Let's break down the expression \( ³\log 9 + ²\log \frac{1}{2} \), which involves logarithms with different bases.

### Expression:
\[
³\log 9 + ²\log \frac{1}{2}
\]

This means:
- \( ³\log 9 \) is the logarithm of 9 with base 3.
- \( ²\log \frac{1}{2} \) is the logarithm of \( \frac{1}{2} \) with base 2.

We will solve each part step-by-step.

### Step 1: \( ³\log 9 \)
The logarithm of 9 with base 3 is:
\[
³\log 9 = \log_3 9
\]
Since \( 9 = 3^2 \), we have:
\[
\log_3 9 = \log_3 (3^2) = 2
\]

### Step 2: \( ²\log \frac{1}{2} \)
The logarithm of \( \frac{1}{2} \) with base 2 is:
\[
²\log \frac{1}{2} = \log_2 \frac{1}{2}
\]
We can express \( \frac{1}{2} \) as \( 2^{-1} \), so:
\[
\log_2 \frac{1}{2} = \log_2 (2^{-1}) = -1
\]

### Step 3: Combine the results
Now, we can add the results of the two logarithms:
\[
³\log 9 + ²\log \frac{1}{2} = 2 + (-1) = 1
\]

### Final Answer:
\[
³\log 9 + ²\log \frac{1}{2} = 1
\]

Would you like more details on this, or do you have any questions?

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Here are 5 related questions to explore this further:
1. What is \( ³\log 27 \)?
2. How do you calculate \( ²\log 4 \)?
3. What is the general formula for converting logarithms with different bases?
4. How do you use properties of exponents in logarithmic expressions?
5. What happens when the logarithmic base equals the argument, like \( \log_b b \)?

**Tip**: To handle logarithms, always try to express the number in terms of powers of the base. This simplifies the calculation.

Understand how to solve the logarithmic expression ³log 9 + ²log 1/2 step by step. This problem involves key logarithmic and exponent concepts, including base 3 and base 2 logarithms. Suitable for Grades 10-12.

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