Question

If $x^2+y^2=14 x y$ and $2 \log (k(x+y))=(\log x+\log y)$, then the value of $k$ is

• A. $\frac{1}{16}$
• B. $\frac{1}{4}$
• C. $2 \log 2$
• D. $\frac{\log 14}{2}$

Answer 1

Answer: (B)

Given $ 2 \log (k(x+y))=(\log x+\log y) $
$\Rightarrow (k(x+y))^2=x y $
$\therefore k^2\left(x^2+y^2+2 x y\right)=x y$
$\Rightarrow k^2(14 x y+2 x y)=x y$
$\therefore k^2=\frac{1}{16} \Rightarrow k=\frac{1}{4} $