sech−1(23)=log((K+1)+K2+1K) Then K=____

sech−1x=log(1+1−x2x)⇒sech−1(23)=log(3+52) ∴K+1=3⇒K=2

sech−1(23)=log((K+1)+K2+1K) Then K=____

sech−1x=log(1+1−x2x)⇒sech−1(23)=log(3+52) ∴K+1=3⇒K=2

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${sech}^{- 1} (\frac{2}{3}) = \log (\frac{(K + 1) + \sqrt{K^{2} + 1}}{K})$ Then $K =____$

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${sech}^{- 1} x = \log (\frac{1 + \sqrt{1 - x^{2}}}{x}) \Rightarrow {sech}^{- 1} (\frac{2}{3}) = \log (\frac{3 + \sqrt{5}}{2}) ∴ K + 1 = 3 \Rightarrow K = 2$

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sech−1(23)=log((K+1)+K2+1K) Then K=____

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${sech}^{- 1} (\frac{2}{3}) = \log (\frac{(K + 1) + \sqrt{K^{2} + 1}}{K})$ Then $K =____$