What are all ordered pairs of real number (x,y) for which 5^(y−x) (x+y) = 1 and (x+y)^(x−y) = 5


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Question Number 115858 by bemath last updated on 29/Sep/20

$W h a t a r e a l l o r d e r e d p a i r s o f r e a l$ $n u m b e r (x, y) f o r w h i c h$ $5^{y - x} (x + y) = 1 a n d {(x + y)}^{x - y} = 5$
	Answered by floor(10²Eta[1]) last updated on 29/Sep/20

	$(x + y) = \frac{1}{5^{y - x}} = 5^{x - y} \Rightarrow {(5^{(x - y)})}^{x - y} = 5$ $\Rightarrow {(x - y)}^{2} = 1 ∴ x - y = \pm 1$ $x = y \pm 1$ $y = 2 \Rightarrow x = 3$ $y = \frac{3}{5} \Rightarrow x = \frac{- 2}{5}$
	Answered by bobhans last updated on 29/Sep/20
	$⇔ (x+y) = 5^(x−y) ∧ (x+y)^(x−y) = 5 ⇒ [ (x+y)^(x−y) ]^(x−y) = 5^(x−y) ⇒ (x+y)^((x−y)^2 ) = (x+y)⇒ (x−y)^2 =1 → { ((x=y+1 )),((x=y−1)) :} case(1) →x=y+1 (2y+1)^1 = 5 ⇒y = 2 ∧x=3 case(2)→x=y−1 (2y−1)^(−1) = 5 ⇒2y−1=(1/5) y = (3/5) ∧ x=−(2/5) solution : (−(2/5),(3/5)) and (3,2)$
	$\Leftrightarrow (x + y) = 5^{x - y} \land {(x + y)}^{x - y} = 5$ $\Rightarrow {[{(x + y)}^{x - y}]}^{x - y} = 5^{x - y}$ $\Rightarrow {(x + y)}^{{(x - y)}^{2}} = (x + y) \Rightarrow {(x - y)}^{2} = 1$ $\to {\begin{cases} x = y + 1 \\ x = y - 1 \end{cases}$ $c a s e (1) \to x = y + 1$ ${(2 y + 1)}^{1} = 5 \Rightarrow y = 2 \land x = 3$ $c a s e (2) \to x = y - 1$ ${(2 y - 1)}^{- 1} = 5 \Rightarrow 2 y - 1 = \frac{1}{5}$ $y = \frac{3}{5} \land x = - \frac{2}{5}$ $s o l u t i o n : (- \frac{2}{5}, \frac{3}{5}) a n d (3, 2)$
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