Question

The solution set of the system of equations ${\log }_{3}x+{\log }_{3}y=2+{\log }_{3}2and{\log }_{27}(x+y)=\frac{2}{3}$ is :

• A. {6,3}
• B. {3,6}
• C. {6,12}
• D. {12,6}

Answer 1

Answer: (A)

{6,3}

Given,

${\log }_3\left(x\right)+{\log }_3\left(y\right)=2+{\log }_3\left(2\right)$

${\log }_3\left(x\right)=2+{\log }_3\left(2\right)-{\log }_3\left(y\right)$

$x=3^{2+{\log }_3\left(2\right)-{\log }_3\left(y\right)}$

$=3^{{\log }_3\left(2\right)}\cdot 3^{-{\log }_3\left(y\right)}\cdot 3^2$

$=2\cdot 3^{-{\log }_3\left(y\right)}\cdot 3^2$

$=2y^{-1}\cdot 3^2$

$\Rightarrow x=\frac{18}{y}$.......(1)

Now,

${\log }_{27}(x+y)=\frac{2}{3}$

from (1)

${\log }_{27}(\frac{18}{y}+y)=\frac{2}{3}$

$\frac{18}{y}+y={27}^{\frac{2}{3}}$

$\Rightarrow 18+y^2=9y$

$y^2-9y+18=0$

$(y-3)(y-6)=0$

$\therefore y=3,6$

$x=\frac{18}{y}=\frac{18}{3}=6$

$x=\frac{18}{y}=\frac{18}{6}=3$

$(x,y)=\{6,3\}or\{3,6\}$