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Question Number 172488 by mathlove last updated on 28/Jun/22

Answered by Rasheed.Sindhi last updated on 28/Jun/22

$$3^y=3^x-19940$$$$y\log 3=\log (3^x-19940)$$$$y=\frac{\log (3^x-19940)}{\log 3}$$$$(x,y)=(x,\frac{\log (3^x-19940)}{\log 3});x>9forrealy$$$$\left(Generalsolution\right)$$$$\Rightarrow Manynumericalsolutions.$$

Answered by aleks041103 last updated on 28/Jun/22

$$19940=2\times 2\times 5\times 997$$$$wewant(x,y)\in {\left({\mathbb{N}}^0\right)}^2.$$$$obvx>y$$$$\Rightarrow 3^y(3^{x-y}-1)=19940$$$$since3\nmid 19940\Rightarrow y=0$$$$\Rightarrow 3^x-1=19940$$$$\Rightarrow 3^x=19941$$$$but19941=3.{17}^2.23$$$$\Rightarrow nosolnsfor(x,y)\in {\mathbb{Z}}^2$$

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