Áp dụng công thức Lôgarit , ta được:

a)  $\log _{30}1350=\log _{30}3^{2}.5.30=\log _{30}3^{2}+\log _{30}5+\log _{30}30=2\log _{30}3+\log _{30}5+1=2a+b+1$

b) $\log _{25}15=\log _{5^{2}}15=\frac{1}{2}\log _{5}3.5=\frac{1}{2}(\log _{5}3+\log _{5}5)$

Mà theo bài ra: $c=\log _{15}3$

<=> $c=\frac{1}{\log _{3}15}=\frac{1}{\log _{3}3.5}=\frac{1}{1+\log _{3}5}$

=> $\log _{3}5=\frac{1}{c}-1$

=> $\log _{5}3=\frac{c}{1-c}$

=> $\log _{25}15=\frac{1}{2}(\frac{c}{1-c}+1)=\frac{1}{2(1-c)}$