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: 1

Answer

(a) $\ln 3-2\ln 2$ (b) $2\ln 2-3\ln 3 $ (c) $-\ln 2 $ (d) $\frac{3}{2}\ln 3$ (e) $\ln 3+\frac{1}{2}\ln 2$ (f) $ \frac{1}{2}\left(3\ln3-\ln 2 \right)$

Work Step by Step

(a) Since \begin{align*} \ln (0.75) &= \ln \frac{3}{4}\\ &=\ln 3-\ln 4\\ &=\ln 3-2\ln 2 \end{align*} (b ) \begin{align*} \ln \frac{4}{9}&=\ln 4-\ln 9\\ &=2\ln 2-3\ln 3 \end{align*} (c) \begin{align*} \ln \frac{1}{2}&=\ln 1-\ln 2\\ &=-\ln 2 \end{align*} (d) \begin{align*} \ln\sqrt[3]{9}&=\ln 3^{2/3}\\ &=\frac{2}{3}\ln 3 \end{align*} (e) \begin{align*} \ln 3\sqrt{2}&=\ln 3+\ln \sqrt{2}\\ &=\ln 3+\frac{1}{2}\ln 2 \end{align*} (f) \begin{align*} \ln \sqrt{13.5}&=\frac{1}{2}\ln \frac{27}{2}\\ &=\frac{1}{2}\left(3\ln3-\ln 2 \right) \end{align*}
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