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Question Number 143906 by mathdanisur last updated on 19/Jun/21

	Answered by TheHoneyCat last updated on 19/Jun/21
	$⇔x^x −y^y =e.ln((y/x))=e(ln(y)−ln(x)) ⇔x^x +elnx=y^y +elny so oviously {(t,t), t∈R_+ ^∗ } is part of the solutions but are there any other considering f: { (R_+ ^∗ ,→,R),(x, ,(x^x +elnx)) :} if x≠y are a solution, f (x)=f(y) thus f is not injective but f is actualy strictly increasing (by a painful study of the derivative) (if anyone has a better method, please send it I′ve had to check derivatives too far to show it here) so it is injective (f(x)=f(y)⇒x=y) The solutions are : (x,y)∈{(t,t) t∈R_+ ^∗ } the equation is equivalent to ′′x=y′′$
	$\Leftrightarrow x^{x} - y^{y} = e . \ln (\frac{y}{x}) = e (\ln (y) - \ln (x))$ $\Leftrightarrow x^{x} + e \ln x = y^{y} + e \ln y$ $so oviously {(t, t), t \in R_{+}^{}} is part of the solutions$ $but are there any other$ $considering f : {\begin{cases} R_{+}^{} & \to & R \\ x & x^{x} + e \ln x \end{cases}$ $if x \neq y are a solution, f (x) = f (y)$ $thus f is not injective$ $but f is actualy strictly increasing$ $(b y a p a i n f u l s t u d y o f t h e d e r i v a t i v e)$ $(i f a n y o n e h a s a b e t t e r m e t h o d, p l e a s e s e n d i t$ $I^{'} v e h a d t o c h e c k d e r i v a t i v e s t o o f a r t o s h o w i t h e r e)$ $so it is injective (f (x) = f (y) \Rightarrow x = y)$ $The solutions are :$ $(x, y) \in {(t, t) t \in R_{+}^{*}}$ $t h e e q u a t i o n i s e q u i v a l e n t t o^{″} x = y^{″}$
	Commented by mathdanisur last updated on 19/Jun/21

	$t h a n k y o u S i r$
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