MCQ
If ${x^a}{y^b} = {e^m},{x^c}{y^d} = {e^n},{\Delta _1} = \left| {\,\begin{array}{*{20}{c}}m&b\\n&d\end{array}\,} \right|\,\,{\Delta _2} = \left| {\,\begin{array}{*{20}{c}}a&m\\c&n\end{array}\,} \right|$ and ${\Delta _3} = \left| {\,\begin{array}{*{20}{c}}a&b\\c&d\end{array}\,} \right|$, then the values of $x$  and $y$  are respectively
  • A
    ${\Delta _1}/{\Delta _3}$ and ${\Delta _2}/{\Delta _3}$
  • B
    ${\Delta _2}/{\Delta _1}$ and ${\Delta _3}/{\Delta _1}$
  • C
    $log$ $({\Delta _1}/{\Delta _3})$ and $log$ $({\Delta _2}/{\Delta _3})$
  • ${e^{{\Delta _1}/{\Delta _3}}}$ and ${e^{{\Delta _2}/{\Delta _3}}}$

Answer

Correct option: D.
${e^{{\Delta _1}/{\Delta _3}}}$ and ${e^{{\Delta _2}/{\Delta _3}}}$
d
(d) Given ${x^a}$${y^b} = {e^m},\,{x^c}{y^d} = {e^n}$

$ \Rightarrow $ $a\log x + b\log y = m$ and $c\log x + d\log y = n$

By Cramer’s rule, $\log x = \frac{{{\Delta _1}}}{{{\Delta _3}}}$ and $\log y = \frac{{{\Delta _2}}}{{{\Delta _3}}}$

==> $x = {e^{{\Delta _1}/{\Delta _3}}}$ and $y = {e^{{\Delta _2}/{\Delta _3}}}$.

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