## Algebra: A Combined Approach (4th Edition)

Published by Pearson

# Chapter 12 - Section 12.6 - Properties of Logarithms - Exercise Set - Page 880: 14

#### Answer

$log_{3}\frac{12}{z}$

#### Work Step by Step

$\div$We know that $log_{b}\frac{x}{y}=log_{b}x-log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$). Therefore, $log_{3}12-log_{3}z=log_{3}\frac{12}{z}$.

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