### 11.54 We have:

\[
\begin{aligned}
& \log _{b c}(a b)=\log _{b c} a+\log _{b c} b \\
& \log _{b} a=\frac{\log _{b c} a}{\log _{b c} b} \Rightarrow \\
& \Rightarrow \log _{b c} a=m \log _{b c} b=m \frac{1}{1+\log _{b} c}=m \frac{1}{1+\frac{1}{n}}=\frac{m n}{n+1} \\
& \log _{c} b=\frac{\log _{b c} b}{\log _{b c} c} \Rightarrow \log _{b c} b=n \log _{b c} c=n \frac{1}{n+1}=\frac{n}{n+1}
\end{aligned}
\]

Substituting expressions (2) and (3) into equality (1), we get \(\log _{b c}(a b)=\frac{n(m+1)}{n+1}\).

Answer: \(\frac{n(m+1)}{n+1}\).
11.54 It is known that $\log _{b} a=m$ and $\log _{c} b=n$. Find $\log _{b c}(a b)$.

保留源文本的换行和格式，直接输出翻译结果如下：

11.54 It is known that $\log _{b} a=m$ and $\log _{c} b=n$. Find $\log _{b c}(a b)$.