1. **Problem Statement:** Evaluate each logarithm expression given. 2. **Formula and Rules:** Recall that $\log_a b = c$ means $a^c = b$. 3. **Step-by-step evaluation:** - a) $\log_6 36$: Since $36 = 6^2$, $\log_6 36 = 2$. - b) $\log_2 32$: Since $32 = 2^5$, $\log_2 32 = 5$. - c) $\log_{10} 0.1$: Since $0.1 = 10^{-1}$, $\log_{10} 0.1 = -1$. - d) $\log_6 \frac{1}{36}$: Since $\frac{1}{36} = 6^{-2}$, $\log_6 \frac{1}{36} = -2$. - e) $\log_3 \sqrt{3}$: Since $\sqrt{3} = 3^{\frac{1}{2}}$, $\log_3 \sqrt{3} = \frac{1}{2}$. - f) $\log_5 1$: Since $5^0 = 1$, $\log_5 1 = 0$. - g) $\log_3 \frac{1}{\sqrt{3}}$: Since $\frac{1}{\sqrt{3}} = 3^{-\frac{1}{2}}$, $\log_3 \frac{1}{\sqrt{3}} = -\frac{1}{2}$. - h) $\log_8 64$: Since $64 = 8^{\frac{2}{3}}$ (because $8 = 2^3$ and $64 = 2^6$, so $64 = 8^{6/3} = 8^2$ is incorrect, correct is $8^{\frac{2}{3}}$?), let's check carefully: $8 = 2^3$, $64 = 2^6$, so $64 = (2^3)^x = 8^x = 2^{3x}$, so $3x = 6 \Rightarrow x=2$, so $\log_8 64 = 2$. - i) $\log_{64} 8$: Since $64 = 8^2$, so $8 = 64^{\frac{1}{2}}$, thus $\log_{64} 8 = \frac{1}{2}$. - j) $\log_9 27$: Since $9 = 3^2$, $27 = 3^3$, so $\log_9 27 = \log_{3^2} 3^3 = \frac{3}{2}$. - k) $\log_{27} \frac{1}{9}$: Since $27 = 3^3$, $9 = 3^2$, so $\frac{1}{9} = 3^{-2}$, thus $\log_{27} \frac{1}{9} = \log_{3^3} 3^{-2} = \frac{-2}{3}$. 4. **Final answers:** - a) 2 - b) 5 - c) -1 - d) -2 - e) $\frac{1}{2}$ - f) 0 - g) $-\frac{1}{2}$ - h) 2 - i) $\frac{1}{2}$ - j) $\frac{3}{2}$ - k) $-\frac{2}{3}$