Question

$\log \left(\frac{x+y}{3}\right)=\frac{1}{2}(\log x+\log y)$ then find the valued $\frac{x^2}{y^2}+\frac{y^2}{x^2}$

Answer 1

$\log \left(\frac{x+y}{3}\right)=\frac{1}{2}(\log x+\log y)$

$\Rightarrow \log \left(\frac{x+y}{3}\right)=\frac{1}{2}\log xy$

$\Rightarrow \log \left(\frac{x+y}{3}\right)=\log \sqrt{xy}$

$\Rightarrow \frac{x+y}{3}=\sqrt{xy}$

$\Rightarrow {(x+y)}^2=9xy$

$\Rightarrow x^2+2xy+y^2=9xy$

$\Rightarrow x^2+y^2=9xy-2xy=7xy$

$\Rightarrow \frac{x^2+y^2}{xy}=7$

$\Rightarrow \frac{x}{y}+\frac{y}{x}=7$

$\Rightarrow {(\frac{x}{y}+\frac{y}{x})}^2=49$

$\Rightarrow \frac{x^2}{y^2}+\frac{y^2}{x^2}+2\frac{x}{y}\frac{y}{x}=49$

$\therefore \frac{x^2}{y^2}+\frac{y^2}{x^2}=49-2=47$

$\therefore \frac{x^2}{y^2}+\frac{y^2}{x^2}=47$