Question

The solution set of the system of equations ${\log }_{12}x\left(\begin{array}{c}\frac{1}{{\log }_{x}2}+{\log }_{2}y\end{array}\right)={\log }_{2}x$ and ${\log }_{2}x.\left({\log }_3\right(x+y\left)\right)=3{\log }_{3}x$ is

• A. $x=6;y=2$
• B. $x=4;y=3$
• C. $x=2;y=6$
• D. $x=3;y=4$

Answer 1

Answer: (A), (C)

The correct options are
A $x=6;y=2$
C $x=2;y=6$

Also, ${\log }_{b}a=\frac{\log a}{\log b},{\log }_{a}b=\frac{1}{{\log }_{a}b}$
Given, ${\log }_{12}x\times (\frac{1}{{\log }_{x}2}+{\log }_{2}y)={\log }_{2}x$---->A
${\log }_{2}x\times {\log }_2(x+y)=3\times {\log }_{3}x$---->B
Consider equation A,
$\Rightarrow {\log }_{12}x\times (\frac{1}{{\log }_{x}2}+{\log }_{2}y)={\log }_{2}x$
$\Rightarrow \left(\frac{1}{{\log }_{2}x+{\log }_{2}y}\right)={\log }_{12}2$
$\Rightarrow ({\log }_{2}x+{\log }_{2}y)={\log }_{12}2$
$\Rightarrow {\log }_{2}xy={\log }_{2}12$
$\Rightarrow xy=12$----->C
Consider equation B,
${\log }_{2}x\times {\log }_3(x+y)=3\times {\log }_{3}x$
$\Rightarrow {\log }_3(x+y)=3\times {\log }_{3}2$
$\Rightarrow {\log }_3(x+y)={\log }_{3}8$
$\Rightarrow x+y=8$--->D
On solving equation C and D,
$\Rightarrow x=6,y=2$ or $x=2,y=6$
Options A,C are True.