Question

If $log\left(\frac{x+y}{6}\right)=\frac{1}{2}(logx+logy)$ show that $\frac{x}{y}+\frac{y}{x}=34$

Answer 1

Given that,

$\log \left(\frac{x+y}{6}\right)=\frac{1}{2}(\log x+\log y)$

$\Rightarrow 2\log \left(\frac{x+y}{6}\right)=(\log x+\log y)$

$\Rightarrow \log {\left(\frac{x+y}{6}\right)}^2=(\log x+\log y)$

$\Rightarrow \log {\left(\frac{x+y}{6}\right)}^2=\log xy$

$\Rightarrow {\left(\frac{x+y}{6}\right)}^2=xy$

$\Rightarrow {(x+y)}^2=36xy$

$\Rightarrow x^2+y^2+2xy=36xy$

$\Rightarrow x^2+y^2=34xy$

$\Rightarrow \frac{x^2+y^2}{xy}=34$

$\Rightarrow \frac{x^2}{xy}+\frac{y^2}{xy}=34$

$\Rightarrow \frac{x}{y}+\frac{y}{x}=34$

Hence, proved.