Correct option is (C) 7
Given that \(log\left(\frac{x+y}{3}\right)=\frac{1}{2} (log\,x+log\,y)\)
\(\Rightarrow\) \(2\,log\,(\frac{x+y}3)\) = log x + log y
\(\Rightarrow\) \(log\,(\frac{x+y}3)^2\) = log xy \((\because\) log A + log B = log AB and n log a \(=log\,a^n)\)
\(\Rightarrow\) \((\frac{x+y}3)^2\) = xy (By comparing)
\(\Rightarrow\) \(x^2+y^2+2xy\) = 9xy
\(\Rightarrow\) \(\frac{x^2}{xy}+\frac{y^2}{xy}+\frac{2xy}{xy}=\frac{9xy}{xy}\) (Dividing both sides by xy)
\(\Rightarrow\) \(\frac xy+\frac yx+2=9\)
\(\Rightarrow\) \(\frac{x}{y}+\frac{y}{x}\) = 9 - 2 = 7
