Calculate: $$\log_4 3 \cdot \log_3 \sqrt[5]{2}$$ $$\log_{81} 5 \cdot \log_5 \sqrt[8]{3}$$

Okay, let's evaluate the expressions. Part 1: $$\log_4 3 \cdot \log_3 \sqrt[5]{2}$$ We will use the change of base formula: $$\log_a b = \frac{\log_c b}{\log_c a}$$. Also remember that $$\sqrt[n]{a} = a^{\frac{1}{n}}$$. So, we have: $$\log_4 3 \cdot \log_3 \sqrt[5]{2} = \frac{\log 3}{\log 4} \cdot \frac{\log \sqrt[5]{2}}{\log 3} = \frac{\log 3}{\log 4} \cdot \frac{\log 2^{\frac{1}{5}}}{\log 3}$$ $$\frac{\log 3}{\log 4} \cdot \frac{\frac{1}{5} \log 2}{\log 3} = \frac{1}{5} \cdot \frac{\log 2}{\log 4} = \frac{1}{5} \cdot \frac{\log 2}{\log 2^2} = \frac{1}{5} \cdot \frac{\log 2}{2 \log 2} = \frac{1}{5} \cdot \frac{1}{2} = \frac{1}{10}$$ Part 2: $$\log_{81} 5 \cdot \log_5 \sqrt[8]{3}$$ $$\log_{81} 5 \cdot \log_5 \sqrt[8]{3} = \frac{\log 5}{\log 81} \cdot \frac{\log \sqrt[8]{3}}{\log 5} = \frac{\log 5}{\log 3^4} \cdot \frac{\log 3^{\frac{1}{8}}}{\log 5}$$ $$\frac{\log 5}{4 \log 3} \cdot \frac{\frac{1}{8} \log 3}{\log 5} = \frac{1}{4} \cdot \frac{1}{8} \cdot \frac{\log 3}{\log 3} = \frac{1}{32}$$ Answer: $$\log_4 3 \cdot \log_3 \sqrt[5]{2} = \frac{1}{10}$$ $$\log_{81} 5 \cdot \log_5 \sqrt[8]{3} = \frac{1}{32}$$