Question

If $x^a{y}^b=e^m,x^c{y}^d=e^n,{△}_1=\left|\begin{array}{cc}m& b\\ n& d\end{array}\right|,{△}_2=\left|\begin{array}{cc}a& m\\ c& n\end{array}\right|$ and ${△}_3=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$ the value of $x$ and $y$ are respectively

• A. $\frac{{△}_1}{{△}_3}$ and $\frac{{△}_2}{{△}_3}$
• B. $\frac{{△}_2}{{△}_1}$ and $\frac{{△}_3}{{△}_1}$
• C. $\log \left(\frac{{△}_1}{{△}_3}\right)and\log \left(\frac{{△}_2}{{△}_3}\right)$
• D. $e^{{△}_1/{△}_3}and{e}^{{△}_2/{△}_3}$

Answer 1

Answer: (D)

$e^{{△}_1/{△}_3}and{e}^{{△}_2/{△}_3}$

$x^a{y}^b=e^m$

Taking ${\log }_e$ on both sides,

${\log }_e\left(x^a{y}^b\right)={\log }_e{e}^m$

$a{\log }_{e}x+b{\log }_{e}y=m\to 1$

Similarly,

$c{\log }_{e}x+d{\log }_{e}y=n\to 2$

Solving $1$ and $2$ by Cramer's rule,

${\log }_{e}x=\frac{\left|\begin{array}{cc}m& b\\ n& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}=\frac{{△}_1}{{△}_3}$

${\log }_{e}y=\frac{\left|\begin{array}{cc}a& m\\ c& n\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}=\frac{{△}_2}{{△}_3}$

$\therefore x=e^{\left({△}_1/{△}_3\right)},y=e^{\left({△}_2/{△}_3\right)}$

Answer: D