1. The problem asks to find the value of $4 \log 8 + 4 \log 32$. 2. Recall the logarithm property: $a \log b = \log b^a$. 3. Apply this property: $4 \log 8 = \log 8^4$ and $4 \log 32 = \log 32^4$. 4. Calculate powers: $8^4 = (2^3)^4 = 2^{12} = 4096$ and $32^4 = (2^5)^4 = 2^{20} = 1048576$. 5. So the expression becomes $\log 4096 + \log 1048576$. 6. Use the logarithm addition rule: $\log a + \log b = \log (a \times b)$. 7. Multiply inside the log: $4096 \times 1048576 = 2^{12} \times 2^{20} = 2^{32}$. 8. Therefore, the expression simplifies to $\log 2^{32}$. 9. Use the power rule: $\log a^b = b \log a$. 10. Since $\log 2$ (base 10) is a constant, the expression is $32 \log 2$. 11. If the logarithm base is 10, $\log 2 \approx 0.3010$, so the value is approximately $32 \times 0.3010 = 9.632$. Final answer: $4 \log 8 + 4 \log 32 = 32 \log 2 \approx 9.632$.