Question

If $\frac{1}{2}\log x+\frac{1}{2}\log y+\log 2=\log (x+y)$, then relation between $x$ and $y$ is

• A. $x+y=0$
• B. $x=y$
• C. $x=2$ and $y=0$
• D. $x=\log y$

Answer 1

Answer: (B)

$x=y$

$\frac{1}{2}\log x+\frac{1}{2}\log y+\log 2=\log (x+y)$

$\Rightarrow \log (x^{\frac{1}{2}}\times y^{\frac{1}{2}}\times 2)=\log (x+y)$

$\Rightarrow 2\sqrt{xy}=(x+y)$
$\Rightarrow 4xy=(x+y{)}^2$
$\Rightarrow (x+y{)}^2-4xy=0$
$\Rightarrow (x-y{)}^2=0$
$\Rightarrow x-y=0$
$\Rightarrow x=y$