10. $\log _{54} 168$, if $\log _{7} 12=a, \log _{12} 24=b$.

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10. $\log _{54} 168=\log _{54} 7+\log _{54} 24=\frac{1}{\log _{12} 54}\left(\log _{12} 7+b\right)=\frac{1}{\log _{12} 54}\left(\frac{1}{a}+b\right)$.

$$
\frac{1}{\log _{12} 54}=\log _{54} 3+\log _{54} 4=\frac{\log _{2} 3}{3 \log _{2} 3+1}+\frac{2}{3 \log _{2} 3+1}=\frac{2+\log _{2} 3}{3 \log _{2} 3+1}
$$

$\log _{12} 24=\frac{3+\log _{2} 3}{2+\log _{2} 3}=b$. Therefore, $\log _{2} 3=\frac{3-2 b}{b-1} ; \quad \frac{1}{\log _{12} 54}=\frac{1}{8-5 b}$.

Thus, $\log _{54} 168=\frac{1+a b}{a(8-5 b)}$.