Question

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

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

Answer 1

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

Given equations are $a{e}^x+b{e}^y=candp{e}^x+q{e}^y=d$
By Cramer's Rule,
$e^x=\frac{{\Delta }_2}{{\Delta }_1}\Rightarrow x=\ln \frac{{\Delta }_2}{{\Delta }_1}$
Similarly,
$e^y=\frac{{\Delta }_3}{{\Delta }_1}\Rightarrow y=\ln \frac{{\Delta }_3}{{\Delta }_1}$