Question

If $log\left(\frac{x+y}{3}\right)=\frac{1}{2}(logx+logy)$, then find the value of $\frac{x}{y}+\frac{y}{x}$.

Answer 1

We have,

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

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

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

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

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

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

On squaring both side and we get,

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

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

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

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

On divide both side by $xy$ and we get,

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

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

Hence, this is the answer.