Answer: Option C
$$\eqalign{
& \frac{1}{2}\left( {\log x + \log y} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr
& \Rightarrow \frac{1}{2}\log \left( {xy} \right) = \log \left( {\frac{{x + y}}{2}} \right) \cr
& \Rightarrow \log {\left( {xy} \right)^{\frac{1}{2}}} = \log \left( {\frac{{x + y}}{2}} \right) \cr
& \Rightarrow {\left( {xy} \right)^{\frac{1}{2}}} = \left( {\frac{{x + y}}{2}} \right) \cr
& \Rightarrow xy = {\left( {\frac{{x + y}}{2}} \right)^2} \cr
& \Rightarrow 4xy = {x^2} + {y^2} + 2xy \cr
& \Rightarrow {x^2} + {y^2} - 2xy = 0 \cr
& \Rightarrow {\left( {x - y} \right)^2} = 0 \cr
& \Rightarrow x - y = 0 \cr
& \Rightarrow x = y\, \cr} $$