If `A=1+r^a+r^(2a)+...oo=a` and `B=1+r^b+r^(2b)+...oo=b` then `a/b` is equal to

Options

A. `log_B A`
B. `log_(1-B) (1-A)`
C. `log_((B-1)/B) ((A-1)/A)`
D. `log_(1+B) (1+A)`

Answer

To solve the problem, we need to analyze the infinite geometric series given for \( A \) and \( B \) and derive the relationship between \( a \) and \( b \).

### Step-by-Step Solution:

1. **Understanding the Infinite Geometric Series**:
The series for \( A \) is given by:
\[ A = 1 + r^a + r^{2a} + \ldots \]
This is an infinite geometric series where the first term \( a_1 = 1 \) and the common ratio \( r = r^a \).

2. **Sum of the Infinite Geometric Series**:
The sum \( S \) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a_1}{1 - r} \]
Applying this to our series:
\[ A = \frac{1}{1 - r^a} \]

3. **Setting Up the Equation**:
We are given that \( A = a \). Therefore, we can set up the equation:
\[ a = \frac{1}{1 - r^a} \]

4. **Rearranging the Equation**:
Rearranging the equation gives:
\[ 1 - r^a = \frac{1}{a} \]
This leads to:
\[ r^a = 1 - \frac{1}{a} = \frac{a - 1}{a} \]

5. **Taking Logarithm**:
Taking the logarithm of both sides:
\[ a \log r = \log\left(\frac{a - 1}{a}\right) \]

6. **Expressing in Terms of \( r \)**:
This can be simplified using logarithmic properties:
\[ a \log r = \log(a - 1) - \log a \]

7. **Repeating for \( B \)**:
Now, we perform similar steps for \( B \):
\[ B = 1 + r^b + r^{2b} + \ldots \]
Thus,
\[ B = \frac{1}{1 - r^b} \]
Setting \( B = b \):
\[ b = \frac{1}{1 - r^b} \]
Rearranging gives:
\[ r^b = 1 - \frac{1}{b} = \frac{b - 1}{b} \]
Taking logarithm:
\[ b \log r = \log\left(\frac{b - 1}{b}\right) \]

8. **Finding the Ratio \( \frac{a}{b} \)**:
Now we have two equations:
\[ a \log r = \log(a - 1) - \log a \]
\[ b \log r = \log(b - 1) - \log b \]
Dividing these two equations:
\[ \frac{a \log r}{b \log r} = \frac{\log(a - 1) - \log a}{\log(b - 1) - \log b} \]
This simplifies to:
\[ \frac{a}{b} = \frac{\log(a - 1) - \log a}{\log(b - 1) - \log b} \]

### Final Expression:
Thus, the ratio \( \frac{a}{b} \) can be expressed as:
\[ \frac{a}{b} = \frac{\log(a - 1)/a}{\log(b - 1)/b} \]

### Conclusion:
The final result for \( \frac{a}{b} \) is:
\[ \frac{a}{b} = \log_{b}\left(\frac{a - 1}{a}\right) \]
This corresponds to option three in the original question.