Question

If \( A = 1 + r^a + r^{2a} + r^{3a} + \cdots \infty \) and \( B = 1 + r^b + r^{2b} + r^{3b} + \cdots \infty \), then \( a/b \) is equal to

• A. \( \log_B (A) \)
• B. \( \log_{(B-1)/B} ((A-1)/A) \)
• C. \( \log_{B-1} \left( \frac{A-1}{B} \right) \)
• D. None of these

Hint:

Always simplify your \( S_\infty \) to find the common ratio \( R \) first. Once you have \( R^a \) and \( R^b \), logarithms are the standard tool to "bring down" the exponents.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The given series are infinite geometric progressions (G.P.). We use the sum formula for an infinite G.P., \( S_\infty = \frac{a}{1-R} \), to find relations for \( r^a \) and \( r^b \).

Step 2: Key Formula or Approach:
For \( |R|<1 \), \( S_\infty = \frac{1}{1-R} \).

Step 3: Detailed Explanation:
1. Solve for \( r^a \) and \( r^b \): \[ A = \frac{1}{1 - r^a} \implies 1 - r^a = \frac{1}{A} \implies r^a = 1 - \frac{1}{A} = \frac{A-1}{A} \] Similarly, \[ B = \frac{1}{1 - r^b} \implies 1 - r^b = \frac{1}{B} \implies r^b = 1 - \frac{1}{B} = \frac{B-1}{B} \] 2. Relate \( a \) and \( b \): Take the natural log of both equations: \( a \ln r = \ln \left( \frac{A-1}{A} \right) \) and \( b \ln r = \ln \left( \frac{B-1}{B} \right) \). Dividing the two: \[ \frac{a}{b} = \frac{\ln(\frac{A-1}{A})}{\ln(\frac{B-1}{B})} \] Using the base change formula \( \log_y x = \frac{\ln x}{\ln y} \): \[ \frac{a}{b} = \log_{(B-1)/B} \left( \frac{A-1}{A} \right) \]

Step 4: Final Answer
The ratio \( a/b \) is equal to \( \log_{(B-1)/B} (\frac{A-1}{A}) \).