If `4^((x+y))=256` and `(256)^((x-y))=4`, what are the value 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 256 in terms of base 4
We know that \(256\) can be expressed as a power of \(4\):
\[ 256 = 4^4 \]
Thus, we can rewrite the first equation:
\[ 4^{(x+y)} = 4^4 \]

### Step 2: Set the exponents equal
Since the bases are the same, we can set the exponents equal to each other:
\[ x + y = 4 \quad \text{(Equation 1)} \]

### Step 3: Rewrite the second equation
Next, we rewrite the second equation \(256^{(x-y)} = 4\) using the same base:
\[ 256 = 4^4 \implies 256^{(x-y)} = (4^4)^{(x-y)} = 4^{4(x-y)} \]
Now, we can rewrite the equation:
\[ 4^{4(x-y)} = 4^1 \]

### Step 4: Set the exponents equal again
Again, since the bases are the same, we can set the exponents equal:
\[ 4(x - y) = 1 \quad \text{(Equation 2)} \]

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

From Equation 2, we can simplify:
\[ x - y = \frac{1}{4} \]

### Step 6: Add the two equations
Now we can add Equation 1 and the simplified Equation 2:
\[ (x + y) + (x - y) = 4 + \frac{1}{4} \]
This simplifies to:
\[ 2x = 4 + \frac{1}{4} = \frac{16}{4} + \frac{1}{4} = \frac{17}{4} \]
Thus,
\[ x = \frac{17}{8} \]

### Step 7: 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:
\[ x = \frac{17}{8}, \quad y = \frac{15}{8} \]

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