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

$t={\mathrm{sech}}^{-1}\left(\frac{1}{2}\right)-{\mathrm{cosech}}^{-1}\left(\frac{3}{k}\right)$

$\Rightarrow t=\ln \left(\frac{1}{\left({\displaystyle \frac{1}{2}}\right)}+\sqrt{\frac{1}{{\left({\displaystyle \frac{1}{2}}\right)}^2}-1}\right)-\ln \left(\frac{1}{\left({\displaystyle \frac{3}{k}}\right)}+\sqrt{\frac{1}{{\left({\displaystyle \frac{3}{k}}\right)}^2}+1}\right)$

$\Rightarrow t=\ln \left(2+\sqrt{3}\right)-\ln \left(\frac{k}{3}+\frac{\sqrt{k^2+9}}{3}\right) .....\left(i\right)$

Given, $3e^t=2+\sqrt{3}$

$\Rightarrow \ln \left(3e^t\right)=\ln \left(2+\sqrt{3}\right)$

$\Rightarrow \ln 3+t=\ln \left(2+\sqrt{3}\right)$

$\Rightarrow \ln 3+\ln \left(2+\sqrt{3}\right)-\ln \left(\frac{k}{3}+\frac{\sqrt{k^2+9}}{3}\right)=\ln \left(2+\sqrt{3}\right)$

$\Rightarrow \ln \left(\frac{k+\sqrt{k+9}}{9}\right)=0$

$\Rightarrow k+\sqrt{k^2+9}=9$

$\Rightarrow k^2+9=81+k^2-18k$

$\Rightarrow k=4$