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: 65



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, $$sech^{-1}\Big(\frac{3}{5}\Big)=\ln\Big(\frac{1+\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}\Big)$$ $$=\ln\Big(\frac{1+\sqrt{\frac{16}{25}}}{\frac{3}{5}}\Big)=\ln\Big(\frac{1+\frac{4}{5}}{\frac{3}{5}}\Big)$$ $$=\ln\frac{\frac{9}{5}}{\frac{3}{5}}=\ln\frac{9}{3}=\ln3$$
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