Question

If ${\log }_{y}x+{\log }_{x}y=1,x^2+y=12$, then the value of xy is:

• A. 9
• B. 12
• C. 15
• D. 21

Answer 1

The correct option is A 9

Let $t={\log }_{y}x(x,y,>0,$ and $\ne 1)$, then $t+\frac{1}{t}=2$ or ${(t-1)}^2=0$.

$\therefore t={\log }_{y}x=1$, i.e.,$x=y$. We get $x^2+x-12=0$ or x=-4,3

x=3 only(-4 rejected)

$y=3$

$\therefore xy=9$