have not mentioned.
The height will be $O(log N)$, please notice that $m$ disappeared, because it effectively is multiplication by a constant.
Now at every node you have at most $m$ sorted elements, so you can perform binary search giving $log_2(m)$, so the proper complexity is $O(log(N) * log(m))$.
Since $m &lt;&lt; N$, and what is more important, is that it does not depend on $N$, so it should not be mixed, or it might be given explicitly (with $m$ not $N$ or appearing $n$).