Question

If $a{e}^x+b{e}^y=c;p{e}^x+q{e}^y=d$ and ${\Delta }_1=\left|\begin{array}{cc}a& b\\ p& q\end{array}\right|;{\Delta }_2=\left|\begin{array}{cc}c& d\\ b& q\end{array}\right|;{\Delta }_3=\left|\begin{array}{cc}a& c\\ p& d\end{array}\right|$ then the value of $(x,y)$ is

• A. $(\frac{{\Delta }_2}{{\Delta }_1},\frac{{\Delta }_3}{{\Delta }_1})$
• B. $(\log \frac{{\Delta }_2}{{\Delta }_1},\log \frac{{\Delta }_3}{{\Delta }_1})$
• C. $(\log \frac{{\Delta }_1}{{\Delta }_3},\log \frac{{\Delta }_1}{{\Delta }_2})$
• D. $(\log \frac{{\Delta }_1}{{\Delta }_2},\log \frac{{\Delta }_1}{{\Delta }_3})$

Answer 1

The correct option is B $(\log \frac{{\Delta }_2}{{\Delta }_1},\log \frac{{\Delta }_3}{{\Delta }_1})$

Given equations are $a{e}^x+b{e}^y=c$ and $p{e}^x+q{e}^y=d$

By determinant method , $e^x=\frac{{\Delta }_2}{{\Delta }_1}$ , which implies $x=\log \frac{{\Delta }_2}{{\Delta }_1}$

By determinant method , $e^y=\frac{{\Delta }_3}{{\Delta }_1}$ , which implies $y=\log \frac{{\Delta }_3}{{\Delta }_1}$

Therefore the correct option is $B$