## College Algebra 7th Edition

(a) $-3$ (b) $\dfrac{1}{2}$ (c) $-1$
RECALL: (1) $\log_b{b}=1$ (2) $\log_b{1}=0$ (3) $\log_b{b^x}=x$ (a) $\frac{1}{27}$ is equivalent to $\frac{1}{3^3}$ so the given expression can be written as: $=\log_3{(\frac{1}{3^3})}$ Since $\dfrac{1}{a^m} = a^{-m}$, the given expression is equivalent to: $=\log_3{(3^{-3})}$ Use rule (3) above to obtain: $=-3$ (b) RECALL: $\sqrt{a} = a^{\frac{1}{2}}$ Thus, the given expression is equivalent to: $=\log_{10}{(10^{\frac{1}{2}})}$ Use rule (3) above to obtain: $=\dfrac{1}{2}$ (c) Note that $0.2 = \dfrac{1}{5}$. Use the rule $\dfrac{1}{a^m} = a^{-m}$ to obtain: $\dfrac{1}{5} = 5^{-1}$ This means that the given expression is equivalent to: $=\log_5{(5^{-1})}$ Use rule (3) above to obtain: $=-1$