x-log-xy-y-log-xy-x-log-x-y-y-log-x-y-0-find-x-y-

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

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

^{x} \log (x y) .^{y} \log (x y) = -^{x} \log (x - y) .^{y} \log (x - y)

^{(x - y)} \log (x y) = -^{(x y)} \log (x - y)

^{(x - y)} \log (x y) =^{\frac{1}{x y}} \log (x - y)

\Rightarrow {\begin{cases} x - y = \frac{1}{x y} \\ x y = x - y \end{cases}

\Rightarrow x y = 1 \land x - y = 1

y = \frac{1}{x} \Rightarrow x - \frac{1}{x} = 1

x^{2} - x - 1 = 0 \Rightarrow x = \frac{1 + \sqrt{5}}{2}

\land y = \frac{2}{\sqrt{5} + 1} = \frac{2 (\sqrt{5} - 1)}{4}

y = \frac{\sqrt{5} - 1}{2} . s o x + y = \sqrt{5}

Commented by jagoll last updated on 25/Mar/20

great sir

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

\frac{\log x y . \log x y}{\log x . \log y} + \frac{\log (x - y) . \log (x - y)}{\log x . \log y} = 0

{(\log x y)}^{2} + {\log (x - y)}^{2} = 0

\Rightarrow \log x y = 0 ∣ \log (x - y) = 0

\Rightarrow x y = 1 ∣ x - y = 1

\Rightarrow x = \frac{1}{y} ∣ x - y = 1

\frac{1}{y} - y = 1 ∣ x = y + 1

\Rightarrow y = \frac{\sqrt{5} - 1}{2} ∣ x = \frac{\sqrt{5} + 1}{2}

x + y = \sqrt{5}

Commented by jagoll last updated on 25/Mar/20

thank you sir

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