Question Number 196576 by sonukgindia last updated on 27/Aug/23
Commented by Frix last updated on 27/Aug/23
$$\mathrm{Infinite}\:\mathrm{solutions}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{R} \\ $$$$\mathrm{No}\:\mathrm{solution}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{Z} \\ $$
Answered by AST last updated on 27/Aug/23
$$\mathrm{2}^{\mathrm{2}{x}} −\mathrm{2}^{{y}} \mathrm{3}^{{y}} =\mathrm{2}^{\mathrm{48}} \mathrm{3}^{\mathrm{12}} =\mathrm{2}^{{y}} \left(\mathrm{2}^{\mathrm{2}{x}−{y}} −\mathrm{3}^{{y}} \right)\Rightarrow{y}=\mathrm{48} \\ $$$$\Rightarrow\mathrm{4}^{{x}} =\mathrm{48}^{\mathrm{12}} +\mathrm{6}^{\mathrm{48}} \Rightarrow{x}={log}_{\mathrm{4}} \mathrm{3}^{\mathrm{12}} +{log}_{\mathrm{4}} \left(\mathrm{3}^{\mathrm{36}} +\mathrm{1}\right)+\mathrm{24} \\ $$$$\Rightarrow{x}+{y}=\mathrm{72}+\mathrm{6}{log}_{\mathrm{2}} \mathrm{3}+{log}_{\mathrm{4}} \left(\mathrm{3}^{\mathrm{36}} +\mathrm{1}\right) \\ $$
Answered by sniper237 last updated on 27/Aug/23
![Such numbers can′t exist!!! 4^x ≡1[3] ; 48^(12) ≡0[3]⇒ 1−6^y ≡0[3] So y=0. Then 4^x =1+48^(12) 48^(12) ≡0[2]⇒ 4^x ≡1[2]⇒x=0 But 4^0 −6^0 ≠48^(12) .](https://www.tinkutara.com/question/Q196593.png)
$${Such}\:{numbers}\:{can}'{t}\:{exist}!!! \\ $$$$\mathrm{4}^{{x}} \equiv\mathrm{1}\left[\mathrm{3}\right]\:;\:\mathrm{48}^{\mathrm{12}} \equiv\mathrm{0}\left[\mathrm{3}\right]\Rightarrow\:\mathrm{1}−\mathrm{6}^{{y}} \equiv\mathrm{0}\left[\mathrm{3}\right] \\ $$$${So}\:\:{y}=\mathrm{0}.\:{Then}\:\:\mathrm{4}^{{x}} =\mathrm{1}+\mathrm{48}^{\mathrm{12}} \\ $$$$\mathrm{48}^{\mathrm{12}} \equiv\mathrm{0}\left[\mathrm{2}\right]\Rightarrow\:\mathrm{4}^{{x}} \equiv\mathrm{1}\left[\mathrm{2}\right]\Rightarrow{x}=\mathrm{0} \\ $$$${But}\:\:\mathrm{4}^{\mathrm{0}} −\mathrm{6}^{\mathrm{0}} \neq\mathrm{48}^{\mathrm{12}} .\: \\ $$
Commented by AST last updated on 27/Aug/23
$${That}'{s}\:{if}\:{x},{y}\in\mathbb{Z} \\ $$