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Question Number 110843 by dw last updated on 31/Aug/20

(x+1)^((x+1)) =(√2)      find all values of x  (Please step by step)

$$\left({x}+\mathrm{1}\right)^{\left({x}+\mathrm{1}\right)} =\sqrt{\mathrm{2}}\:\:\:\:\:\:{find}\:{all}\:{values}\:{of}\:{x} \\ $$$$\left({Please}\:{step}\:{by}\:{step}\right) \\ $$

Answered by john santu last updated on 31/Aug/20

→ln (x+1)^((x+1)) =ln ((√2))  (x+1)ln (x+1)=ln ((√2))  e^(ln (x+1)) .ln (x+1)=ln ((√2))  →W(e^(ln (x+1)) .ln (x+1))=W(ln (√2))  ⇒ x+1 = W(ln (√2))  ⇒x = −1+W(ln (√2) )  ⇒x=−1+W(((ln 2)/2))  ⇒x=−1+0.70710678

$$\rightarrow\mathrm{ln}\:\left({x}+\mathrm{1}\right)^{\left({x}+\mathrm{1}\right)} =\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\right) \\ $$$$\left({x}+\mathrm{1}\right)\mathrm{ln}\:\left({x}+\mathrm{1}\right)=\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\right) \\ $$$${e}^{\mathrm{ln}\:\left({x}+\mathrm{1}\right)} .\mathrm{ln}\:\left({x}+\mathrm{1}\right)=\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\right) \\ $$$$\rightarrow{W}\left({e}^{\mathrm{ln}\:\left({x}+\mathrm{1}\right)} .\mathrm{ln}\:\left({x}+\mathrm{1}\right)\right)={W}\left(\mathrm{ln}\:\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow\:{x}+\mathrm{1}\:=\:{W}\left(\mathrm{ln}\:\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}\:=\:−\mathrm{1}+{W}\left(\mathrm{ln}\:\sqrt{\mathrm{2}}\:\right) \\ $$$$\Rightarrow{x}=−\mathrm{1}+{W}\left(\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{x}=−\mathrm{1}+\mathrm{0}.\mathrm{70710678} \\ $$

Commented by dw last updated on 31/Aug/20

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

Answered by Kerly last updated on 08/Jan/21

(x+1)^(x+1) =((1/(x+1)))^(−x−1) =(√2)=2^(1/2) =(2^(1/( (√2))) )^(1/( (√2))) =

$$\left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} =\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{−{x}−\mathrm{1}} =\sqrt{\mathrm{2}}=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} =\left(\mathrm{2}^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \right)^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} = \\ $$

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