If `4^(x+y)` = 256 and `(256)^(x-y)` = 4. what are the values of x and y ?

Options

A. `17/8, 15/8`
B. `17/4,15/4`
C. `9/17,15/17`
D. `8/17,8/15`

Answer

To solve the equations \(4^{(x+y)} = 256\) and \( (256)^{(x-y)} = 4\), we will follow these steps:

### Step 1: Rewrite the equations in terms of powers of 2
We know that:
- \(4 = 2^2\)
- \(256 = 2^8\)

Thus, we can rewrite the first equation:
\[ 4^{(x+y)} = (2^2)^{(x+y)} = 2^{2(x+y)} = 256 = 2^8 \]
This gives us:
\[ 2(x+y) = 8 \]
Dividing both sides by 2:
\[ x+y = 4 \quad \text{(Equation 1)} \]

### Step 2: Rewrite the second equation
Now, we rewrite the second equation:
\[ (256)^{(x-y)} = (2^8)^{(x-y)} = 2^{8(x-y)} = 4 = 2^2 \]
This gives us:
\[ 8(x-y) = 2 \]
Dividing both sides by 8:
\[ x-y = \frac{1}{4} \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have a system of linear equations:
1. \(x + y = 4\) (Equation 1)
2. \(x - y = \frac{1}{4}\) (Equation 2)

We can solve these equations by adding them together:
\[ (x+y) + (x-y) = 4 + \frac{1}{4} \]
This simplifies to:
\[ 2x = 4 + \frac{1}{4} = \frac{16}{4} + \frac{1}{4} = \frac{17}{4} \]
Thus:
\[ x = \frac{17}{8} \]

### Step 4: Substitute to find y
Now, we substitute \(x\) back into Equation 1 to find \(y\):
\[ \frac{17}{8} + y = 4 \]
Subtracting \(\frac{17}{8}\) from both sides:
\[ y = 4 - \frac{17}{8} = \frac{32}{8} - \frac{17}{8} = \frac{15}{8} \]

### Final Values
Thus, the values of \(x\) and \(y\) are:
\[ x = \frac{17}{8}, \quad y = \frac{15}{8} \]