\begin{tabular}{|ll} Assessment activity 3.7 \\ Calculate the following without using a calculator: \\ \( \begin{array}{ll}\text { 1. } \log _{5} 125 & \text { 2. } \log _{3} 9 \\ \text { 3. } \log _{9} 3 & \text { 4. } \log _{2} \sqrt{2} \\ \text { 5. } \log _{0,3} 0,027 & \text { 6. } \log _{4} \frac{1}{64} \\ \text { 7. } \log 20+\log 50 & \text { 8. } \log _{250} 25 \log 2-\log 5 \\ \text { 9. } \log 0,001+\log 1+2^{\log 0.1} & \text { 10. } \log _{2} \sqrt{32}-\log _{3} \sqrt{27} \\ \text { 11. } \log _{2} \frac{1}{2}-\log _{2} \frac{1}{16}+\log _{2} \frac{1}{4} & \text { 12. } \log _{2}\left(\log _{3} 81\right) \\ \text { 13. } \log _{5} 2,5-\log _{5} 0,5 & \text { 14. } \log _{3} 6,75+\log _{3} 4\end{array} \)\end{tabular}

Let's solve each logarithmic expression step by step. ### 1. \( \log_{5} 125 \) We know that \( 125 = 5^3 \). Therefore: \[ \log_{5} 125 = \log_{5} (5^3) = 3 \] ### 2. \( \log_{3} 9 \) We know that \( 9 = 3^2 \). Therefore: \[ \log_{3} 9 = \log_{3} (3^2) = 2 \] ### 3. \( \log_{9} 3 \) We can express \( 9 \) as \( 3^2 \): \[ \log_{9} 3 = \log_{3^2} 3 = \frac{1}{2} \log_{3} 3 = \frac{1}{2} \cdot 1 = \frac{1}{2} \] ### 4. \( \log_{2} \sqrt{2} \) We know that \( \sqrt{2} = 2^{1/2} \). Therefore: \[ \log_{2} \sqrt{2} = \log_{2} (2^{1/2}) = \frac{1}{2} \] ### 5. \( \log_{0.3} 0.027 \) We can express \( 0.027 \) as \( 27^{-1} = (3^3)^{-1} = 3^{-3} \) and \( 0.3 = 3^{-1} \cdot 10^{-1} \): \[ \log_{0.3} 0.027 = \log_{3^{-1} \cdot 10^{-1}} 3^{-3} \] Using the change of base formula: \[ \log_{0.3} 0.027 = \frac{\log_{3} 3^{-3}}{\log_{3} (3^{-1} \cdot 10^{-1})} = \frac{-3}{-1 + \log_{3} 10} \] This requires numerical approximation for \( \log_{3} 10 \), but we can simplify it as: \[ \log_{0.3} 0.027 = \frac{3}{1 - \log_{3} 10} \] ### 6. \( \log_{4} \frac{1}{64} \) We know that \( 64 = 4^3 \). Therefore: \[ \log_{4} \frac{1}{64} = \log_{4} (4^{-3}) = -3 \] ### 7. \( \log 20 + \log 50 \) Using the property of logarithms: \[ \log 20 + \log 50 = \log (20 \cdot 50) = \log 1000 = 3 \] ### 8. \( \log_{250} 25 \log 2 - \log 5 \) We know that \( 25 = 5^2 \) and \( 250 = 5^3 \): \[ \log_{250} 25 = \frac{\log 25}{\log 250} = \frac{2 \log 5}{3 \log 5} = \frac{2}{3} \] Thus: \[ \log_{250} 25 \log 2 - \log 5 = \frac{2}{3} \log 2 - \log 5 \] ### 9. \( \log 0.001 + \log 1 + 2^{\log 0.1} \) We know that \( \log 0.001 = -3 \) and \( \log 1 = 0 \): \[ 2^{\log 0.1} = 2^{-1} = \frac{1}{2} \] Thus: \[ \log 0.001 + \log 1 + 2^{\log 0.1} = -3 + 0 + \frac{1}{2} = -3 + 0.5 = -2.5 \] ### 10. \( \log_{2} \sqrt{32} - \log_{3} \sqrt{27} \) We know that \( \sqrt{32} = 32^{1/2} = 2^{5/2} \) and \( \sqrt{27} = 27^{1/2} = 3^{3/2} \): \[ \log_{2} \sqrt{32} = \frac{5}{2}, \quad \log_{3} \sqrt{27} = \frac{3}{2} \] Thus: \[ \log_{2} \sqrt{32} - \log_{3} \sqrt{27} = \frac{5}{2} - \frac{3}{2} = 1 \] ### 11. \( \log_{2} \frac{1}{2} - \log_{2} \frac{1}{16} + \log_{2} \frac{1}{4} \) We know that: \[ \log_{2} \frac{1}{2} = -1, \quad \log_{2} \frac{1}{16} = -4, \quad \log_{2} \frac{1}{4} = -2 \] Thus: \[ -1 - (-4) - 2 = -1 + 4 - 2 = 1 \] ### 12. \( \log_{2}(\log_{3} 81) \) We know that \( 81 = 3^4 \): \[ \log_{3} 81 = 4 \quad \Rightarrow \quad \log_{2}(\log_{3} 81) = \log_{2} 4 = 2 \] ### 13. \( \log_{5} 2.5 - \log_{5} 0.5 \) Using the property of logarithms: \[ \log_{5} 2.5 - \log_{5} 0.5 = \log_{5} \left(\frac{2.5}{0.5}\right) = \log_{5} 5 = 1 \] ### 14. \( \log_{3} 6.75 + \log_{3} 4 \) Using the property of logarithms: \[ \log_{3} 6.75 + \log_{3} 4 = \log_{3} (6.75 \cdot 4) = \log_{3} 27 = 3 \] ### Summary of Results 1. \( 3 \) 2. \( 2 \) 3. \( \frac{1}{2} \) 4. \( \frac{1}{2