1. Stating the problem: Calculate the values of the following expressions using logarithm properties. 2. Recall important logarithm properties: - $\log_a 1 = 0$ for any base $a$. - $\log_a (a^k) = k$. - $\log_a (xy) = \log_a x + \log_a y$. - $\log_a \left(\frac{x}{y}\right) = \log_a x - \log_a y$. - $k \log_a x = \log_a (x^k)$. - Change of base formula: $\log_a b = \frac{\log_c b}{\log_c a}$. 3. Solve each: a) $\log_3 \left(\frac{1}{81}\right) = \log_3 (3^{-4}) = -4$. b) $2^{3 \log_2 3} = 2^{\log_2 3^3} = 2^{\log_2 27} = 27$. c) $6^{1 + \log_6 2} = 6^1 \cdot 6^{\log_6 2} = 6 \cdot 2 = 12$. d) $\lg 8 + \lg 125 = \lg (8 \times 125) = \lg 1000 = 3$. e) $\frac{1}{2} \lg 16 + 2 \lg 5 = \lg 16^{\frac{1}{2}} + \lg 5^2 = \lg 4 + \lg 25 = \lg (4 \times 25) = \lg 100 = 2$. f) $\log_7 196 - 2 \log_7 2 = \log_7 (196) - \log_7 (2^2) = \log_7 \left(\frac{196}{4}\right) = \log_7 49 = 2$. g) $\frac{1}{\log_5 7} = \log_7 5$ (by change of base formula). h) $\frac{\lg 27 + \lg 12}{\lg 2 + 2 \lg 3} = \frac{\lg (27 \times 12)}{\lg 2 + \lg 3^2} = \frac{\lg 324}{\lg (2 \times 9)} = \frac{\lg 324}{\lg 18}$. Since $324 = 18^2$, $\lg 324 = \lg (18^2) = 2 \lg 18$, so the expression equals $\frac{2 \lg 18}{\lg 18} = 2$. Final answers: a) $-4$ b) $27$ c) $12$ d) $3$ e) $2$ f) $2$ g) $\log_7 5$ h) $2$