Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 67: 35


a)$\log_2{32}=5$ b)$\log_8{2}=\frac{1}{3}$

Work Step by Step

a)$\log_2{32}=\log_2{2^5}=5\log_2{2}=5$ b)$\log_8{2}=\log_8{8^\frac{1}{3}}=\frac{1}{3}\log_8{8}=\frac{1}{3}$
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