University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 418: 62



Work Step by Step

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