The intensity fo a light pulse traveling along a communication channel decreases exponentially with distance x according to the relation `I=I_(0)e^(-alphax)`where `I_(0)` is the intensity at x=0 and `alpha` is the attenuation constant. The percentage decrease in intensity after a distance of `((In5)/(alpha))` is ........%.

Explanation

To solve the problem, we need to find the percentage decrease in intensity after a distance of \( \frac{\ln 5}{\alpha} \).

### Step-by-Step Solution:

1. **Understand the intensity formula**:
The intensity of the light pulse at a distance \( x \) is given by:
\[ I = I_0 e^{-\alpha x} \]
where \( I_0 \) is the initial intensity at \( x = 0 \) and \( \alpha \) is the attenuation constant.

2. **Substitute the distance**:
We need to find the intensity at \( x = \frac{\ln 5}{\alpha} \):
\[ I = I_0 e^{-\alpha \left(\frac{\ln 5}{\alpha}\right)} \]
Simplifying this, we get:
\[ I = I_0 e^{-\ln 5} \]

3. **Use the property of logarithms**:
Recall that \( e^{-\ln a} = \frac{1}{a} \):
\[ I = I_0 \cdot \frac{1}{5} = \frac{I_0}{5} \]

4. **Calculate the decrease in intensity**:
The decrease in intensity is given by:
\[ \text{Decrease} = I_0 - I = I_0 - \frac{I_0}{5} = I_0 \left(1 - \frac{1}{5}\right) = I_0 \left(\frac{4}{5}\right) \]

5. **Calculate the percentage decrease**:
The percentage decrease in intensity is calculated as:
\[ \text{Percentage Decrease} = \left(\frac{\text{Decrease}}{I_0}\right) \times 100 = \left(\frac{I_0 \left(\frac{4}{5}\right)}{I_0}\right) \times 100 = \frac{4}{5} \times 100 = 80\% \]

### Final Answer:
The percentage decrease in intensity after a distance of \( \frac{\ln 5}{\alpha} \) is **80%**.