Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises - Page 325: 18

Answer

a.$\ \ \ -1.3862$ b.$\ \ \ 3.1779$ c.$\ \ \ 0.8283$ d.$\ \ \ -4.2765$

Work Step by Step

a. $0.25=\displaystyle \frac{1}{4}=\frac{1}{2^{2}}=2^{-2}$ $\ln 2^{-2}=(-2)\ln 2\approx-2(0.6931)=-1.3862$ b. $24=8\cdot 3=2^{3}\cdot 3$ $\ln(2^{3}\cdot 3)=\ln(2^{3})+\ln(3)=3\ln 2+\ln 3$ $\approx 3(0.6931)+1.0986=3.1779$ c. $\sqrt[3]{12}=12^{1/3}=(4\cdot 3)^{1/3}=(2^{2}\cdot 3)^{1/3}=2^{2/3}\cdot 3^{1/3}$ $\displaystyle \ln(2^{2/3}\cdot 3^{1/3})=\ln(2^{2/3} )+\ln(3^{1/3})=\frac{2}{3}\ln 2+\frac{1}{3}\ln 3$ $\displaystyle \approx\frac{2}{3}(0.6931)+\frac{1}{3}(1.0986)\approx 0.8283$ d. $\displaystyle \frac{1}{72}=72^{-1}=(8\cdot 9)^{-1}=(2^{3}\cdot 3^{2})^{-1}=2^{-3}\cdot 3^{-2}$ $\ln(2^{-3}\cdot 3^{-2})=-3\ln 2+(-2)\ln 3$ $\approx-3(0.6931)-2(1.0986)=-4.2765$
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