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: 32



Work Step by Step

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