If logx+y3=12(logx+logy), then xy+yx=

log(x+y3)=12(logx+logy) x+y3=xy (x+y)2=9xy x2+y2+2xy=9xy x2+y2=7xy xy+yx=7

If logx+y3=12(logx+logy), then xy+yx=

log(x+y3)=12(logx+logy) x+y3=xy (x+y)2=9xy x2+y2+2xy=9xy x2+y2=7xy xy+yx=7

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If $\log \frac{x + y}{3} = \frac{1}{2} (\log x + \log y)$ , then $\frac{x}{y} + \frac{y}{x} =$

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$\log (\frac{x + y}{3}) = \frac{1}{2} (\log x + \log y) \frac{x + y}{3} = \sqrt{x y} {(x + y)}^{2} = 9 x y x^{2} + y^{2} + 2 x y = 9 x y x^{2} + y^{2} = 7 x y \frac{x}{y} + \frac{y}{x} = 7$

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If $\log \frac{x + y}{3} = \frac{1}{2} (\log x + \log y)$ , then $\frac{x}{y} + \frac{y}{x} =$