## Precalculus (10th Edition)

Published by Pearson

# Chapter 5 - Exponential and Logarithmic Functions - 5.5 Properties of Logarithms - 5.5 Assess Your Understanding - Page 305: 21

#### Answer

$1$

#### Work Step by Step

We know that $\log_a {x}-\log_a {y}=\log_a {\left(\frac{x}{y}\right)}$. Hence, $\log_6 {18}-\log_6 {3} \\=\log_6{\left(\frac{18}{3}\right)} \\=\log_6 {6} \\=\log_6 {6^1}$. We know that $\log_a {a^x}=x$. Thus, $\log_6 {6^1}=1$

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