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

{(x + 1)}^{(x + 1)} = \sqrt{2} f i n d a l l v a l u e s o f x

(P l e a s e s t e p b y s t e p)

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

\to \ln {(x + 1)}^{(x + 1)} = \ln (\sqrt{2})

(x + 1) \ln (x + 1) = \ln (\sqrt{2})

e^{\ln (x + 1)} . \ln (x + 1) = \ln (\sqrt{2})

\to W (e^{\ln (x + 1)} . \ln (x + 1)) = W (\ln \sqrt{2})

\Rightarrow x + 1 = W (\ln \sqrt{2})

\Rightarrow x = - 1 + W (\ln \sqrt{2})

\Rightarrow x = - 1 + W (\frac{\ln 2}{2})

\Rightarrow x = - 1 + 0.70710678

Commented by dw last updated on 31/Aug/20

T h a n k y o u S i r

Answered by Kerly last updated on 08/Jan/21

{(x + 1)}^{x + 1} = {(\frac{1}{x + 1})}^{- x - 1} = \sqrt{2} = 2^{\frac{1}{2}} = {(2^{\frac{1}{\sqrt{2}}})}^{\frac{1}{\sqrt{2}}} =

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