1. **Evaluate the logarithms:** (1) Find $\log_{81} \frac{1}{3}$. Recall that $\log_a b = c$ means $a^c = b$. Since $81 = 3^4$, rewrite the base: $$\log_{3^4} 3^{-1} = x \implies (3^4)^x = 3^{-1}$$ Simplify the left side: $$3^{4x} = 3^{-1}$$ Equate exponents: $$4x = -1 \implies x = -\frac{1}{4}$$ So, $\log_{81} \frac{1}{3} = -\frac{1}{4}$. (2) Find $\log_{125} 5$. Rewrite $125 = 5^3$: $$\log_{5^3} 5 = y \implies (5^3)^y = 5$$ Simplify: $$5^{3y} = 5^1$$ Equate exponents: $$3y = 1 \implies y = \frac{1}{3}$$ So, $\log_{125} 5 = \frac{1}{3}$. (3) Find $\log_{13} \frac{1}{13}$. Rewrite $\frac{1}{13} = 13^{-1}$: $$\log_{13} 13^{-1} = z \implies 13^z = 13^{-1}$$ Equate exponents: $$z = -1$$ So, $\log_{13} \frac{1}{13} = -1$. 2. **Condense the expressions and find their values:** (4) Simplify $6 \log 2 + 2 \log 12 - 3 \log 4$. Use the power rule: $a \log b = \log b^a$: $$6 \log 2 = \log 2^6$$ $$2 \log 12 = \log 12^2$$ $$3 \log 4 = \log 4^3$$ Rewrite expression: $$\log 2^6 + \log 12^2 - \log 4^3$$ Use product and quotient rules: $$\log \left( \frac{2^6 \times 12^2}{4^3} \right)$$ Calculate powers: $$2^6 = 64, \quad 12^2 = 144, \quad 4^3 = 64$$ Substitute: $$\log \left( \frac{64 \times 144}{64} \right) = \log 144$$ Since $\log 144$ is base 10, approximate or leave as is. (5) Simplify $2(\log_3 10 - \log_3 5) + \frac{1}{2} \log_3 36$. Use difference rule: $$\log_3 10 - \log_3 5 = \log_3 \frac{10}{5} = \log_3 2$$ Multiply by 2: $$2 \log_3 2 = \log_3 2^2 = \log_3 4$$ Simplify $\frac{1}{2} \log_3 36$: $$\frac{1}{2} \log_3 36 = \log_3 36^{1/2} = \log_3 6$$ Sum: $$\log_3 4 + \log_3 6 = \log_3 (4 \times 6) = \log_3 24$$ (6) Simplify $\frac{1}{2} \log_7 25 + \frac{1}{4} \log_7 16$. Apply power rule: $$\frac{1}{2} \log_7 25 = \log_7 25^{1/2} = \log_7 5$$ $$\frac{1}{4} \log_7 16 = \log_7 16^{1/4} = \log_7 2$$ Sum: $$\log_7 5 + \log_7 2 = \log_7 (5 \times 2) = \log_7 10$$ **Final answers:** (1) $-\frac{1}{4}$ (2) $\frac{1}{3}$ (3) $-1$ (4) $\log 144$ (5) $\log_3 24$ (6) $\log_7 10$