If `(4)^(x+y)=1` and `(4)^(x-y)=4` the the value of x and y will be respectively

Options

A. `1/2` and `-1/2`
B. `1/2` and `1/2`
C. `-1/2` and `-1/2`
D. `-1/2` and `1/2`

Answer

To solve the equations \( (4)^{(x+y)} = 1 \) and \( (4)^{(x-y)} = 4 \), we can follow these steps:

### Step 1: Analyze the first equation
The first equation is:
\[ (4)^{(x+y)} = 1 \]
We know that \( 4^0 = 1 \). Therefore, we can equate the exponents:
\[ x + y = 0 \quad \text{(Equation 1)} \]

### Step 2: Analyze the second equation
The second equation is:
\[ (4)^{(x-y)} = 4 \]
We know that \( 4^1 = 4 \). Therefore, we can equate the exponents:
\[ x - y = 1 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have a system of two equations:
1. \( x + y = 0 \)
2. \( x - y = 1 \)

We can solve these equations simultaneously.

#### Adding the two equations:
\[ (x + y) + (x - y) = 0 + 1 \]
This simplifies to:
\[ 2x = 1 \]
Dividing both sides by 2 gives:
\[ x = \frac{1}{2} \]

### Step 4: Substitute the value of \( x \) back into one of the equations
Now, we can substitute \( x = \frac{1}{2} \) back into Equation 1:
\[ \frac{1}{2} + y = 0 \]
Solving for \( y \):
\[ y = -\frac{1}{2} \]

### Final Result
Thus, the values of \( x \) and \( y \) are:
\[ x = \frac{1}{2}, \quad y = -\frac{1}{2} \]