The value of (x,y) if `5^x+3^y=8, 5^(x-1)+3^(y-1) = 2` is :

Options

A. 2, 3
B. 3, 2
C. 1, 1
D. 0, 1

Answer

To solve the equations \(5^x + 3^y = 8\) and \(5^{x-1} + 3^{y-1} = 2\), we will follow these steps:

### Step 1: Rewrite the second equation
The second equation can be rewritten in terms of the first equation. We know that:
\[ 5^{x-1} = \frac{5^x}{5} \quad \text{and} \quad 3^{y-1} = \frac{3^y}{3} \]
Thus, the second equation becomes:
\[ \frac{5^x}{5} + \frac{3^y}{3} = 2 \]
Multiplying through by 15 (the least common multiple of 5 and 3) gives:
\[ 3 \cdot 5^x + 5 \cdot 3^y = 30 \]

### Step 2: Set up a system of equations
Now we have a system of equations:
1. \(5^x + 3^y = 8\) (Equation 1)
2. \(3 \cdot 5^x + 5 \cdot 3^y = 30\) (Equation 2)

### Step 3: Solve for one variable
From Equation 1, we can express \(3^y\) in terms of \(5^x\):
\[ 3^y = 8 - 5^x \]
Substituting this into Equation 2 gives:
\[ 3 \cdot 5^x + 5(8 - 5^x) = 30 \]
Expanding this results in:
\[ 3 \cdot 5^x + 40 - 5 \cdot 5^x = 30 \]
Combining like terms:
\[ -2 \cdot 5^x + 40 = 30 \]
Subtracting 40 from both sides:
\[ -2 \cdot 5^x = -10 \]
Dividing by -2:
\[ 5^x = 5 \]
Taking logarithm base 5:
\[ x = 1 \]

### Step 4: Substitute back to find \(y\)
Now substitute \(x = 1\) back into Equation 1 to find \(y\):
\[ 5^1 + 3^y = 8 \]
This simplifies to:
\[ 5 + 3^y = 8 \]
Subtracting 5 from both sides:
\[ 3^y = 3 \]
Taking logarithm base 3:
\[ y = 1 \]

### Final Answer
Thus, the values of \(x\) and \(y\) are:
\[ (x, y) = (1, 1) \]