Why is $O(\log_{M/B} N/M)$ the same as $O(\log_{M/B} N/B)$?
Your construction will work just fine:
$$\begin{align}
\log_{M/B}\frac{N}{M}&amp;=\log_{M/B}\frac{NB}{BM}=\log_{M/B}\frac{N}{B}\cdot\frac{B}{M}\\
 &amp;= \log_{M/B}\frac{N}{B}+\log_{M/B}\frac{B}{M}\\
 &amp;= \log_{M/B}\frac{N}{B}-1
\end{align}$$
Since the two terms differ by 1, they're asymptotically equal.