Find the domain of the following functions.
`f(x)=log_(e)=(2+x)/(2-x)`

Answer

To find the domain of the function \( f(x) = \log_e\left(\frac{2+x}{2-x}\right) \), we need to ensure that the argument of the logarithm is positive and that the denominator is not zero.

### Step-by-Step Solution:

1. **Identify the Argument of the Logarithm**:
The function is defined as \( f(x) = \log_e\left(\frac{2+x}{2-x}\right) \). The argument \( \frac{2+x}{2-x} \) must be greater than zero for the logarithm to be defined.

2. **Set Up the Inequality**:
We need to solve the inequality:
\[ \frac{2+x}{2-x} > 0 \]

3. **Determine Where the Fraction is Positive**:
The fraction \( \frac{2+x}{2-x} \) is positive when both the numerator and denominator are either both positive or both negative.

- **Numerator**: \( 2+x > 0 \) implies \( x > -2 \).
- **Denominator**: \( 2-x > 0 \) implies \( x < 2 \).

4. **Find Critical Points**:
The critical points from the inequalities are \( x = -2 \) and \( x = 2 \).

5. **Test Intervals**:
We will test intervals determined by the critical points:
- Interval 1: \( (-\infty, -2) \)
- Interval 2: \( (-2, 2) \)
- Interval 3: \( (2, \infty) \)

- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ \frac{2 + (-3)}{2 - (-3)} = \frac{-1}{5} < 0 \quad \text{(not in domain)} \]
- For \( -2 < x < 2 \) (e.g., \( x = 0 \)):
\[ \frac{2 + 0}{2 - 0} = \frac{2}{2} = 1 > 0 \quad \text{(in domain)} \]
- For \( x > 2 \) (e.g., \( x = 3 \)):
\[ \frac{2 + 3}{2 - 3} = \frac{5}{-1} < 0 \quad \text{(not in domain)} \]

6. **Combine Results**:
The function is defined for \( x \) in the interval \( (-2, 2) \). However, we must also ensure that the denominator does not equal zero. The denominator \( 2-x \) is zero when \( x = 2 \), which we already excluded.

7. **Final Domain**:
Therefore, the domain of the function \( f(x) \) is:
\[ (-2, 2) \]

### Summary of the Domain:
The domain of \( f(x) = \log_e\left(\frac{2+x}{2-x}\right) \) is \( x \in (-2, 2) \).