Question Number 172488 by mathlove last updated on 28/Jun/22
Answered by Rasheed.Sindhi last updated on 28/Jun/22
$$\mathrm{3}^{{y}} =\mathrm{3}^{{x}} −\mathrm{19940} \\ $$$${y}\mathrm{log}\:\mathrm{3}=\mathrm{log}\left(\mathrm{3}^{{x}} −\mathrm{19940}\right)\: \\ $$$${y}=\frac{\mathrm{log}\left(\mathrm{3}^{{x}} −\mathrm{19940}\right)}{\mathrm{log}\:\mathrm{3}} \\ $$$$\left({x},{y}\right)=\left({x},\frac{\mathrm{log}\left(\mathrm{3}^{{x}} −\mathrm{19940}\right)}{\mathrm{log}\:\mathrm{3}}\right)\:;{x}>\mathrm{9}\:{for}\:{real}\:{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({General}\:{solution}\right) \\ $$$$\Rightarrow{Many}\:{numerical}\:{solutions}. \\ $$
Answered by aleks041103 last updated on 28/Jun/22
$$\mathrm{19940}=\mathrm{2}×\mathrm{2}×\mathrm{5}×\mathrm{997} \\ $$$${we}\:{want}\:\left({x},{y}\right)\in\left(\mathbb{N}^{\mathrm{0}} \right)^{\mathrm{2}} . \\ $$$${obv}\:{x}>{y} \\ $$$$\Rightarrow\mathrm{3}^{{y}} \left(\mathrm{3}^{{x}−{y}} −\mathrm{1}\right)=\mathrm{19940} \\ $$$${since}\:\mathrm{3}\nmid\mathrm{19940}\Rightarrow{y}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}^{{x}} −\mathrm{1}=\mathrm{19940} \\ $$$$\Rightarrow\mathrm{3}^{{x}} =\mathrm{19941} \\ $$$${but}\:\mathrm{19941}=\mathrm{3}.\mathrm{17}^{\mathrm{2}} .\mathrm{23} \\ $$$$\Rightarrow{no}\:{solns}\:{for}\:\left({x},{y}\right)\in\mathbb{Z}^{\mathrm{2}} \\ $$