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Question Number 102418 by Akeyz last updated on 09/Jul/20

	Answered by Rasheed.Sindhi last updated on 09/Jul/20
	$^(★1:) x,y∈W(={0,1,2,3,...}) 2^x +3^y ∈E⇒2^x ,3^y ∈E ∨ 2^x ,3^y ∈O 3^y ∈O⇒2^x ∈O⇒x=0 2^x +3^y =72⇒2^0 +3^y =72 ⇒3^y =71⇒y∉W ∴The system has no solution in W ^(★2:) When only one of x , y is negative whole number. x∈Z^− ⇒ 0<2^x <1(or y∈Z^− ⇒ 0<2^y <1) ⇒3^y <2^x +3^y <3^y +1 ⇒ 2^x +3^y ∉W (Similar case when y∈Z^− ) ^(★3) When both x & y are negative whole numbers 0<2^x <1 ∧ 0<3^y <1 ⇒ 0<2^x +3^y <2 ∴ 2^x +3^y =72 is impossible This proves that the given system has no solution in Z ......... .....$
	$^{★ 1 :} x, y \in W (= {0, 1, 2, 3, . . .})$ $2^{x} + 3^{y} \in E \Rightarrow 2^{x}, 3^{y} \in E \lor 2^{x}, 3^{y} \in O$ $3^{y} \in O \Rightarrow 2^{x} \in O \Rightarrow x = 0$ $2^{x} + 3^{y} = 72 \Rightarrow 2^{0} + 3^{y} = 72$ $\Rightarrow 3^{y} = 71 \Rightarrow y \notin W$ $∴ T h e s y s t e m h a s n o s o l u t i o n i n W$ $^{★ 2 :} W h e n o n l y o n e o f x, y i s n e g a t i v e$ $w h o l e n u m b e r .$ $x \in Z^{-} \Rightarrow 0 < 2^{x} < 1 (o r y \in Z^{-} \Rightarrow 0 < 2^{y} < 1)$ $\Rightarrow 3^{y} < 2^{x} + 3^{y} < 3^{y} + 1$ $\Rightarrow 2^{x} + 3^{y} \notin W$ $(S i m i l a r c a s e w h e n y \in Z^{-})$ $^{★ 3} W h e n b o t h x & y a r e n e g a t i v e$ $w h o l e n u m b e r s$ $0 < 2^{x} < 1 \land 0 < 3^{y} < 1$ $\Rightarrow 0 < 2^{x} + 3^{y} < 2$ $∴ 2^{x} + 3^{y} = 72 i s i m p o s s i b l e$ $T h i s p r o v e s t h a t t h e g i v e n s y s t e m$ $h a s n o s o l u t i o n i n Z$ $. . . . . . . . .$ $. . . . .$
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