Logarithmic identity calculation

We are given:

1. $\log_a 30 = A \implies \log_a 2 + \log_a 3 + \log_a 5 = A$
2. $\log_a \frac{5}{3} = -B \implies \log_a 5 - \log_a 3 = -B$
3. $\log_2 a = \frac{1}{3} \implies \log_a 2 = 3$

Subtract (2) from (1): $(\log_a 2 + \log_a 3 + \log_a 5) - (\log_a 5 - \log_a 3) = A - (-B)$ $\log_a 2 + 2 \log_a 3 = A + B$ Substitute $\log_a 2 = 3$: $3 + 2 \log_a 3 = A + B \implies 2 \log_a 3 = A + B - 3 \implies \log_a 3 = \frac{A + B - 3}{2}$

Therefore, $\log_3 a = \frac{1}{\log_a 3} = \frac{2}{A + B - 3}$.