The value of `log_(3) e- log_(9) e + log_(27) e- log_(81) e+…infty` is

Options

A. `log_(2)3`
B. `log_(3)2`
C. `log_(10)e`
D. `log_(e)2`

Answer

To solve the series \( \log_{3} e - \log_{9} e + \log_{27} e - \log_{81} e + \ldots \) up to infinity, we can follow these steps:

### Step 1: Rewrite the logarithms
We start by rewriting the logarithms in terms of a common base. We know that \( \log_{a^b} c = \frac{1}{b} \log_{a} c \).

- \( \log_{9} e = \log_{3^2} e = \frac{1}{2} \log_{3} e \)
- \( \log_{27} e = \log_{3^3} e = \frac{1}{3} \log_{3} e \)
- \( \log_{81} e = \log_{3^4} e = \frac{1}{4} \log_{3} e \)

Thus, we can rewrite the series as:
\[ \log_{3} e - \frac{1}{2} \log_{3} e + \frac{1}{3} \log_{3} e - \frac{1}{4} \log_{3} e + \ldots \]

### Step 2: Factor out \( \log_{3} e \)
Now, we can factor out \( \log_{3} e \) from the series:
\[ \log_{3} e \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \right) \]

### Step 3: Identify the remaining series
The remaining series \( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \) is known as the alternating harmonic series, which converges to \( \ln(2) \).

### Step 4: Combine the results
Thus, we have:
\[ \log_{3} e \cdot \ln(2) \]

### Step 5: Use the change of base formula
We can use the change of base formula for logarithms:
\[ \log_{a} b = \frac{\log_{c} b}{\log_{c} a} \]
Applying this, we get:
\[ \log_{3} e = \frac{\ln(e)}{\ln(3)} = \frac{1}{\ln(3)} \]

### Step 6: Substitute back into the equation
Now substituting this back into our expression:
\[ \log_{3} e \cdot \ln(2) = \frac{1}{\ln(3)} \cdot \ln(2) \]

### Step 7: Final result
Thus, we can write:
\[ \frac{\ln(2)}{\ln(3)} = \log_{3} 2 \]

### Conclusion
The value of the original series is:
\[ \log_{3} 2 \]