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: 20



Work Step by Step

We know that $\log_a {x}+\log_a {y}=\log_a {(x\cdot y)}$. Hence, $\log_6 {4}+\log_6 {9} \\=\log_6{(4\cdot 9)} \\=\log_6{36} \\=log_6 {(6^2)}$. We know that $\log_a {a^x}=x$. Thus, $\log_6 {6^2}=2$
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