College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.4 - Logarithmic Functions - 6.4 Assess Your Understanding - Page 449: 30



Work Step by Step

Since $9=3^2$, the given expression is equivalent to: $\log_3{(\frac{1}{3^2})}$ Using the rule $\frac{1}{a^m} = a^{-m}, a\ne0$ gives: $\log_3{(\frac{1}{3^2})}=\log_3{(3^{-2})}$ RECALL: $\log_a{(a^n)} = n, a \gt 0, a \ne1$ Using the property above gives: $\log_3{(3^{-2})} = -2$ Thus, $\log_3{(\frac{1}{9})} = -2$.
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