Statement-I :
`((x-y)/(x))+(1)/(2)((x-y)/(x))^(2)+(1)/(3)((x-y)/(x))^(3)+….=log_(e )x-log_(e )y`
Statement-II :
`(a-1)-((a-1)^(2))/(2)+((a-1)^(3))/(3)-((a-1)^(4))/(4 )+….=log_(e )a`
Which of the above statements true

Options

A. Only I
B. Only II
C. both I & II
D. neither I nor II

Answer

To solve the question, we need to analyze both statements provided and verify their correctness.

### Statement I:
The first statement is:
\[ \frac{x-y}{x} + \frac{1}{2}\left(\frac{x-y}{x}\right)^2 + \frac{1}{3}\left(\frac{x-y}{x}\right)^3 + \ldots = \log_e x - \log_e y \]

1. **Let** \( z = \frac{x-y}{x} \). Then we can rewrite the left-hand side as:
\[ z + \frac{1}{2}z^2 + \frac{1}{3}z^3 + \ldots \]

2. **Recognize the series**: The series \( z + \frac{1}{2}z^2 + \frac{1}{3}z^3 + \ldots \) is the Taylor series expansion for \( -\log(1-z) \) when \( |z| < 1 \).

3. **Substituting back**: Since \( z = \frac{x-y}{x} \), we have:
\[ -\log\left(1 - \frac{x-y}{x}\right) = -\log\left(\frac{y}{x}\right) = \log\left(\frac{x}{y}\right) = \log_e x - \log_e y \]

Thus, Statement I is **true**.

### Statement II:
The second statement is:
\[ (a-1) - \frac{(a-1)^2}{2} + \frac{(a-1)^3}{3} - \frac{(a-1)^4}{4} + \ldots = \log_e a \]

1. **Let** \( x = a - 1 \). Then we can rewrite the left-hand side as:
\[ x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

2. **Recognize the series**: The series \( x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \) is the Taylor series expansion for \( \log(1+x) \) when \( |x| < 1 \).

3. **Substituting back**: Since \( x = a - 1 \), we have:
\[ \log(1 + (a - 1)) = \log(a) \]

Thus, Statement II is also **true**.

### Conclusion:
Both statements are correct. Therefore, the answer is:
**Option C: Both 1 and 2 are true.**