## University Calculus: Early Transcendentals (3rd Edition)

$$\cosh^{-1}\Big(\frac{5}{3}\Big)=\ln3$$
We have $$\cosh^{-1}x=\ln(x+\sqrt{x^2-1})$$ for $x\in[1,\infty)$ Therefore, $$\cosh^{-1}\Big(\frac{5}{3}\Big)=\ln\Big(\frac{5}{3}+\sqrt{\frac{25}{9}-1}\Big)$$ $$=\ln\Big(\frac{5}{3}+\sqrt{\frac{16}{9}}\Big)$$ $$=\ln\Big(\frac{5}{3}+\frac{4}{3}\Big)$$ $$=\ln\frac{9}{3}=\ln3$$