If the area bounded by `y le e -|x-e| and y ge 0` is A sq. units, then `log_(e)(A)` is equal is

Answer

To solve the problem, we need to find the area \( A \) bounded by the curves \( y \leq e - |x - e| \) and \( y \geq 0 \), and then compute \( \log_e(A) \).

### Step-by-Step Solution:

1. **Understand the function**:
The function \( y = e - |x - e| \) can be rewritten as:
\[ y = \begin{cases} e - (x - e) & \text{if } x \geq e \\ e + (x - e) & \text{if } x < e \end{cases} \]
This simplifies to:
\[ y = \begin{cases} 2e - x & \text{if } x \geq e \\ x & \text{if } x < e \end{cases} \]

2. **Find the points of intersection**:
We need to find where \( y = e - |x - e| \) intersects with \( y = 0 \).
- For \( x < e \): Set \( x = 0 \) to find the intersection:
\[ 0 = e - (e - x) \implies x = 0 \]
- For \( x \geq e \): Set \( y = 0 \):
\[ 0 = e - (x - e) \implies x = 2e \]
Thus, the points of intersection are \( (0, 0) \) and \( (2e, 0) \).

3. **Determine the vertex of the triangle**:
The vertex of the triangle formed by the lines occurs at \( (e, e) \), where the two lines meet.

4. **Calculate the area of the triangle**:
The triangle has a base along the x-axis from \( (0, 0) \) to \( (2e, 0) \) with a height reaching up to \( (e, e) \).
- The base length is \( 2e \) (from \( 0 \) to \( 2e \)).
- The height is \( e \) (from the x-axis to the point \( (e, e) \)).
- The area \( A \) of the triangle is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2e \times e = e^2 \]

5. **Calculate \( \log_e(A) \)**:
Now we need to find \( \log_e(A) \):
\[ \log_e(A) = \log_e(e^2) = 2 \]

### Final Answer:
Thus, \( \log_e(A) = 2 \).