If `3^(x+y) =81` and `81^(x-y) = 3`, then the value of x and y are:

Options

A. `17/8, 9/8`
B. `17/8, 15/8`
C. `17/8, 11/8`
D. `15/8, 11/8`

Answer

To solve the equations \(3^{(x+y)} = 81\) and \(81^{(x-y)} = 3\), we will follow these steps:

### Step 1: Rewrite the equations in terms of powers of 3
The first equation can be rewritten using the fact that \(81 = 3^4\):
\[ 3^{(x+y)} = 3^4 \]
Since the bases are the same, we can equate the exponents:
\[ x + y = 4 \quad \text{(Equation 1)} \]

For the second equation, we can also rewrite \(81\) as \(3^4\):
\[ (3^4)^{(x-y)} = 3 \]
Using the power of a power property, we have:
\[ 3^{4(x-y)} = 3^1 \]
Again, since the bases are the same, we can equate the exponents:
\[ 4(x - y) = 1 \quad \text{(Equation 2)} \]

### Step 2: 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 it:
\[ x - y = \frac{1}{4} \quad \text{(Equation 3)} \]

### Step 3: Add Equation 1 and Equation 3
Now we can add Equation 1 and Equation 3:
\[ (x + y) + (x - y) = 4 + \frac{1}{4} \]
This simplifies to:
\[ 2x = 4 + \frac{1}{4} \]
To add \(4\) and \(\frac{1}{4}\), we convert \(4\) to a fraction:
\[ 4 = \frac{16}{4} \]
So,
\[ 2x = \frac{16}{4} + \frac{1}{4} = \frac{17}{4} \]
Dividing both sides by 2 gives:
\[ x = \frac{17}{8} \]

### Step 4: Substitute \(x\) back 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:
\[ y = 4 - \frac{17}{8} \]
Again, converting \(4\) to a fraction:
\[ 4 = \frac{32}{8} \]
So,
\[ y = \frac{32}{8} - \frac{17}{8} = \frac{15}{8} \]

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

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