If ` 8^(x) xx 4^(y) = 32 and 81 ^(x) xx 27 ^(y) = 3 `, find the value of x and y

Answer

To solve the equations \( 8^x \cdot 4^y = 32 \) and \( 81^x \cdot 27^y = 3 \), we will follow these steps:

### Step 1: Rewrite the bases in terms of powers of 2 and 3
1. Rewrite \( 8 \), \( 4 \), \( 32 \), \( 81 \), and \( 27 \) in terms of their prime bases:
- \( 8 = 2^3 \)
- \( 4 = 2^2 \)
- \( 32 = 2^5 \)
- \( 81 = 3^4 \)
- \( 27 = 3^3 \)
- \( 3 = 3^1 \)

### Step 2: Substitute the rewritten values into the equations
2. Substitute these values into the equations:
- For the first equation:
\[ (2^3)^x \cdot (2^2)^y = 2^5 \]
This simplifies to:
\[ 2^{3x} \cdot 2^{2y} = 2^5 \]
Combining the exponents gives:
\[ 2^{3x + 2y} = 2^5 \]
Therefore, we have:
\[ 3x + 2y = 5 \quad \text{(Equation 1)} \]

- For the second equation:
\[ (3^4)^x \cdot (3^3)^y = 3^1 \]
This simplifies to:
\[ 3^{4x} \cdot 3^{3y} = 3^1 \]
Combining the exponents gives:
\[ 3^{4x + 3y} = 3^1 \]
Therefore, we have:
\[ 4x + 3y = 1 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
3. We now have a system of linear equations:
- Equation 1: \( 3x + 2y = 5 \)
- Equation 2: \( 4x + 3y = 1 \)

4. To eliminate one variable, we can multiply Equation 1 by 4 and Equation 2 by 3:
- Multiply Equation 1 by 4:
\[ 12x + 8y = 20 \quad \text{(Equation 3)} \]
- Multiply Equation 2 by 3:
\[ 12x + 9y = 3 \quad \text{(Equation 4)} \]

5. Now, subtract Equation 4 from Equation 3:
\[ (12x + 8y) - (12x + 9y) = 20 - 3 \]
This simplifies to:
\[ -y = 17 \]
Therefore:
\[ y = -17 \]

### Step 4: Substitute back to find x
6. Substitute \( y = -17 \) back into Equation 1:
\[ 3x + 2(-17) = 5 \]
Simplifying gives:
\[ 3x - 34 = 5 \]
Adding 34 to both sides:
\[ 3x = 39 \]
Dividing by 3:
\[ x = 13 \]

### Final Result
Thus, the values of \( x \) and \( y \) are:
\[ x = 13, \quad y = -17 \]