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 - Page 463: 32



Work Step by Step

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