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If $\log_{a} 30 = \text{A}, \log_{a} (5/3) = – \text{B}$ and $\log_{2} a = 1/3,$ then $\log_{3}a$ equals

1. $\frac{2}{\text{A + B}} \;– 3$
2. $\frac{\text{A + B} - 3}{2}$
3. $\frac{2}{\text{A + B} – 3}$
4. $\frac{\text{A + B}}{2}\; – 3$

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$: \text{C}$

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