Question

If $\frac{1}{2}\log x+\frac{1}{2}\log y+\log 2=\log (x+y)$, then:

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

Answer 1

The correct option is B $x=y$

$\frac{1}{2}logx+\frac{1}{2}logy+log2=log(x+y)$

$\Rightarrow log\sqrt{x}+log\sqrt{y}+log2=log(x+y)$

$\Rightarrow log\left(2\sqrt{x}\sqrt{y}\right)=log(x+y)$

Take antilog

$\Rightarrow x+y=2\sqrt{x}\sqrt{y}=0$

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