Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 2

Answer

(a) $-3\ln 5$ (b) $2\ln 7 -\ln 5$ (c) $\frac{3}{2} \ln7$ (d) $2\ln5 + 2\ln 7$ (e) $\ln 7 - 3\ln 5$ (f) $\frac{1}{2}$

Work Step by Step

(a) $\ln \frac{1}{125}= -\ln 125$ $= -\ln 5^{3}$ $= -3\ln5$ (b) $\ln 9.8=\ln \frac{49}{5}$ $= \ln 49-\ln 5$ $=2\ln 7 - \ln 5$ (c) $\ln 7\sqrt{7}$ $= \ln 7 + \ln \sqrt 7$ $= \ln 7+\ln 7^{1/2}$ $=\ln 7+\frac{1}{2}\ln 7$ $=\frac{3}{2} \ln7$ (d) $\ln 1225= \ln 25 + \ln 49$ $= 2\ln 5 + 2\ln 7$ (e) $\ln 0.056= \ln \frac{7}{125}$ $= \ln 7 - \ln 125$ $=\ln 7-\ln 5^3$ $=\ln 7-3\ln 5$ (f) $\frac{\ln 35 + \ln \frac{1}{7}}{\ln 25}= \frac{\ln 5 + \ln 7 - \ln 7}{2\ln 5}$ $=\frac{\ln 5}{2\ln 5}$ $=\frac{1}{2}$
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