1. Find $\log _{35} 28$, if $\log _{14} 7=a$ and $\log _{14} 5=b$.
Solution. We have

$$
\log _{35} 28=\frac{\log _{14} 28}{\log _{14} 35}=\frac{\log _{14} 14+\log _{14} 2}{\log _{14} 5+\log _{14} 7}=\frac{1+\log _{14} \frac{14}{7}}{a+b}=\frac{1+\log _{14} 14-\log _{14} 7}{a+b}=\frac{2-a}{a+b}
$$