Question

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