Q4: Write \( \log _{4} 2-\log _{4} 5+\log _{4} 40 \) as a single logarithm.

To combine the expression \( \log_{4} 2 - \log_{4} 5 + \log_{4} 40 \) into a single logarithm, we can use the properties of logarithms. 1. **Use the Quotient Rule**: The difference of logarithms can be expressed as the logarithm of a quotient: \[ \log_{4} 2 - \log_{4} 5 = \log_{4} \left( \frac{2}{5} \right) \] 2. **Use the Product Rule**: The sum of logarithms can be expressed as the logarithm of a product: \[ \log_{4} \left( \frac{2}{5} \right) + \log_{4} 40 = \log_{4} \left( \frac{2}{5} \cdot 40 \right) \] 3. **Calculate the product**: \[ \frac{2}{5} \cdot 40 = \frac{2 \cdot 40}{5} = \frac{80}{5} = 16 \] 4. **Combine the logarithms**: \[ \log_{4} \left( \frac{2}{5} \cdot 40 \right) = \log_{4} 16 \] 5. **Express \( 16 \) as a power of \( 4 \)**: \[ 16 = 4^2 \] 6. **Use the Power Rule**: This allows us to simplify: \[ \log_{4} 16 = \log_{4} (4^2) = 2 \] Thus, the expression \( \log_{4} 2 - \log_{4} 5 + \log_{4} 40 \) simplifies to: \[ \log_{4} 16 = 2 \] Therefore, the final answer is: \[ \log_{4} 2 - \log_{4} 5 + \log_{4} 40 = \log_{4} 16 = 2 \]