Question

If ${\log }_{y}x+{\log }_{x}y=2,x^2+y=12$, then value of $xy$ is

• A. $9$
• B. $12$
• C. $15$
• D. $21$

Answer 1

The correct option is A $9$

${\log }_{y}x+{\log }_{x}y=2$
And ${\log }_{y}x=\frac{1}{{\log }_{x}y}$
So ${\log }_{y}x+\frac{1}{{\log }_{y}x}=2$
Solving, we get ${\log }_{y}x=1$
Hence, $x=y$
Now, $x^2+y=12$
$⟹y^2+y-12=0$
$⟹(y+4)(y-3)=0$
$⟹y=-4$ or $3$
But, $y>0$, hence $x=y=3$
$xy=9$