Question

If $x$ and $y$ hold good with the equations
${\log }_{10}(x-2)+{\log }_{10}y=0$ and $\sqrt{x}+\sqrt{y-2}=\sqrt{x+y},$ then which of the following option is correct?

• A. $x>y$
• B. ${\log }_{y}x=2+{\log }_{x}y$
• C. ${\log }_{y}x+{\log }_{x}y=2$
• D. $y>x$

Answer 1

The correct option is C ${\log }_{y}x+{\log }_{x}y=2$

${\log }_{10}(x-2)+{\log }_{10}y=0$
Where $x>2$ and $y>0$
$\Rightarrow {\log }_{10}y(x-2)=0\Rightarrow y(x-2)=1\cdots \left(1\right)$

Now,
$\sqrt{x}+\sqrt{y-2}=\sqrt{x+y}$
Squaring on both sides,
$x+y-2+2\sqrt{x(y-2)}=x+y\Rightarrow x(y-2)=1\cdots \left(2\right)$
Where $y>2$

From $\left(1\right)\&\left(2\right)$,
$y(x-2)=x(y-2)\Rightarrow xy-2y=xy-2x\Rightarrow x=y$

Hence, ${\log }_{y}x+{\log }_{x}y=2$