已知 $\log _{2} 3=a, \log _{3} 7=b$, 则 $\log _{1456}=$
由 $\log _{2} 3=a, \log _{3} 7=b$, 得 $\log _{2} 7=a b$,

则 $\log _{14} 56=\frac{\log _{2} 56}{\log _{2} 14}=\frac{\log _{2} 8+\log _{2} 7}{\log _{2} 2+\log _{2} 7}=\frac{3+\log _{2} 7}{1+\log _{2} 7}=\frac{3+a b}{1+a b}$.

答案: $\frac{3+a b}{1+a b}$