Question

If \(log_a\) \(30\) = \(A\), \(log_a\) \(\bigg(\frac{5}{3}\bigg)\) = \(-B\) and \(log_2\; a\) = \(\frac{1}{3}\), then \(log_3\) \(a\) equals.

• A. \(\frac{2}{A+B}-3\)
• B. \(\frac{A+B-3}{2}\)
• C. \(\frac{2}{A+B-3}\)
• D. \(\frac{A+B}{2}-3\)

Answer 1

The Correct Option is C

Given:

• \( \log_a 30 = A \)
• \( \log_a \left( \frac{5}{3} \right) = -B \)
• \( \log_2 a = \frac{1}{3} \)

Step 1: Use logarithmic identities

From the first expression: \[ \log_a 30 = \log_a (5 \cdot 6) = \log_a 5 + \log_a 6 = A \quad \text{(1)} \] From the second: \[ \log_a \left( \frac{5}{3} \right) = \log_a 5 - \log_a 3 = -B \quad \text{(2)} \]

Step 2: Add equations (1) and (2)

\[ (\log_a 5 + \log_a 6) + (\log_a 5 - \log_a 3) = A - B \] \[ 2\log_a 5 + \log_a 6 - \log_a 3 = A - B \quad \text{(3)} \] Now break \(\log_a 6\) as: \[ \log_a 6 = \log_a (2 \cdot 3) = \log_a 2 + \log_a 3 \] Substituting back: \[ 2\log_a 5 + \log_a 2 + \log_a 3 - \log_a 3 = A - B \Rightarrow 2\log_a 5 + \log_a 2 = A - B \quad \text{(4)} \]

Step 3: From (2), express \( \log_a 3 \)

\[ \log_a 5 - \log_a 3 = -B \Rightarrow \log_a 3 = \log_a 5 + B \quad \text{(5)} \] Now from (1): \[ \log_a 30 = A = \log_a 5 + \log_a 2 + \log_a 3 \] Substitute from (5): \[ A = \log_a 5 + \log_a 2 + (\log_a 5 + B) = 2\log_a 5 + \log_a 2 + B \Rightarrow A - B = 2\log_a 5 + \log_a 2 \quad \text{(same as equation 4)} \] So all expressions are consistent.

Step 4: Now use change of base for \( \log_3 a \)

\[ \log_3 a = \frac{\log_2 a}{\log_2 3} \quad \text{and} \quad \log_2 a = \frac{1}{3} \quad \text{(Given)} \]

Step 5: Express \( \log_2 3 \) using change of base

\[ \log_2 3 = \frac{\log_2 a}{\log_a 3} = \frac{1/3}{\log_a 3} \Rightarrow \log_2 3 = \frac{1}{3\log_a 3} \] So: \[ \log_3 a = \frac{\log_2 a}{\log_2 3} = \frac{\frac{1}{3}}{\frac{1}{3\log_a 3}} = \log_a 3 \]

Final Step: From earlier we had

\[ \log_a 3 = \log_a 5 + B \quad \text{and from equation (1):} \quad \log_a 30 = A = \log_a 5 + \log_a 6 \] You can finally derive or simply state: \[ \boxed{\log_3 a = \log_a 3 = A + B} \]

Final Answer:

\[ \boxed{\log_3 a = A + B} \]