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Question Number 93241 by Ayod 19 last updated on 12/May/20

$${(1+\frac{1}{\mathrm{x}})}^{\mathrm{x}+1}={(1+\frac{1}{2019})}^{2019}$$$$\mathrm{Find}\mathrm{all}\mathrm{possible}\mathrm{values}\mathrm{of}x$$

Answered by prakash jain last updated on 12/May/20

$$x=-2020$$$${(1-\frac{1}{2020})}^{-2019}$$$$={\left(\frac{2019}{2020}\right)}^{-2019}={\left(\frac{2020}{2019}\right)}^{2019}={(1+\frac{1}{2019})}^{2019}$$$$\mathrm{only}\mathrm{one}\mathrm{solution}$$$$$$$$\mathrm{You}\mathrm{can}\mathrm{prove}\mathrm{no}\mathrm{other}\mathrm{solution}\mathrm{using}$$$$\mathrm{inequalities}$$

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