Question Number 110843 by dw last updated on 31/Aug/20
$$\left({x}+\mathrm{1}\right)^{\left({x}+\mathrm{1}\right)} =\sqrt{\mathrm{2}}\:\:\:\:\:\:{find}\:{all}\:{values}\:{of}\:{x} \\ $$$$\left({Please}\:{step}\:{by}\:{step}\right) \\ $$
Answered by john santu last updated on 31/Aug/20
$$\rightarrow\mathrm{ln}\:\left({x}+\mathrm{1}\right)^{\left({x}+\mathrm{1}\right)} =\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\right) \\ $$$$\left({x}+\mathrm{1}\right)\mathrm{ln}\:\left({x}+\mathrm{1}\right)=\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\right) \\ $$$${e}^{\mathrm{ln}\:\left({x}+\mathrm{1}\right)} .\mathrm{ln}\:\left({x}+\mathrm{1}\right)=\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\right) \\ $$$$\rightarrow{W}\left({e}^{\mathrm{ln}\:\left({x}+\mathrm{1}\right)} .\mathrm{ln}\:\left({x}+\mathrm{1}\right)\right)={W}\left(\mathrm{ln}\:\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow\:{x}+\mathrm{1}\:=\:{W}\left(\mathrm{ln}\:\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}\:=\:−\mathrm{1}+{W}\left(\mathrm{ln}\:\sqrt{\mathrm{2}}\:\right) \\ $$$$\Rightarrow{x}=−\mathrm{1}+{W}\left(\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}=−\mathrm{1}+\mathrm{0}.\mathrm{70710678} \\ $$
Commented by dw last updated on 31/Aug/20
$${Thank}\:{you}\:{Sir} \\ $$
Answered by Kerly last updated on 08/Jan/21
$$\left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} =\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{−{x}−\mathrm{1}} =\sqrt{\mathrm{2}}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} =\left(\mathrm{2}^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \right)^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} = \\ $$