Given that,
- $ \log_{a}30 = A \quad \longrightarrow (1) $
- $ \log_{a} \left(\frac{5}{3} \right) = \;– B \quad \longrightarrow (2) $
- $ \log_{2}a = \frac{1}{3} \quad \longrightarrow (3) $
Subtract equation $(2)$ from equation $(1).$
$ A – ( – B) = \log_{a}30 – \log_{a} \left(\frac{5}{3} \right) $
$ \Rightarrow A + B = \log_{a} \left( \dfrac{30}{\frac{5}{3}} \right) \quad \left[ \because \log_{a}m – \log_{a}n = \log_{a} \left( \dfrac{m}{n} \right)\right] $
$ \Rightarrow A + B = \log_{a} 18 $
$ \Rightarrow A + B = \log_{a} (9 \times 2) $
$ \Rightarrow A + B = \log_{a}9 + \log_{a}2 \quad [ \because \log_{a}(mn) = \log_{a}m + \log_{a}n] $
$ \Rightarrow A + B = \log_{a}3^{2} + 3 \quad [ \because \log_{2}a = \frac{1}{3} \Rightarrow \log_{a}2 = 3, \text{from equation $(3)$}] $
$ \Rightarrow A + B = 2 \log_{a}3 + 3 \quad [ \because \log_{a} m^{n} = n \log_{a}m] $
$ \Rightarrow \log_{a}3 = \dfrac{A+B-3}{2} $
$ \Rightarrow \boxed{\log_{3}a = \frac{2}{A+B-3}} \quad \left[ \because \log_{a}b = \frac{1}{\log_{b}a}\right] $
Correct Answer$: \text{C}$