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Question Number 85826 by jagoll last updated on 25/Mar/20

^x log (xy).^y log (xy) +^x log (x−y).^y log (x−y)=0 find x+y

$$\:^{\mathrm{x}} \mathrm{log}\:\left(\mathrm{xy}\right).\:^{\mathrm{y}} \mathrm{log}\:\left(\mathrm{xy}\right)\:+\:^{\mathrm{x}} \mathrm{log}\:\left(\mathrm{x}−\mathrm{y}\right).^{\mathrm{y}} \mathrm{log}\:\left(\mathrm{x}−\mathrm{y}\right)=\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{x}+\mathrm{y}\: \\ $$

Commented by john santu last updated on 25/Mar/20

^x log (xy).^y log (xy) = −^x log (x−y).^y log (x−y) ^((x−y) ) log (xy) = −^((xy)) log (x−y) ^((x−y)) log (xy) = ^(1/(xy)) log (x−y) ⇒ { ((x−y = (1/(xy)))),((xy = x−y )) :} ⇒xy = 1 ∧ x−y = 1 y = (1/x) ⇒ x−(1/x) = 1 x^2 −x−1 = 0 ⇒ x = ((1+ (√5))/2) ∧ y = (2/((√5) +1)) = ((2((√5)−1))/4) y = (((√5)−1)/2) . so x+y = (√5)

$$\:^{{x}} \mathrm{log}\:\left({xy}\right).\:^{{y}} \mathrm{log}\:\left({xy}\right)\:=\:−\:^{{x}} \mathrm{log}\:\left({x}−{y}\right).^{{y}} \mathrm{log}\:\left({x}−{y}\right) \\ $$$$\:^{\left({x}−{y}\right)\:} \:\mathrm{log}\:\left({xy}\right)\:=\:−\:^{\left({xy}\right)} \mathrm{log}\:\left({x}−{y}\right) \\ $$$$\:^{\left({x}−{y}\right)} \mathrm{log}\:\left({xy}\right)\:=\:\:^{\frac{\mathrm{1}}{{xy}}} \:\mathrm{log}\:\left({x}−{y}\right) \\ $$$$\Rightarrow\begin{cases}{{x}−{y}\:=\:\frac{\mathrm{1}}{{xy}}}\\{{xy}\:=\:{x}−{y}\:}\end{cases} \\ $$$$\Rightarrow{xy}\:=\:\mathrm{1}\:\wedge\:{x}−{y}\:=\:\mathrm{1} \\ $$$${y}\:=\:\frac{\mathrm{1}}{{x}}\:\Rightarrow\:{x}−\frac{\mathrm{1}}{{x}}\:=\:\mathrm{1} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}\:=\:\mathrm{0}\:\Rightarrow\:{x}\:=\:\frac{\mathrm{1}+\:\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\wedge\:{y}\:=\:\frac{\mathrm{2}}{\sqrt{\mathrm{5}}\:+\mathrm{1}}\:=\:\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)}{\mathrm{4}}\: \\ $$$${y}\:=\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:.\:{so}\:{x}+{y}\:=\:\sqrt{\mathrm{5}}\: \\ $$

Commented by jagoll last updated on 25/Mar/20

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

Answered by $@ty@m123 last updated on 07/Apr/20

((log xy.log xy)/(log x.log y))+((log (x−y).log (x−y))/(log x.log y))=0 (log xy)^2 +{log (x−y)}^2 =0 ⇒log xy=0 ∣ log (x−y)=0 ⇒xy=1 ∣ x−y=1 ⇒x=(1/y) ∣ x−y=1 (1/y)−y=1 ∣ x=y+1 ⇒y=(((√5) −1)/2) ∣ x=(((√5)+1)/2) x+y=(√5)

$$\frac{\mathrm{log}\:{xy}.\mathrm{log}\:{xy}}{\mathrm{log}\:{x}.\mathrm{log}\:{y}}+\frac{\mathrm{log}\:\left({x}−{y}\right).\mathrm{log}\:\left({x}−{y}\right)}{\mathrm{log}\:{x}.\mathrm{log}\:{y}}=\mathrm{0} \\ $$$$\left(\mathrm{log}\:{xy}\right)^{\mathrm{2}} +\left\{\mathrm{log}\:\left({x}−{y}\right)\right\}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{log}\:{xy}=\mathrm{0}\:\:\:\:\mid\:\:\:\mathrm{log}\:\left({x}−{y}\right)=\mathrm{0} \\ $$$$\Rightarrow{xy}=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:{x}−{y}=\mathrm{1}\: \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{{y}}\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:{x}−{y}=\mathrm{1}\: \\ $$$$\frac{\mathrm{1}}{{y}}−{y}=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mid\:{x}={y}+\mathrm{1} \\ $$$$\Rightarrow{y}=\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\mid\:{x}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$$${x}+{y}=\sqrt{\mathrm{5}} \\ $$

Commented by jagoll last updated on 25/Mar/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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