If ` 2^(x + y ) = 2^(x - y ) = sqrt8` then the value of y is

Options

A. ` (1)/(2)`
B. `(3)/(2) `
C. `0 `
D. none of these

Explanation

To solve the equation \( 2^{(x+y)} = 2^{(x-y)} = \sqrt{8} \), we will follow these steps:

### Step 1: Express \(\sqrt{8}\) in terms of powers of 2
We know that:
\[ \sqrt{8} = \sqrt{2^3} = 2^{3/2} \]
Thus, we can rewrite the equation as:
\[ 2^{(x+y)} = 2^{(x-y)} = 2^{3/2} \]

### 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. \( x + y = \frac{3}{2} \) (Equation 1)
2. \( x - y = \frac{3}{2} \) (Equation 2)

### Step 3: Solve the system of equations
Now we have a system of two equations:
1. \( x + y = \frac{3}{2} \)
2. \( x - y = \frac{3}{2} \)

We can solve these equations by adding them together:
\[ (x + y) + (x - y) = \frac{3}{2} + \frac{3}{2} \]
This simplifies to:
\[ 2x = 3 \]
Dividing both sides by 2 gives:
\[ x = \frac{3}{2} \]

### Step 4: Substitute \(x\) back to find \(y\)
Now that we have \(x\), we can substitute it back into either Equation 1 or Equation 2 to find \(y\). Let's use Equation 1:
\[ \frac{3}{2} + y = \frac{3}{2} \]
Subtracting \(\frac{3}{2}\) from both sides gives:
\[ y = 0 \]

### Final Answer
Thus, the value of \(y\) is:
\[ \boxed{0} \]

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