Question

If \( x^a y^b = e^m, \)

and

\[ x^c y^d = e^n, \]

and

\[ \Delta_1 = \begin{vmatrix} m & b \\ n & d \\ \end{vmatrix}, \quad \Delta_2 = \begin{vmatrix} a & m \\ c & n \\ \end{vmatrix}, \quad \Delta_3 = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix} \]

Then the values of \( x \) and \( y \) respectively (where \( e \) is the base of the natural logarithm) are:

• A. \( \frac{\Delta_1}{\Delta_3} \) and \( \frac{\Delta_2}{\Delta_3} \)
• B. \( \frac{\Delta_2}{\Delta_1} \) and \( \frac{\Delta_3}{\Delta_1} \)
• C. \( \log\left( \frac{\Delta_1}{\Delta_3} \right) \) and \( \log\left( \frac{\Delta_2}{\Delta_3} \right) \)
• D. \( \frac{\Delta_1}{e^{\Delta_3}} \) and \( \frac{\Delta_2}{e^{\Delta_3}} \)

Hint:

When solving exponential systems with variables in powers, take natural logarithms and apply linear system methods like Cramer's Rule.

Answer 1

The Correct Option is D

Take natural log on both equations: \[ \ln(x^a y^b) = \ln(e^m) \Rightarrow a \ln x + b \ln y = m
\text{(1)} \] \[ \ln(x^c y^d) = \ln(e^n) \Rightarrow c \ln x + d \ln y = n
\text{(2)} \]
Solve the system of equations: \[ \begin{aligned} a \ln x + b \ln y &= m \\ c \ln x + d \ln y &= n \end{aligned} \] Let \( \ln x = u, \ln y = v \). Then it becomes: \[ \begin{aligned} a u + b v &= m \\ c u + d v &= n \end{aligned} \] Use Cramer's Rule to solve: \[ \Delta_3 = \begin{vmatrix} a & b \\ c & d \\ \end{vmatrix},
\Delta_1 = \begin{vmatrix} m & b \\ n & d \\ \end{vmatrix},
\Delta_2 = \begin{vmatrix} a & m \\ c & n \\ \end{vmatrix} \] Then: \[ u = \ln x = \frac{\Delta_1}{\Delta_3},
v = \ln y = \frac{\Delta_2}{\Delta_3} \] So: \[ x = e^{\frac{\Delta_1}{\Delta_3}} = \frac{\Delta_1}{e^{\Delta_3}},
y = e^{\frac{\Delta_2}{\Delta_3}} = \frac{\Delta_2}{e^{\Delta_3}} \] \[ \boxed{x = \frac{\Delta_1}{e^{\Delta_3}},
y = \frac{\Delta_2}{e^{\Delta_3}}} \] % Tip