The given expression is:

$4 \log(\sqrt{128}) + 4 \log(8)$

We can simplify this expression step by step.

Step 1: Use the square root property.

$\sqrt{128} = 128^{1/2}$ Thus: $\log(\sqrt{128}) = \log(128^{1/2}) = \frac{1}{2} \log(128)$

Step 2: Rewrite the expression.

Now, the original expression becomes: $4 \log(128^{1/2}) + 4 \log(8) = 4 \times \frac{1}{2} \log(128) + 4 \log(8)$ Simplify: $2 \log(128) + 4 \log(8)$

Step 3: Simplify $\log(128)$ and $\log(8)$.

Express 128 and 8 in terms of powers of 2: $128 = 2^7 \quad \text{and} \quad 8 = 2^3$ So: $\log(128) = \log(2^7) = 7 \log(2)$ $\log(8) = \log(2^3) = 3 \log(2)$

Step 4: Substitute into the expression.

Now substitute these back into the expression: $2 \log(128) + 4 \log(8) = 2 \times 7 \log(2) + 4 \times 3 \log(2)$ This simplifies to: $14 \log(2) + 12 \log(2) = 26 \log(2)$

Thus, the simplified expression is: $26 \log(2)$

Do you want more details or have any questions about this? Here are some related questions you could explore:

1. How do logarithmic properties simplify expressions like this?
2. Why is $\log(2^x) = x \log(2)$ a key rule in logarithms?
3. How would you approach solving an expression with natural logarithms ($\ln$) instead of $\log$?
4. What happens if you encounter logarithms with different bases?
5. How would the answer change if the argument of the logarithm was a fraction?

Tip: Always check if you can express numbers as powers of the same base when dealing with logarithms—it greatly simplifies calculations.