Question

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

Answer 1

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

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

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

$logx+logy=log\left(xy\right)$

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

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

$x^2+y^2+2xy-36xy=0$

$x^2+y^2-34xy=0$

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

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

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