If ` 2^(2x - y) = 16 and 2^(x + y) = 32 ` the value of xy is

Options

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

Answer

To solve the equations \( 2^{2x - y} = 16 \) and \( 2^{x + y} = 32 \), we can follow these steps:

### Step 1: Rewrite the equations in terms of powers of 2
We know that:
- \( 16 = 2^4 \)
- \( 32 = 2^5 \)

So we can rewrite the equations as:
1. \( 2^{2x - y} = 2^4 \)
2. \( 2^{x + y} = 2^5 \)

### Step 2: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal to each other:
1. \( 2x - y = 4 \) (Equation 1)
2. \( x + y = 5 \) (Equation 2)

### Step 3: Solve the system of equations
We have a system of two equations:
1. \( 2x - y = 4 \)
2. \( x + y = 5 \)

We can solve this system using the elimination method.

### Step 4: Add the equations to eliminate \( y \)
Adding Equation 1 and Equation 2:
\[ (2x - y) + (x + y) = 4 + 5 \]
This simplifies to:
\[ 3x = 9 \]
So, we find:
\[ x = 3 \]

### Step 5: Substitute \( x \) back to find \( y \)
Now, substitute \( x = 3 \) into Equation 2:
\[ 3 + y = 5 \]
Solving for \( y \):
\[ y = 5 - 3 = 2 \]

### Step 6: Calculate \( xy \)
Now that we have \( x \) and \( y \):
\[ xy = 3 \times 2 = 6 \]

### Final Answer
The value of \( xy \) is \( 6 \).

---