If `(4)^(x+y)=1and(4)^(x-y)=4`, then 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 given in the question, we will follow these steps:

### Step 1: Write down the equations
We are given two equations:
1. \( 4^{(x+y)} = 1 \)
2. \( 4^{(x-y)} = 4 \)

### Step 2: Rewrite the first equation
Since \( 4^{(x+y)} = 1 \), we can express 1 as \( 4^0 \). Therefore, we have:
\[ 4^{(x+y)} = 4^0 \]

### Step 3: Equate the exponents from the first equation
Since the bases are the same, we can equate the exponents:
\[ x + y = 0 \]
This is our first equation (Equation 1).

### Step 4: Rewrite the second equation
For the second equation, \( 4^{(x-y)} = 4 \), we can express 4 as \( 4^1 \). Thus, we have:
\[ 4^{(x-y)} = 4^1 \]

### Step 5: Equate the exponents from the second equation
Again, since the bases are the same, we can equate the exponents:
\[ x - y = 1 \]
This is our second equation (Equation 2).

### Step 6: Solve the system of equations
Now we have a system of equations:
1. \( x + y = 0 \) (Equation 1)
2. \( x - y = 1 \) (Equation 2)

We can add both equations to eliminate \( y \):
\[ (x + y) + (x - y) = 0 + 1 \]
This simplifies to:
\[ 2x = 1 \]

### Step 7: Solve for \( x \)
Dividing both sides by 2 gives us:
\[ x = \frac{1}{2} \]

### Step 8: Substitute \( x \) back to find \( y \)
Now, we can substitute \( x \) back into Equation 1 to find \( y \):
\[ \frac{1}{2} + y = 0 \]
Subtracting \( \frac{1}{2} \) from both sides gives:
\[ y = -\frac{1}{2} \]

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

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