If `x^(2a)y^(3b)=e^(5m), x^(3c)y^(4d)=e^(2n), Delta_(1)=|(5m, 3b),(2n, 4d)|, Delta_(2)=|(2a, 5m),(3c, 2n)| and Delta_(3)=|(2a, 3b),(3c, 4d)|`, then the values of x and y are

Options

A. `(Delta_(1))/(Delta_(3)), (Delta_(2))/(Delta_(3))`
B. `(Delta_(2))/(Delta_(1)),(Delta_(3))/(Delta_(1))`
C. `log((Delta_(1))/(Delta_(3))), log((Delta_(2))/(Delta_(3)))`
D. `e^((Delta_(1))/(Delta_(3))`,`e^((Delta_(2))/(Delta_(3)))`

Answer

To solve the problem step by step, we will analyze the given equations and determinants.

### Step 1: Write the equations in logarithmic form
We start with the equations:
1. \( x^{2a} y^{3b} = e^{5m} \)
2. \( x^{3c} y^{4d} = e^{2n} \)

Taking the natural logarithm (log base e) of both sides, we have:
1. \( 2a \log x + 3b \log y = 5m \) (Equation 1)
2. \( 3c \log x + 4d \log y = 2n \) (Equation 2)

### Step 2: Set up the determinants
We define the determinants as follows:
- \( \Delta_1 = \begin{vmatrix} 5m & 3b \\ 2n & 4d \end{vmatrix} \)
- \( \Delta_2 = \begin{vmatrix} 2a & 5m \\ 3c & 2n \end{vmatrix} \)
- \( \Delta_3 = \begin{vmatrix} 2a & 3b \\ 3c & 4d \end{vmatrix} \)

### Step 3: Calculate the determinants
1. **Calculate \( \Delta_1 \)**:
\[ \Delta_1 = (5m)(4d) - (3b)(2n) = 20md - 6bn \]

2. **Calculate \( \Delta_2 \)**:
\[ \Delta_2 = (2a)(2n) - (5m)(3c) = 4an - 15mc \]

3. **Calculate \( \Delta_3 \)**:
\[ \Delta_3 = (2a)(4d) - (3b)(3c) = 8ad - 9bc \]

### Step 4: Express \( \log x \) and \( \log y \) in terms of the determinants
Using Cramer's rule, we can express \( \log x \) and \( \log y \) as:
- \( \log x = \frac{\Delta_1}{\Delta_3} \)
- \( \log y = \frac{\Delta_2}{\Delta_3} \)

### Step 5: Solve for \( x \) and \( y \)
Now, we can find \( x \) and \( y \) by exponentiating:
1. \( x = e^{\log x} = e^{\frac{\Delta_1}{\Delta_3}} \)
2. \( y = e^{\log y} = e^{\frac{\Delta_2}{\Delta_3}} \)

### Final Step: Substitute the determinants back into the equations
Substituting the values of \( \Delta_1 \), \( \Delta_2 \), and \( \Delta_3 \):
1. \( x = e^{\frac{20md - 6bn}{8ad - 9bc}} \)
2. \( y = e^{\frac{4an - 15mc}{8ad - 9bc}} \)

### Conclusion
Thus, the values of \( x \) and \( y \) are:
- \( x = e^{\frac{20md - 6bn}{8ad - 9bc}} \)
- \( y = e^{\frac{4an - 15mc}{8ad - 9bc}} \)