1. **State the problem:** Find the values of the logarithms for the given expressions. 2. **Recall logarithm rules:** - $\log_a a = 1$ - $\log_a 1 = 0$ - $\log_a (a^k) = k$ - Change of base and reciprocal properties apply. 3. **Calculate each:** (a) $\log_3 3 = 1$ because the base and argument are equal. (b) $\log_3 1 = 0$ because any log base of 1 is zero. (c) $\log_3 3^4 = 4$ by the power rule. (d) $\log_{1/27} 3$: Note $1/27 = 3^{-3}$, so $$\log_{3^{-3}} 3 = \frac{\log_3 3}{\log_3 3^{-3}} = \frac{1}{-3} = -\frac{1}{3}$$ (e) $\log_{\sqrt{2}} 16$: Note $\sqrt{2} = 2^{1/2}$ and $16 = 2^4$, so $$\log_{2^{1/2}} 2^4 = \frac{4}{1/2} = 8$$ (f) $\log_4 (1/4)$: Note $1/4 = 4^{-1}$, so $$\log_4 4^{-1} = -1$$ (g) $\log_{1/5} 5$: Note $1/5 = 5^{-1}$, so $$\log_{5^{-1}} 5 = \frac{\log_5 5}{\log_5 5^{-1}} = \frac{1}{-1} = -1$$ (h) $\log_{1/16} 64$: Note $1/16 = 16^{-1} = (2^4)^{-1} = 2^{-4}$ and $64 = 2^6$, so $$\log_{2^{-4}} 2^6 = \frac{6}{-4} = -\frac{3}{2}$$ (i) $\log_{1/3} 9$: Note $1/3 = 3^{-1}$ and $9 = 3^2$, so $$\log_{3^{-1}} 3^2 = \frac{2}{-1} = -2$$ (j) $\log_5 0.2$: Note $0.2 = \frac{1}{5} = 5^{-1}$, so $$\log_5 5^{-1} = -1$$ (k) $\log_{0.001} 0.1$: Note $0.001 = 10^{-3}$ and $0.1 = 10^{-1}$, so $$\log_{10^{-3}} 10^{-1} = \frac{-1}{-3} = \frac{1}{3}$$ (l) $\log_{10} \sqrt{10}$: Note $\sqrt{10} = 10^{1/2}$, so $$\log_{10} 10^{1/2} = \frac{1}{2}$$ (m) $\log_5 20 + \log_5 \frac{125}{4} - \log_5 \frac{1}{25}$: Use log addition/subtraction rules: $$= \log_5 \left(20 \times \frac{125}{4} \times 25\right)$$ Calculate inside: $$20 \times \frac{125}{4} = 20 \times 31.25 = 625$$ Then multiply by 25: $$625 \times 25 = 15625$$ Note $15625 = 5^6$ (since $5^6 = 15625$), so $$\log_5 5^6 = 6$$ **Final answers:** (a) 1 (b) 0 (c) 4 (d) $-\frac{1}{3}$ (e) 8 (f) $-1$ (g) $-1$ (h) $-\frac{3}{2}$ (i) $-2$ (j) $-1$ (k) $\frac{1}{3}$ (l) $\frac{1}{2}$ (m) 6