Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 7 - Functions - Exercise Set 7.2 - Page 415: 32

Answer

$a-\\ \log_{8}27=\frac{\log_{2}27}{\log_{2}8}=\frac{\log_{2}3^{3}}{\log_{2}2^{3}}=\frac{3\log_{2}3}{3\log_{2}2} =\frac{\log_{2}3}{\log_{2}2}=\frac{\log_{2}3}{1}=\log_{2}3 \\ b-\\ \log_{16}9=\frac{\log_{4}9}{\log_{4}16}=\frac{\log_{4}3^{2}}{\log_{4}4^{2}}=\frac{2\log_{4}3}{2\log_{4}4} =\frac{\log_{4}3}{\log_{4}4}=\frac{\log_{4}3}{1}=\log_{4}3 \\ (we\,use\,\log_{y}x=\frac{\log_{a}x}{\log_{a}y}\\\log_{y}x^{n}=n\log_{y}x \\\log_{x}x=1) $

Work Step by Step

$a-\\ \log_{8}27=\frac{\log_{2}27}{\log_{2}8}=\frac{\log_{2}3^{3}}{\log_{2}2^{3}}=\frac{3\log_{2}3}{3\log_{2}2} =\frac{\log_{2}3}{\log_{2}2}=\frac{\log_{2}3}{1}=\log_{2}3 \\ b-\\ \log_{16}9=\frac{\log_{4}9}{\log_{4}16}=\frac{\log_{4}3^{2}}{\log_{4}4^{2}}=\frac{2\log_{4}3}{2\log_{4}4} =\frac{\log_{4}3}{\log_{4}4}=\frac{\log_{4}3}{1}=\log_{4}3 \\ (we\,use\,\log_{y}x=\frac{\log_{a}x}{\log_{a}y}\\\log_{y}x^{n}=n\log_{y}x \\\log_{x}x=1) $
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