Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - 5.5 Properties of Logarithms - 5.5 Assess Your Understanding - Page 305: 19



Work Step by Step

Recall: $\log _{a}m+\log _{a}n=\log _{a}\left( m\times n\right)$ Thus, $\log _{8}2+\log _{8}4=\log _{8}\left( 2\times 4\right) =\log _{8}8$ Since $\log_a{a}=1$, then $\log_8{8}=1$
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