The sum of the series `log_(9)3+log_(27)3-log_(81)3+log_(243)3- . . .` is

Options

A. `1-log_(e)2`
B. `1+log_(e)2`
C. `log_(e)2`
D. `log_(e)3`

Answer

To find the sum of the series \( \log_{9}3 + \log_{27}3 - \log_{81}3 + \log_{243}3 - \ldots \), we can follow these steps:

### Step 1: Rewrite each term using the change of base formula
We can express each logarithm in the series using the change of base formula:
\[ \log_{a}b = \frac{\log b}{\log a} \]
Thus, we have:
- \( \log_{9}3 = \frac{\log 3}{\log 9} \)
- \( \log_{27}3 = \frac{\log 3}{\log 27} \)
- \( \log_{81}3 = \frac{\log 3}{\log 81} \)
- \( \log_{243}3 = \frac{\log 3}{\log 243} \)

### Step 2: Simplify each logarithm
Next, we can simplify each logarithm:
- \( \log 9 = \log 3^2 = 2 \log 3 \) so \( \log_{9}3 = \frac{\log 3}{2 \log 3} = \frac{1}{2} \)
- \( \log 27 = \log 3^3 = 3 \log 3 \) so \( \log_{27}3 = \frac{\log 3}{3 \log 3} = \frac{1}{3} \)
- \( \log 81 = \log 3^4 = 4 \log 3 \) so \( \log_{81}3 = \frac{\log 3}{4 \log 3} = \frac{1}{4} \)
- \( \log 243 = \log 3^5 = 5 \log 3 \) so \( \log_{243}3 = \frac{\log 3}{5 \log 3} = \frac{1}{5} \)

### Step 3: Write the series with simplified terms
Now we can rewrite the series:
\[ \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots \]

### Step 4: Identify the pattern
The series can be expressed as:
\[ S = \left( \frac{1}{2} + \frac{1}{3} \right) + \left( -\frac{1}{4} + \frac{1}{5} \right) + \left( -\frac{1}{6} + \frac{1}{7} \right) + \ldots \]
This series alternates in sign and can be grouped in pairs.

### Step 5: Sum the series
The series can be recognized as a logarithmic series. The sum can be derived from the known series:
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
For \( x = 1 \):
\[ \log(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \]
Thus, the sum of our series is:
\[ S = \log(2) \]

### Step 6: Final result
Since the base of the logarithm was not specified, we assume it is base \( 10 \) or natural logarithm, hence:
\[ S = \log_{10}(2) \text{ or } S = \ln(2) \]

### Conclusion
The sum of the series \( \log_{9}3 + \log_{27}3 - \log_{81}3 + \log_{243}3 - \ldots \) is \( \log(2) \).

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