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

$$3^x=2^{x}y+1$$$$\{\mathit{x}:\mathit{y}\}\epsilon \mathbb{N}.$$

Answered by MathsFan last updated on 25/Oct/21

$$3^x-2^{x}y=1$$$$3^x-2^{x}y=3-2$$$$bycomparison$$$$\{\begin{array}{l}{3}^x=3\Rightarrow x=1\\ 2^{x}y=2\Rightarrow y=1\end{array}$$

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

$$\mathrm{According}\mathrm{to}\mathrm{your}\mathrm{way}...$$$$\underset{―}{3^x-2^{x}y=1-0}=3^0-0$$$$x=0\wedge 2^{0}y=0\Rightarrow y=0$$$$\underset{―}{3^x-2^{x}y=9-8}=3^2-8$$$$x=2\wedge 2^{2}y=8\Rightarrow y=2$$$$\underset{―}{3^x-2^{x}y=81-80}=3^4-80$$$$x=4\wedge 2^{4}y=80\Rightarrow y=5$$$$(0,0),(2,2)\mathrm{\&}(4,5)arealsosolutions.$$

Commented by MathsFan last updated on 26/Oct/21

$$thankssir$$

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

$${You}^{\prime }rewelcome!$$

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

$$3^x=2^{x}y+1;x,y\in \mathbb{N}$$$$\frac{3^x}{2^x}=y+\frac{1}{2^x}$$$$\frac{3^x}{2^x}-\frac{1}{2^x}=y$$$${\left(\frac{3}{2}\right)}^x-{\left(\frac{1}{2}\right)}^x=y$$$$$$$$x=0:{\left(\frac{3}{2}\right)}^0-{\left(\frac{1}{2}\right)}^0=y\Rightarrow y=1-1=0$$$$x=1:\frac{3}{2}-\frac{1}{2}=y\Rightarrow y=1$$$$x=2:\frac{9}{4}-\frac{1}{4}=y\Rightarrow y=2$$$$x=3:\frac{27}{8}-\frac{1}{8}=y\Rightarrow y=\frac{13}{4}\notin \mathbb{N}$$$$x=4:\frac{81}{16}-\frac{1}{16}=y\Rightarrow y=\frac{80}{16}=5$$$$x=5:\frac{243}{32}-\frac{1}{32}=y\Rightarrow y=\frac{242}{32}\notin \mathbb{N}$$$$x=6:\frac{729}{64}-\frac{1}{64}=y\Rightarrow y=\frac{728}{64}\notin \mathbb{N}$$$$\mathrm{If}\mathbb{N}=\{0,1,2,3,...\}$$$$(x,y)=(0,0),(1,1),(2,2),(4,5)$$$$Notsureforothersolutions.$$

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