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: 22



Work Step by Step

We know that $\log_a {x}-\log_a {y}=\log_a {\left(\frac{x}{y}\right)}$. Hence, $\log_8 {16}-\log_8 {2} \\=\log_8{\left(\frac{16}{2}\right)}\\ =\log_8 {8}\\ =\log_8 {8^1}$. We know that $\log_a {a^x}=x$. Thus, $\log_8 {8^1}=1.$
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