Directions: Condense, then use the change of base formula to evaluate the logarithm. 9. \( \log _{9} 35-\log _{9} 7 \) & \( 10 \cdot 2 \cdot \log _{3} 8-4 \cdot \log _{3} 2 \) 11. \( \frac{1}{3} \cdot \log _{4} 8+\log _{4} 15 \) & 12. \( \log _{2} 27+\log _{2} 4-2 \cdot \log _{2} 3 \)

To begin with the first expression \( \log_{9} 35 - \log_{9} 7 \), we can apply the property of logarithms to condense it: \[ \log_{9} \left( \frac{35}{7} \right) = \log_{9} 5. \] Using the change of base formula, we get: \[ \log_{9} 5 = \frac{\log_{10} 5}{\log_{10} 9}. \] For the second expression \( 10 \cdot 2 \cdot \log_{3} 8 - 4 \cdot \log_{3} 2 \), we can simplify it as follows: \[ 20 \cdot \log_{3} 8 - 4 \cdot \log_{3} 2 = 20 \cdot \log_{3} (2^3) - 4 \cdot \log_{3} 2 = 20 \cdot 3\log_{3} 2 - 4\log_{3} 2 = (60 - 4)\log_{3} 2 = 56\log_{3} 2. \] Now, let's look at the third expression \( \frac{1}{3}\log_{4} 8 + \log_{4} 15 \): First, condense the first logarithm: \[ \frac{1}{3}\log_{4} 8 = \log_{4} (8^{1/3}) = \log_{4} 2. \] Thus, the entire expression becomes: \[ \log_{4} 2 + \log_{4} 15 = \log_{4} (2 \cdot 15) = \log_{4} 30. \] Lastly, for \( \log_{2} 27 + \log_{2} 4 - 2\log_{2} 3 \): Condense the first two logarithms: \[ \log_{2} (27 \cdot 4) - 2\log_{2} 3 = \log_{2} 108 - \log_{2} (3^2) = \log_{2} \left( \frac{108}{9} \right) = \log_{2} 12. \] So, here are the evaluated logarithmic expressions: 1. \( \log_{9} 5 \) 2. \( 56\log_{3} 2 \) 3. \( \log_{4} 30 \) 4. \( \log_{2} 12 \)