CAT 2020 Set-3 | Question: 51

Question

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

• A. $ \dfrac{2}{\text{A + B}} \;– 3 $
   
• B. $ \dfrac{\text{A + B} - 3}{2} $
   
• C. $ \dfrac{2}{\text{A + B} – 3} $
   
• D. $ \dfrac{\text{A + B}}{2}\; – 3 $

Answer 1

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$: \text{C}$