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.

Learn how to simplify the logarithmic expression 4 log(√128) + 4 log(8) with step-by-step instructions. This problem covers logarithmic properties, such as exponents and square roots, and is suitable for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation