1. **State the problem:** Simplify the expression $$\left(81 \frac{1}{4} - \frac{1}{2} \log_9 4 + 25 \log_{125} 8\right) \cdot 49 \log_7 2$$. 2. **Rewrite mixed number:** Convert $$81 \frac{1}{4}$$ to an improper fraction or decimal. $$81 \frac{1}{4} = 81 + \frac{1}{4} = \frac{324}{4} + \frac{1}{4} = \frac{325}{4} = 81.25$$ 3. **Use logarithm change of base formula:** $$\log_a b = \frac{\log b}{\log a}$$ where the base of the logarithm on the right side can be any convenient base (e.g., 10 or e). 4. **Simplify each logarithm:** - $$\log_9 4 = \frac{\log 4}{\log 9}$$ - $$\log_{125} 8 = \frac{\log 8}{\log 125}$$ - $$\log_7 2 = \frac{\log 2}{\log 7}$$ 5. **Express numbers as powers of primes:** - $$9 = 3^2$$ - $$4 = 2^2$$ - $$125 = 5^3$$ - $$8 = 2^3$$ - $$7$$ and $$2$$ are primes. 6. **Rewrite logarithms using prime factorization:** $$\log_9 4 = \frac{\log 2^2}{\log 3^2} = \frac{2 \log 2}{2 \log 3} = \frac{\log 2}{\log 3}$$ $$\log_{125} 8 = \frac{\log 2^3}{\log 5^3} = \frac{3 \log 2}{3 \log 5} = \frac{\log 2}{\log 5}$$ 7. **Substitute back:** $$81.25 - \frac{1}{2} \cdot \frac{\log 2}{\log 3} + 25 \cdot \frac{\log 2}{\log 5}$$ 8. **Simplify the multiplication outside:** $$49 \log_7 2 = 49 \cdot \frac{\log 2}{\log 7}$$ 9. **Combine entire expression:** $$\left(81.25 - \frac{1}{2} \cdot \frac{\log 2}{\log 3} + 25 \cdot \frac{\log 2}{\log 5}\right) \cdot 49 \cdot \frac{\log 2}{\log 7}$$ 10. **Approximate logarithms (base 10):** $$\log 2 \approx 0.3010$$ $$\log 3 \approx 0.4771$$ $$\log 5 \approx 0.6990$$ $$\log 7 \approx 0.8451$$ 11. **Calculate each term:** $$- \frac{1}{2} \cdot \frac{0.3010}{0.4771} = - \frac{1}{2} \cdot 0.631 = -0.3155$$ $$25 \cdot \frac{0.3010}{0.6990} = 25 \cdot 0.4305 = 10.7625$$ 12. **Sum inside parentheses:** $$81.25 - 0.3155 + 10.7625 = 91.697$$ 13. **Multiply by outside term:** $$49 \cdot \frac{0.3010}{0.8451} = 49 \cdot 0.356 = 17.444$$ 14. **Final multiplication:** $$91.697 \times 17.444 \approx 1599.5$$ **Final answer:** $$\boxed{1599.5}$$