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Question Number 99410 by Harasanemanabrandah last updated on 20/Jun/20

Commented by PRITHWISH SEN 2 last updated on 20/Jun/20

$one of the soln . set is for every positive real number$ $greater than 0, x = y$
Commented by Rasheed.Sindhi last updated on 20/Jun/20
$2^4 =4^2 ; {x,y}={2,4}$
$2^{4} = 4^{2}; {x, y} = {2, 4}$
	Answered by mr W last updated on 20/Jun/20
	$for x,y∈R there are infinite many solutions. x=y=t∈R^+ is always a solution. for x≠y: METHOD I let x=t ∈R^+ x^y =y^t y=t^(y/t) =e^((y/t)ln t) ye^(−(y/t)ln t) =1 (−(y/t)ln t)e^(−(y/t)ln t) =−((ln t)/t) −(y/t)ln t=W(−((ln t)/t)) y=−(t/(ln t))W(−((ln t)/t)) ⇒general solution { ((x=t)),((y=−(t/(ln t))W(−((ln t)/t)))) :} METHOD II let y=tx with t∈R^+ x^(tx) =(tx)^x x^t =tx x^(t−1) =t ⇒x=t^(1/(t−1)) ⇒y=t^(1+(1/(t−1))) =t^(t/(t−1)) ⇒general solution { ((x=t^(1/(t−1)) )),((y=t^(t/(t−1)) )) :}$
	$f o r x, y \in R t h e r e a r e i n f i n i t e m a n y$ $s o l u t i o n s .$ $x = y = t \in R^{+} i s a l w a y s a s o l u t i o n .$ $f o r x \neq y :$ $M E T H O D I$ $l e t x = t \in R^{+}$ $x^{y} = y^{t}$ $y = t^{\frac{y}{t}} = e^{\frac{y}{t} \ln t}$ ${y e}^{- \frac{y}{t} \ln t} = 1$ $(- \frac{y}{t} \ln t) e^{- \frac{y}{t} \ln t} = - \frac{\ln t}{t}$ $- \frac{y}{t} \ln t = W (- \frac{\ln t}{t})$ $y = - \frac{t}{\ln t} W (- \frac{\ln t}{t})$ $\Rightarrow g e n e r a l s o l u t i o n {\begin{cases} x = t \\ y = - \frac{t}{\ln t} W (- \frac{\ln t}{t}) \end{cases}$ $M E T H O D I I$ $l e t y = t x w i t h t \in R^{+}$ $x^{t x} = {(t x)}^{x}$ $x^{t} = t x$ $x^{t - 1} = t$ $\Rightarrow x = t^{\frac{1}{t - 1}}$ $\Rightarrow y = t^{1 + \frac{1}{t - 1}} = t^{\frac{t}{t - 1}}$ $\Rightarrow g e n e r a l s o l u t i o n {\begin{cases} x = t^{\frac{1}{t - 1}} \\ y = t^{\frac{t}{t - 1}} \end{cases}$
	Commented by 1549442205 last updated on 21/Jun/20

	$Thank you, sir . I like second way, it is$ $explicit$
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