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: 23



Work Step by Step

We know that $\log_a {b}=\dfrac{\log_c {b}}{\log_c {a}}$. Hence, $\log_6 {8}=\dfrac{\log_2 {8}}{\log_2 {6}}$. Thus, $\log_2 {6}\cdot \log_6 {8} \\=\log_2 {6}\cdot \dfrac{\log_2 {8}}{\log_2 {6}} \\=\log_2 {8} \\=\log_2 {2^3}$ Use the rule $\log_a{a^x}=x$ to obtain: $\log_2{2^3} = 3$
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