1. Problem: Simplify expression 8.1: $\frac{(\log 3 - \log 5)(\log 2 + \log 5)}{\log 9 - \log 25}$. 2. Use logarithm properties: $\log a - \log b = \log \frac{a}{b}$ and $\log a + \log b = \log (ab)$. 3. Simplify numerator: $$(\log 3 - \log 5)(\log 2 + \log 5) = \log \frac{3}{5} \cdot \log (2 \times 5) = \log \frac{3}{5} \cdot \log 10$$ 4. Simplify denominator: $$\log 9 - \log 25 = \log \frac{9}{25}$$ 5. Write the entire expression: $$\frac{\log \frac{3}{5} \cdot \log 10}{\log \frac{9}{25}}$$ 6. Note that $\log 10 = 1$ (assuming common logarithm base 10), so simplify numerator to: $$\log \frac{3}{5}$$ 7. The expression is now: $$\frac{\log \frac{3}{5}}{\log \frac{9}{25}}$$ 8. Note that $\frac{9}{25} = \left(\frac{3}{5}\right)^2$, so: $$\log \frac{9}{25} = \log \left(\frac{3}{5}\right)^2 = 2 \log \frac{3}{5}$$ 9. Substitute back: $$\frac{\log \frac{3}{5}}{2 \log \frac{3}{5}} = \frac{1}{2}$$ 10. Therefore, simplified expression 8.1 equals $\boxed{\frac{1}{2}}$. --- 1. Problem: Simplify expression 8.2: $$\frac{3 \log 5 \cdot \log 8 + \log 8^2}{\log 64 - \log 4}$$ 2. Recognize $\log 8^2 = 2 \log 8$. 3. Rewrite numerator: $$3 \log 5 \cdot \log 8 + 2 \log 8 = \log 8 (3 \log 5 + 2)$$ 4. Simplify denominator using the subtraction property: $$\log 64 - \log 4 = \log \frac{64}{4} = \log 16$$ 5. Now expression is: $$\frac{\log 8 (3 \log 5 + 2)}{\log 16}$$ 6. Write the logs as powers of 2 since $8 = 2^3$, $16 = 2^4$: $$\log 8 = \log 2^3 = 3 \log 2$$ $$\log 16 = \log 2^4 = 4 \log 2$$ 7. Substitute these: $$\frac{3 \log 2 (3 \log 5 + 2)}{4 \log 2}$$ 8. Cancel $\log 2$: $$\frac{3 (3 \log 5 + 2)}{4}$$ 9. Expand numerator: $$3 (3 \log 5 + 2) = 9 \log 5 + 6$$ 10. Final simplified expression: $$\frac{9 \log 5 + 6}{4}$$ 11. If desired, leave as is or compute numerical value depending on base. Final answers: - 8.1 simplified: $\frac{1}{2}$ - 8.2 simplified: $\frac{9 \log 5 + 6}{4}$