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Question Number 174291 by akolade last updated on 29/Jul/22
Find  (x/y)+(y/x)  if log(((x+y)/3))=((logx+logy)/2)
$$\mathrm{Find} \\ $$$$\frac{\mathrm{x}}{\mathrm{y}}+\frac{\mathrm{y}}{\mathrm{x}} \\ $$$$\mathrm{if}\:\mathrm{log}\left(\frac{\mathrm{x}+\mathrm{y}}{\mathrm{3}}\right)=\frac{\mathrm{logx}+\mathrm{logy}}{\mathrm{2}} \\ $$
Answered by Rasheed.Sindhi last updated on 29/Jul/22
log(((x+y)/3))=((logx+logy)/2) ; (x/y)+(y/x)=?  log(((x+y)/3))=((logxy)/2)=log((xy)^(1/2) )  ((x+y)/3)=(√(xy))  (((x+y)/3))^2 =xy  x^2 +2xy+y^2 =9xy  x^2 +y^2 =7xy  ((x^2 +y^2 )/(xy))=7  (x/y)+(y/x)=7
$$\mathrm{log}\left(\frac{\mathrm{x}+\mathrm{y}}{\mathrm{3}}\right)=\frac{\mathrm{logx}+\mathrm{logy}}{\mathrm{2}}\:;\:\frac{\mathrm{x}}{\mathrm{y}}+\frac{\mathrm{y}}{\mathrm{x}}=? \\ $$$$\mathrm{log}\left(\frac{\mathrm{x}+\mathrm{y}}{\mathrm{3}}\right)=\frac{\mathrm{logxy}}{\mathrm{2}}=\mathrm{log}\left(\left(\mathrm{xy}\right)^{\mathrm{1}/\mathrm{2}} \right) \\ $$$$\frac{\mathrm{x}+\mathrm{y}}{\mathrm{3}}=\sqrt{\mathrm{xy}} \\ $$$$\left(\frac{\mathrm{x}+\mathrm{y}}{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{xy} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} =\mathrm{9xy} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{7}{xy} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{xy}}=\mathrm{7} \\ $$$$\frac{{x}}{{y}}+\frac{{y}}{{x}}=\mathrm{7} \\ $$

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