The given expression appears to be:

$\frac{1}{2} \log(32) + \frac{1}{2} \log\left(\frac{1}{128}\right) + 2 \log(\sqrt{8})$

Let me solve it step by step.

Step 1: Simplify the logarithmic terms

We will simplify each logarithmic term individually.

• $\frac{1}{2} \log(32)$: Using the property $\log(a^b) = b \log(a)$, we get:

$\frac{1}{2} \log(32) = \log(\sqrt{32})$

Now, $\sqrt{32} = 4\sqrt{2}$, so:

$\log(\sqrt{32}) = \log(4\sqrt{2}) = \log(4) + \log(\sqrt{2}) = \log(4) + \frac{1}{2} \log(2)$

Thus, $\frac{1}{2} \log(32)$ simplifies to:

$2 \log(2) + \frac{1}{2} \log(2) = 2.5 \log(2)$

• $\frac{1}{2} \log\left(\frac{1}{128}\right)$: Using the quotient property $\log\left(\frac{a}{b}\right) = \log(a) - \log(b)$ and simplifying $\frac{1}{128}$ as $128^{-1}$, we get:

$\frac{1}{2} \log\left(\frac{1}{128}\right) = \frac{1}{2} \log(128^{-1}) = -\frac{1}{2} \log(128)$

Now, $128 = 2^7$, so:

$-\frac{1}{2} \log(128) = -\frac{1}{2} \log(2^7) = -\frac{1}{2} \cdot 7 \log(2) = -\frac{7}{2} \log(2)$

• $2 \log(\sqrt{8})$: We know $\sqrt{8} = 2\sqrt{2}$, so:

$2 \log(\sqrt{8}) = 2 \log(2\sqrt{2}) = 2 (\log(2) + \log(\sqrt{2})) = 2 \log(2) + 2 \cdot \frac{1}{2} \log(2)$

This simplifies to:

$2 \log(2) + \log(2) = 3 \log(2)$

Step 2: Combine all terms

Now, we add all the simplified terms together:

$2.5 \log(2) - \frac{7}{2} \log(2) + 3 \log(2)$

This simplifies to:

$2.5 \log(2) - 3.5 \log(2) + 3 \log(2) = 2.0 \log(2)$

Step 3: Final answer

So, the entire expression simplifies to:

$2 \log(2)$

This is the final simplified result.

Would you like further details or any clarification?

Here are five related questions to deepen your understanding:

1. How would the result change if the base of the logarithms was not 10?
2. Can you apply logarithmic properties to solve this expression using a different method?
3. What is the numeric value of the simplified expression $2 \log(2)$?
4. How would you solve this if the numbers involved were not powers of 2?
5. Can you graph the logarithmic function for the components of the expression?

Tip: Always check if you can use properties of logarithms (like the power rule or quotient rule) to simplify expressions before performing calculations!