The `8^(th)` term of `log_(e )2` is

Options

A. `(1)/(8)`
B. `-(1)/(8)`
C. `(1)/(7)`
D. `-(1)/(7)`

Answer

To find the 8th term of the logarithmic series for \( \log_e 2 \), we will use the series expansion of \( \log(1+x) \). The series expansion is given by:

\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]

In this case, we can express \( \log_e 2 \) as \( \log_e(1 + 1) \). Therefore, we can substitute \( x = 1 \) into the series:

\[ \log(1 + 1) = 1 - \frac{1^2}{2} + \frac{1^3}{3} - \frac{1^4}{4} + \frac{1^5}{5} - \frac{1^6}{6} + \frac{1^7}{7} - \frac{1^8}{8} + \cdots \]

This simplifies to:

\[ \log(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \cdots \]

Now, we need to find the 8th term of this series. The terms of the series can be identified as follows:

1. First term: \( 1 \)
2. Second term: \( -\frac{1}{2} \)
3. Third term: \( \frac{1}{3} \)
4. Fourth term: \( -\frac{1}{4} \)
5. Fifth term: \( \frac{1}{5} \)
6. Sixth term: \( -\frac{1}{6} \)
7. Seventh term: \( \frac{1}{7} \)
8. Eighth term: \( -\frac{1}{8} \)

Thus, the 8th term of the series is:

\[ -\frac{1}{8} \]

Now, let's summarize the solution:

**Final Answer:**
The 8th term of \( \log_e 2 \) is \( -\frac{1}{8} \).

---