Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 203424 by Calculusboy last updated on 18/Jan/24

	Answered by Rasheed.Sindhi last updated on 19/Jan/24

	$x + y = 2^{x - y} . . . (i)$ ${(x + y)}^{x - y} = 2 . . . (i i)$ $(i i) / (i) : {(x + y)}^{x - y - 1} = 2^{1 - x + y}$ $\Rightarrow x + y = 2 \land x - y - 1 = 1 - x + y$ $\Rightarrow x + y = 2 \land 2 x - 2 y = 2 \Rightarrow x - y = 1$ $x = 3 / 2 \land y = 1 / 2$
	Commented by mr W last updated on 19/Jan/24

	$f o r {(x + y)}^{x - y - 1} = 2^{1 - x + y} = 2^{- (x - y - 1)}$ $t h e r e i s a n o t h e r p o s s i b i l i t y :$ $x + y = 2^{- 1} = \frac{1}{2}$
	Commented by Rasheed.Sindhi last updated on 19/Jan/24

	$N i c e s i r, T h a n k s!$
	Commented by Rasheed.Sindhi last updated on 19/Jan/24
	${ ((x+y= 2^(x−y) ....i)),(( 2 =(x+y)^(x−y) ...ii)) :} i/ii: ((x+y)/2)=((2/(x+y)))^(x−y) ((x+y)/2)=(((x+y)/2))^(y−x) (((x+y)/2))^(y−x−1) =1=(((x+y)/2))^0 x−y=−1 { ((((x+y)/2)=1⇒x+y=2)),(( or)),((x−y=−1)) :} i×ii: 2(x+y)=2^(x−y) (x+y)^(x−y) (2/2^(x−y) )=(((x+y)^(x−y) )/(x+y)) (x+y)^(x−y−1) =2^(1−x+y) (x+y)^(x−y−1) =2^(−(x−y−1)) { ((x+y=2^(−1) =(1/2))),((x+y=2 ∧ x−y−1=−x+y+1)) :}$
	${\begin{cases} x + y = 2^{x - y} . . . . i \\ 2 = {(x + y)}^{x - y} . . . i i \end{cases}$ $i / i i : \frac{x + y}{2} = {(\frac{2}{x + y})}^{x - y}$ $\frac{x + y}{2} = {(\frac{x + y}{2})}^{y - x}$ ${(\frac{x + y}{2})}^{y - x - 1} = 1 = {(\frac{x + y}{2})}^{0}$ $x - y = - 1$ ${\begin{cases} \frac{x + y}{2} = 1 \Rightarrow x + y = 2 \\ or \\ x - y = - 1 \end{cases}$ $i \times i i : 2 (x + y) = 2^{x - y} {(x + y)}^{x - y}$ $\frac{2}{2^{x - y}} = \frac{{(x + y)}^{x - y}}{x + y}$ ${(x + y)}^{x - y - 1} = 2^{1 - x + y}$ ${(x + y)}^{x - y - 1} = 2^{- (x - y - 1)}$ ${\begin{cases} x + y = 2^{- 1} = \frac{1}{2} \\ x + y = 2 \land x - y - 1 = - x + y + 1 \end{cases}$
	Commented by Calculusboy last updated on 21/Jan/24

	$thanks sir$
	Answered by mr W last updated on 19/Jan/24

	$2^{{(x - y)}^{2}} = 2$ ${(x - y)}^{2} = 1$ $x - y = \pm 1$ $x + y = 2^{\pm 1} = 2 o r \frac{1}{2}$ $\Rightarrow x = \frac{\pm 1 + 2^{\pm 1}}{2} = \frac{3}{2} o r - \frac{1}{4}$ $y = \frac{3}{2} - 1 = \frac{1}{2} o r - \frac{1}{4} + 1 = \frac{3}{4}$ $\Rightarrow (x, y) = (\frac{3}{2}, \frac{1}{2}) o r (- \frac{1}{4}, \frac{3}{4})$
	Commented by Calculusboy last updated on 21/Jan/24

	$t h a n k s s i r$
	Answered by esmaeil last updated on 18/Jan/24

	${(2 (x + y))}^{x - y} = 2 (x + y) \to$ $x = y + 1 \to$ $2 y + 1 = 2 \to y = \frac{1}{2} \to$ $x = \frac{3}{2}$
	Commented by Calculusboy last updated on 21/Jan/24

	$n i c e s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum