(c) \( 2^{\log _{2 \sqrt{2}} 27} \cdot L \)

To simplify the expression \( 2^{\log_{2 \sqrt{2}} 27} \cdot L \), we can start by simplifying the logarithmic part. 1. **Change of Base Formula**: We can use the change of base formula for logarithms, which states that: \[ \log_a b = \frac{\log_c b}{\log_c a} \] In this case, we will change the base from \( 2 \sqrt{2} \) to \( 2 \). 2. **Calculate \( \log_{2 \sqrt{2}} 27 \)**: \[ \log_{2 \sqrt{2}} 27 = \frac{\log_2 27}{\log_2 (2 \sqrt{2})} \] 3. **Calculate \( \log_2 (2 \sqrt{2}) \)**: \[ \log_2 (2 \sqrt{2}) = \log_2 2 + \log_2 \sqrt{2} = 1 + \frac{1}{2} = \frac{3}{2} \] 4. **Calculate \( \log_2 27 \)**: \[ \log_2 27 = \log_2 (3^3) = 3 \log_2 3 \] 5. **Substituting back**: \[ \log_{2 \sqrt{2}} 27 = \frac{3 \log_2 3}{\frac{3}{2}} = 2 \log_2 3 \] 6. **Substituting into the original expression**: \[ 2^{\log_{2 \sqrt{2}} 27} = 2^{2 \log_2 3} \] 7. **Using the property of exponents**: \[ 2^{2 \log_2 3} = (2^{\log_2 3})^2 = 3^2 = 9 \] 8. **Final expression**: \[ 2^{\log_{2 \sqrt{2}} 27} \cdot L = 9 \cdot L \] Thus, the simplified expression is: \[ 9L \]