If `x^yxxy^x=800`, then find `x-y` (where x and y are positive integers and `xgt y`).

Options

A. 2
B. 5
C. 4
D. 3

Explanation

To solve the equation \( x^y \cdot y^x = 800 \) and find \( x - y \) where \( x \) and \( y \) are positive integers and \( x > y \), we can follow these steps:

### Step 1: Prime Factorization of 800
First, we need to factor 800 into its prime factors.

\[ 800 = 8 \times 100 \]

Next, we can break down 8 and 100 further:

\[ 8 = 2^3 \quad \text{and} \quad 100 = 10^2 = (2 \times 5)^2 = 2^2 \times 5^2 \]

Now, combining these:

\[ 800 = 2^3 \times (2^2 \times 5^2) = 2^{3+2} \times 5^2 = 2^5 \times 5^2 \]

### Step 2: Setting up the Equation
Now we have:

\[ x^y \cdot y^x = 2^5 \cdot 5^2 \]

This means we need to express \( x^y \) and \( y^x \) in terms of these prime factors.

### Step 3: Assigning Values to \( x \) and \( y \)
We can assume \( x \) and \( y \) take the forms of these prime factors. Since \( x > y \), we can try:

- Let \( x = 5 \) and \( y = 2 \)

Now we check:

\[ x^y = 5^2 = 25 \quad \text{and} \quad y^x = 2^5 = 32 \]

Now calculate \( x^y \cdot y^x \):

\[ x^y \cdot y^x = 25 \cdot 32 \]

Calculating this gives:

\[ 25 \cdot 32 = 800 \]

### Step 4: Finding \( x - y \)
Now that we have found valid values for \( x \) and \( y \):

\[ x = 5 \quad \text{and} \quad y = 2 \]

Thus,

\[ x - y = 5 - 2 = 3 \]

### Final Answer
The value of \( x - y \) is:

\[ \boxed{3} \]