What-are-all-ordered-pairs-of-real-number-x-y-for-which-5-y-x-x-y-1-and-x-y-x-y-5-

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

\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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