Question Number 120284 by Algoritm last updated on 30/Oct/20
Commented by Algoritm last updated on 30/Oct/20
$$\mathrm{x}=? \\ $$
Answered by TITA last updated on 30/Oct/20
$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{{x}+\mathrm{1}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2017}}\right)^{\mathrm{2018}} \Rightarrow\left({x}+\frac{\mathrm{1}}{{x}}\right)^{{x}+\mathrm{1}} ={x}^{{x}+\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{{x}+\mathrm{1}} \\ $$$${by}\:{equating}\:{coefficient}\: \\ $$$${x}^{{x}+\mathrm{1}} =\mathrm{1}\:\:\:\:\Rightarrow\left({x}+\mathrm{1}\right)\mathrm{ln}\:{x}=\mathrm{ln}\:\mathrm{1} \\ $$$$\Rightarrow\mathrm{ln}\:{x}=\mathrm{0}\:\:{hence}\:\:{x}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\mathrm{2017}\Rightarrow{x}=\:\underset{−} {+}\sqrt{\mathrm{2017}} \\ $$$${x}+\mathrm{1}=\mathrm{2018}\:\Rightarrow\:{x}=\mathrm{2017} \\ $$$${hence}\:{x}=\left\{\mathrm{1},\mathrm{2017},\:\underset{−} {+}\sqrt{\mathrm{2017}\:}\right\} \\ $$
Commented by JDamian last updated on 30/Oct/20
$${Would}\:{you}\:{mind}\:{explaining}\:{how} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}}\right)^{\mathrm{1}+\mathrm{1}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2017}}\right)^{\mathrm{2018}} ? \\ $$