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Question-102418




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  .........  .....
$$\:^{\bigstar\mathrm{1}:} {x},{y}\in\mathbb{W}\left(=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},…\right\}\right) \\ $$$$\mathrm{2}^{{x}} +\mathrm{3}^{{y}} \in\mathbb{E}\Rightarrow\mathrm{2}^{{x}} ,\mathrm{3}^{{y}} \in\mathbb{E}\:\vee\:\mathrm{2}^{{x}} ,\mathrm{3}^{{y}} \in\mathbb{O} \\ $$$$\mathrm{3}^{{y}} \in\mathbb{O}\Rightarrow\mathrm{2}^{{x}} \in\mathbb{O}\Rightarrow{x}=\mathrm{0} \\ $$$$\mathrm{2}^{{x}} +\mathrm{3}^{{y}} =\mathrm{72}\Rightarrow\mathrm{2}^{\mathrm{0}} +\mathrm{3}^{{y}} =\mathrm{72} \\ $$$$\Rightarrow\mathrm{3}^{{y}} =\mathrm{71}\Rightarrow{y}\notin\mathbb{W} \\ $$$$\therefore\mathcal{T}{he}\:{system}\:{has}\:{no}\:{solution}\:{in}\:\mathbb{W}\: \\ $$$$\:^{\bigstar\mathrm{2}:} {When}\:{only}\:{one}\:{of}\:{x}\:,\:{y}\:{is}\:{negative} \\ $$$${whole}\:{number}. \\ $$$$\:\:{x}\in\mathbb{Z}^{−} \Rightarrow\:\mathrm{0}<\mathrm{2}^{{x}} <\mathrm{1}\left({or}\:{y}\in\mathbb{Z}^{−} \Rightarrow\:\mathrm{0}<\mathrm{2}^{{y}} <\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{3}^{{y}} <\mathrm{2}^{{x}} +\mathrm{3}^{{y}} <\mathrm{3}^{{y}} +\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{2}^{{x}} +\mathrm{3}^{{y}} \:\notin\mathbb{W} \\ $$$$\left({Similar}\:{case}\:{when}\:{y}\in\mathbb{Z}^{−} \right) \\ $$$$\:^{\bigstar\mathrm{3}} {When}\:{both}\:{x}\:\&\:{y}\:{are}\:{negative} \\ $$$${whole}\:{numbers} \\ $$$$\:\:\:\:\:\:\mathrm{0}<\mathrm{2}^{{x}} <\mathrm{1}\:\wedge\:\mathrm{0}<\mathrm{3}^{{y}} <\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{0}<\mathrm{2}^{{x}} +\mathrm{3}^{{y}} <\mathrm{2} \\ $$$$\:\:\:\therefore\:\mathrm{2}^{{x}} +\mathrm{3}^{{y}} =\mathrm{72}\:{is}\:{impossible} \\ $$$${This}\:{proves}\:{that}\:{the}\:{given}\:{system} \\ $$$${has}\:{no}\:{solution}\:{in}\:\mathbb{Z} \\ $$$$……… \\ $$$$….. \\ $$

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