Let k0 and t=operatorname{sech}^{-1}left(fra | vedclass

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(B) Given $t=\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{k}\right)$.Using the logarithmic forms $\operat

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$4$

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(B) Given $t=\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{k}\right)$.Using the logarithmic forms $\operatorname{sech}^{-1}(x)=\ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ and $\operatorname{cosech}^{-1}(x)=\ln\left(\frac{1+\sqrt{1+x^2}}{x}\right)$:$t=\ln\left(\frac{1+\sqrt{1-(1/2)^2}}{1/2}\right)-\ln\left(\frac{1+\sqrt{1+(3/k)^2}}{3/k}\right)$$t=\ln(2+\sqrt{3})-\ln\left(\frac{k+\sqrt{k^2+9}}{3}\right)$.Given $3e^t=2+\sqrt{3}$,we have $e^t=\frac{2+\sqrt{3}}{3}$.Taking the natural logarithm on both sides: $t=\ln\left(\frac{2+\sqrt{3}}{3}\right)$.Equating the two expressions for $t$:$\ln\left(\frac{2+\sqrt{3}}{3}\right)=\ln\left(\frac{2+\sqrt{3}}{\frac{k+\sqrt{k^2+9}}{3}}\right)$.This implies $\frac{k+\sqrt{k^2+9}}{3}=3$,so $k+\sqrt{k^2+9}=9$.$\sqrt{k^2+9}=9-k$.Squaring both sides: $k^2+9=81+k^2-18k$.$18k=72$,which gives $k=4$.

Let $k>0$ and $t=\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{k}\right)$. If $3 e^t=2+\sqrt{3}$,then $k=$

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