Answer
$$sech^{-1}\Big(\frac{3}{5}\Big)=\ln3$$
Work Step by Step
We have $$sech^{-1}x=\ln\Big(\frac{1+\sqrt{1-x^2}}{x}\Big)$$ for $0\lt x\le1$
Therefore, $$sech^{-1}\Big(\frac{3}{5}\Big)=\ln\Big(\frac{1+\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}\Big)$$ $$=\ln\Big(\frac{1+\sqrt{\frac{16}{25}}}{\frac{3}{5}}\Big)=\ln\Big(\frac{1+\frac{4}{5}}{\frac{3}{5}}\Big)$$ $$=\ln\frac{\frac{9}{5}}{\frac{3}{5}}=\ln\frac{9}{3}=\ln3$$
