If cosh ^{-1} x = 2 log _e(sqrt{2}+1),then x= | vedclass

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(D) We have,$\cosh ^{-1} x = 2 \log _e(\sqrt{2}+1)$.Using the logarithmic property $n \log a = \log a^n$,we get:$\cosh ^{-1} x = \log _e(\sqrt{2}+1)^2

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$3$

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(D) We have,$\cosh ^{-1} x = 2 \log _e(\sqrt{2}+1)$.Using the logarithmic property $n \log a = \log a^n$,we get:$\cosh ^{-1} x = \log _e(\sqrt{2}+1)^2$.We know that $\cosh ^{-1} x = \log _e(x + \sqrt{x^2-1})$.Therefore,$\log _e(x + \sqrt{x^2-1}) = \log _e(2 + 1 + 2\sqrt{2}) = \log _e(3 + 2\sqrt{2})$.Comparing both sides,we have $x + \sqrt{x^2-1} = 3 + 2\sqrt{2}$.Let $x = 3$,then $\sqrt{x^2-1} = \sqrt{9-1} = \sqrt{8} = 2\sqrt{2}$.Thus,$3 + 2\sqrt{2} = 3 + 2\sqrt{2}$,which satisfies the equation.Hence,$x = 3$.

If $\cosh ^{-1} x = 2 \log _e(\sqrt{2}+1)$,then $x=$

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