## College Algebra 7th Edition

$-3$
Note that $27=3^3$. Thus, $\log_3{(\frac{1}{27})}=\log_3{(\frac{1}{3^3})}$ Use the rule $\dfrac{1}{a^m} = a^{-m}$ to obtain: $\log_3{(\frac{1}{3^3})} = \log_3{(3^{-3})}$ RECALL: $\log_a{(a^n)} = n$ Use the rule above to obtain: $\log_3{(3^{-3})} = -3$