Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Section 3.2 - Exponential Functions - Exercise Set: 28

Answer

$-2$

Work Step by Step

RECALL: $\log_b{b^x} = x.$ Note that $\dfrac{1}{9}=\dfrac{1}{3^3} = 3^{-2}.$ Thus, the given expression can be written as $\log_3{3^{-2}}.$ Use the rule above to obtain $-2.$
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