ABC26GN2067 · Logarithms

$\frac{1}{2}(\log x+\log y)=\log \left(\frac{x+y}{2}\right) \Rightarrow \frac{1}{2} \log (x y)=\log \left(\frac{x+y}{2}\right)$ $$\begin{aligned} \Rightarrow & \log (x y)^{\frac{1}{2}}=\log \left(\frac{x+y}{2}\right) \Rightarrow(x y)^{\frac{1}{2}}=\left(\frac{x+y}{2}\right) \Rightarrow x y=\left(\frac{x+y}{2}\right)^{2} \\ & \Rightarrow 4 x y=x^{2}+y^{2}+2 x y \\ & \Rightarrow x^{2}+y^{2}-2 x y=0 \\ & \Rightarrow(x-y)^{2}=0 \Rightarrow x-y=0 \Rightarrow x=y . \end{aligned}$$