If {x^a}{y^b} = {e^m},{x^c}{y^d} = {e^n},{Del | vedclass

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(D) Given the equations: ${x^a}{y^b} = {e^m}$ and ${x^c}{y^d} = {e^n}$ Taking the natural logarithm on both sides of both equations: $a \ln x + b \ln

Vedclass · Accepted Answer

${e^{{\Delta _1}/{\Delta _3}}}$ and ${e^{{\Delta _2}/{\Delta _3}}}$

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(D) Given the equations: ${x^a}{y^b} = {e^m}$ and ${x^c}{y^d} = {e^n}$ Taking the natural logarithm on both sides of both equations: $a \ln x + b \ln y = m$ $c \ln x + d \ln y = n$ This is a system of linear equations in terms of $\ln x$ and $\ln y$. Using Cramer's rule,we define the determinants: ${\Delta _3} = \left| {\begin{array}{*{20}{c}} a & b \ c & d \end{array}} ight|$ ${\Delta _1} = \left| {\begin{array}{*{20}{c}} m & b \ n & d \end{array}} ight|$ ${\Delta _2} = \left| {\begin{array}{*{20}{c}} a & m \ c & n \end{array}} ight|$ According to Cramer's rule: $\ln x = \frac{{{\Delta _1}}}{{{\Delta _3}}}$ and $\ln y = \frac{{{\Delta _2}}}{{{\Delta _3}}}$ Therefore,solving for $x$ and $y$: $x = {e^{{\Delta _1}/{\Delta _3}}}$ and $y = {e^{{\Delta _2}/{\Delta _3}}}$ Thus,the correct option is $(d)$.

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