To simplify the expression \frac{5 \log 16 + 2 \log 25 - 3 \log \frac{27}{81}}, we can follow these steps:

1. Simplify Each Logarithm:

   • $\log 16 = \log (4^2) = 2 \log 4$
   • $\log 25 = \log (5^2) = 2 \log 5$
   • $\log \frac{27}{81} = \log 27 - \log 81 = \log (3^3) - \log (3^4) = 3 \log 3 - 4 \log 3 = -\log 3$

2. Substituting Back: $5 \log 16 = 5(2 \log 4) = 10 \log 4$ $2 \log 25 = 2(2 \log 5) = 4 \log 5$ $-3 \log \frac{27}{81} = -3(-\log 3) = 3 \log 3$

3. Combine the Logs: $\frac{10 \log 4 + 4 \log 5 + 3 \log 3}{1}$

4. Using Logarithm Properties: Combine the logs: $\log(4^{10} \cdot 5^4 \cdot 3^3)$

5. Final Result: Thus, the expression simplifies to: $\log(4^{10} \cdot 5^4 \cdot 3^3)$

If you need a numerical evaluation or more details about any step, feel free to ask!

Questions for Further Exploration:

1. What are the properties of logarithms that were used in this simplification?
2. How would you evaluate the final logarithmic expression numerically?
3. Can you explain why we can manipulate logarithms this way?
4. What would happen if the bases of the logarithms were different?
5. How can you apply these logarithmic properties in solving exponential equations?

Tip:

Always remember the basic properties of logarithms: $\log(a \cdot b) = \log a + \log b$ and $\log\left(\frac{a}{b}\right) = \log a - \log b$. They are crucial for simplifying expressions.