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Question Number 157637 by quvonch3737 last updated on 25/Oct/21

3^x =2^x y+1 {x:y} εN.

$$\:\:\:\:\:\:\mathrm{3}^{{x}} =\mathrm{2}^{{x}} {y}+\mathrm{1} \\ $$$$\:\:\:\:\:\left\{\boldsymbol{{x}}:\boldsymbol{{y}}\right\}\:\varepsilon\mathbb{N}.\: \\ $$

Answered by MathsFan last updated on 25/Oct/21

3^x −2^x y=1 3^x −2^x y=3−2 by comparison { ((3^x =3 ⇒ x=1)),((2^x y=2 ⇒ y=1)) :}

$$\mathrm{3}^{{x}} −\mathrm{2}^{{x}} {y}=\mathrm{1} \\ $$$$\mathrm{3}^{{x}} −\mathrm{2}^{{x}} {y}=\mathrm{3}−\mathrm{2} \\ $$$${by}\:{comparison} \\ $$$$\:\begin{cases}{\mathrm{3}^{{x}} =\mathrm{3}\:\:\Rightarrow\:{x}=\mathrm{1}}\\{\mathrm{2}^{{x}} {y}=\mathrm{2}\:\Rightarrow\:{y}=\mathrm{1}}\end{cases} \\ $$

Commented by Rasheed.Sindhi last updated on 26/Oct/21

According to your way... 3^x −2^x y=1−0=3^0 −0 x=0 ∧ 2^0 y=0⇒y=0 3^x −2^x y=9−8=3^2 −8 x=2 ∧ 2^2 y=8⇒y=2 3^x −2^x y=81−80 =3^4 −80 x=4∧2^4 y=80⇒y=5 (0,0),(2,2) &(4,5) are also solutions.

$$\mathrm{According}\:\mathrm{to}\:\mathrm{your}\:\mathrm{way}... \\ $$$$\underline{\mathrm{3}^{{x}} −\mathrm{2}^{{x}} {y}=\mathrm{1}−\mathrm{0}}=\mathrm{3}^{\mathrm{0}} −\mathrm{0} \\ $$$${x}=\mathrm{0}\:\wedge\:\mathrm{2}^{\mathrm{0}} {y}=\mathrm{0}\Rightarrow{y}=\mathrm{0} \\ $$$$\underline{\mathrm{3}^{{x}} −\mathrm{2}^{{x}} {y}=\mathrm{9}−\mathrm{8}}=\mathrm{3}^{\mathrm{2}} −\mathrm{8} \\ $$$${x}=\mathrm{2}\:\wedge\:\mathrm{2}^{\mathrm{2}} {y}=\mathrm{8}\Rightarrow{y}=\mathrm{2} \\ $$$$\underline{\mathrm{3}^{{x}} −\mathrm{2}^{{x}} {y}=\mathrm{81}−\mathrm{80}}\:=\mathrm{3}^{\mathrm{4}} −\mathrm{80} \\ $$$${x}=\mathrm{4}\wedge\mathrm{2}^{\mathrm{4}} {y}=\mathrm{80}\Rightarrow{y}=\mathrm{5} \\ $$$$\left(\mathrm{0},\mathrm{0}\right),\left(\mathrm{2},\mathrm{2}\right)\:\&\left(\mathrm{4},\mathrm{5}\right)\:{are}\:{also}\:{solutions}. \\ $$

Commented by MathsFan last updated on 26/Oct/21

thanks sir

$${thanks}\:{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 26/Oct/21

You′re welcome!

$${You}'{re}\:{welcome}! \\ $$

Answered by Rasheed.Sindhi last updated on 26/Oct/21

3^x =2^x y+1; x,y∈N (3^x /2^x )=y+(1/2^x ) (3^x /2^x )−(1/2^x )=y ((3/2))^x −((1/2))^x =y x=0:((3/2))^0 −((1/2))^0 =y⇒y=1−1=0 x=1: (3/2)−(1/2)=y⇒y=1 x=2:(9/4)−(1/4)=y⇒y=2 x=3:((27)/8)−(1/8)=y⇒y=((13)/4)∉N x=4:((81)/(16))−(1/(16))=y⇒y=((80)/(16))=5 x=5:((243)/(32))−(1/(32))=y⇒y=((242)/(32))∉N x=6:((729)/(64))−(1/(64))=y⇒y=((728)/(64))∉N If N={0,1,2,3,...} (x,y)=(0,0),(1,1),(2,2),(4,5) Not sure for other solutions.

$$\:\:\:\:\:\:\mathrm{3}^{{x}} =\mathrm{2}^{{x}} {y}+\mathrm{1};\:{x},{y}\in\mathbb{N} \\ $$$$\:\:\:\frac{\mathrm{3}^{{x}} }{\mathrm{2}^{{x}} }={y}+\frac{\mathrm{1}}{\mathrm{2}^{{x}} } \\ $$$$\:\:\:\frac{\mathrm{3}^{{x}} }{\mathrm{2}^{{x}} }−\frac{\mathrm{1}}{\mathrm{2}^{{x}} }={y} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{{x}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{x}} ={y} \\ $$$$ \\ $$$${x}=\mathrm{0}:\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{0}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{0}} ={y}\Rightarrow{y}=\mathrm{1}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\mathrm{1}:\:\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}={y}\Rightarrow{y}=\mathrm{1} \\ $$$${x}=\mathrm{2}:\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}={y}\Rightarrow{y}=\mathrm{2} \\ $$$${x}=\mathrm{3}:\frac{\mathrm{27}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}={y}\Rightarrow{y}=\frac{\mathrm{13}}{\mathrm{4}}\notin\mathbb{N} \\ $$$${x}=\mathrm{4}:\frac{\mathrm{81}}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{16}}={y}\Rightarrow{y}=\frac{\mathrm{80}}{\mathrm{16}}=\mathrm{5} \\ $$$${x}=\mathrm{5}:\frac{\mathrm{243}}{\mathrm{32}}−\frac{\mathrm{1}}{\mathrm{32}}={y}\Rightarrow{y}=\frac{\mathrm{242}}{\mathrm{32}}\notin\mathbb{N} \\ $$$${x}=\mathrm{6}:\frac{\mathrm{729}}{\mathrm{64}}−\frac{\mathrm{1}}{\mathrm{64}}={y}\Rightarrow{y}=\frac{\mathrm{728}}{\mathrm{64}}\notin\mathbb{N} \\ $$$$\mathrm{If}\:\mathbb{N}=\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},...\right\} \\ $$$$\left({x},{y}\right)=\left(\mathrm{0},\mathrm{0}\right),\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{4},\mathrm{5}\right) \\ $$$${Not}\:{sure}\:{for}\:{other}\:{solutions}. \\ $$

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