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

(1+(1/x))^(x+1) =(1+(1/(2019)))^(2019)   Find all possible values of x

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

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

x=−2020  (1−(1/(2020)))^(−2019)   =(((2019)/(2020)))^(−2019) =(((2020)/(2019)))^(2019) =(1+(1/(2019)))^(2019)   only one solution    You can prove no other solution using  inequalities

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

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