If `x^(y)=y^(x)=` where x andy are distinct natural numbers, then find x+y.

Options

A. 6
B. 8
C. 4
D. 2

Answer

To solve the equation \( x^y = y^x \) where \( x \) and \( y \) are distinct natural numbers, we can follow these steps:

### Step 1: Understand the Equation
We start with the equation \( x^y = y^x \). This means that the value of \( x \) raised to the power of \( y \) is equal to the value of \( y \) raised to the power of \( x \).

### Step 2: Analyze Possible Values
Since \( x \) and \( y \) are distinct natural numbers, we can try different pairs of natural numbers to see if they satisfy the equation.

### Step 3: Test Different Pairs
Let’s test a few pairs of distinct natural numbers:

1. **Testing \( x = 2 \) and \( y = 4 \)**:
\[ 2^4 = 16 \quad \text{and} \quad 4^2 = 16 \]
This pair satisfies the equation.

2. **Testing \( x = 4 \) and \( y = 2 \)**:
\[ 4^2 = 16 \quad \text{and} \quad 2^4 = 16 \]
This pair also satisfies the equation.

### Step 4: Calculate \( x + y \)
Now we can calculate \( x + y \) using either pair:
- Using \( x = 2 \) and \( y = 4 \):
\[ x + y = 2 + 4 = 6 \]
- Using \( x = 4 \) and \( y = 2 \):
\[ x + y = 4 + 2 = 6 \]

### Conclusion
In both cases, we find that \( x + y = 6 \).

Thus, the final answer is:
\[ \boxed{6} \]