If x, y satisfy the equation, `y^(x)=x^(y) and x=2y`, then `x^(2)+y^(2)=`

Answer

To solve the problem where \( x \) and \( y \) satisfy the equations \( y^x = x^y \) and \( x = 2y \), we will follow these steps:

### Step 1: Write down the equations
We have two equations:
1. \( y^x = x^y \)
2. \( x = 2y \)

### Step 2: Take logarithm of both sides of the first equation
Taking logarithm of both sides of the first equation \( y^x = x^y \):
\[ \log(y^x) = \log(x^y) \]

### Step 3: Apply the power rule of logarithms
Using the power rule of logarithms, we can rewrite the equation:
\[ x \log y = y \log x \]

### Step 4: Substitute \( x = 2y \) into the equation
Now, substitute \( x = 2y \) into the equation \( x \log y = y \log x \):
\[ (2y) \log y = y \log (2y) \]

### Step 5: Simplify the equation
Since \( y \) is common on both sides, we can divide both sides by \( y \) (assuming \( y \neq 0 \)):
\[ 2 \log y = \log (2y) \]

### Step 6: Expand the right-hand side
Using the property of logarithms, we can expand the right-hand side:
\[ 2 \log y = \log 2 + \log y \]

### Step 7: Rearrange the equation
Now, rearranging gives:
\[ 2 \log y - \log y = \log 2 \]
\[ \log y = \log 2 \]

### Step 8: Solve for \( y \)
By exponentiating both sides, we find:
\[ y = 2 \]

### Step 9: Substitute \( y \) back to find \( x \)
Now that we have \( y \), we can substitute it back into the second equation \( x = 2y \):
\[ x = 2 \times 2 = 4 \]

### Step 10: Calculate \( x^2 + y^2 \)
Now we need to find \( x^2 + y^2 \):
\[ x^2 + y^2 = 4^2 + 2^2 = 16 + 4 = 20 \]

### Final Answer
Thus, the value of \( x^2 + y^2 \) is \( \boxed{20} \).

---