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 given problem step-by-step, we will use logarithmic properties and determinants. Let's break down the solution:

### Step 1: Set Up the Equations
We start with the equations given in the problem:

1. \( x^{2a} y^{3b} = e^{5m} \)
2. \( x^{3c} y^{4d} = e^{2n} \)

### Step 2: Take Logarithms
Taking the natural logarithm on both sides of each equation:

1. From \( x^{2a} y^{3b} = e^{5m} \):
\[ 2a \ln x + 3b \ln y = 5m \]

2. From \( x^{3c} y^{4d} = e^{2n} \):
\[ 3c \ln x + 4d \ln y = 2n \]

### Step 3: Write in Matrix Form
We can express these equations in matrix form:

\[ \begin{pmatrix} 2a & 3b \\ 3c & 4d \end{pmatrix} \begin{pmatrix} \ln x \\ \ln y \end{pmatrix} = \begin{pmatrix} 5m \\ 2n \end{pmatrix} \]

### Step 4: Apply Cramer’s Rule
Using Cramer's rule, we can find \( \ln x \) and \( \ln y \):

1. **For \( \ln x \)**:
\[ \ln x = \frac{\Delta_1}{\Delta} \]
where \( \Delta_1 \) is the determinant of the matrix formed by replacing the first column with the constants on the right side:
\[ \Delta_1 = \begin{vmatrix} 5m & 3b \\ 2n & 4d \end{vmatrix} = (5m)(4d) - (3b)(2n) = 20md - 6bn \]

And \( \Delta \) is the determinant of the original coefficients:
\[ \Delta = \begin{vmatrix} 2a & 3b \\ 3c & 4d \end{vmatrix} = (2a)(4d) - (3b)(3c) = 8ad - 9bc \]

Thus,
\[ \ln x = \frac{20md - 6bn}{8ad - 9bc} \]

2. **For \( \ln y \)**:
\[ \ln y = \frac{\Delta_2}{\Delta} \]
where \( \Delta_2 \) is the determinant formed by replacing the second column:
\[ \Delta_2 = \begin{vmatrix} 2a & 5m \\ 3c & 2n \end{vmatrix} = (2a)(2n) - (5m)(3c) = 4an - 15mc \]

Thus,
\[ \ln y = \frac{4an - 15mc}{8ad - 9bc} \]

### Step 5: Exponentiate to Find \( x \) and \( y \)
Now we can find \( x \) and \( y \) by exponentiating the logarithmic results:

1. \( x = e^{\ln x} = e^{\frac{20md - 6bn}{8ad - 9bc}} \)
2. \( y = e^{\ln y} = e^{\frac{4an - 15mc}{8ad - 9bc}} \)

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