Let's simplify the expression \(\frac{\log 4}{\log \frac{1}{2}}\).

### Step 1: Understand the logarithm properties
First, recall that:
\[
\log \frac{1}{2} = \log \left(2^{-1}\right)
\]
Using the logarithm power rule, \(\log a^b = b \log a\), this becomes:
\[
\log \frac{1}{2} = -\log 2
\]

### Step 2: Substitute into the original expression
Now substitute \(\log \frac{1}{2} = -\log 2\) into the original expression:
\[
\frac{\log 4}{\log \frac{1}{2}} = \frac{\log 4}{-\log 2}
\]

### Step 3: Simplify the expression
Next, simplify \(\log 4\) using \(\log 4 = \log (2^2) = 2\log 2\):
\[
\frac{\log 4}{-\log 2} = \frac{2\log 2}{-\log 2}
\]

Cancel out \(\log 2\) in the numerator and denominator:
\[
\frac{2\log 2}{-\log 2} = -2
\]

So, the final answer is:
\[
\frac{\log 4}{\log \frac{1}{2}} = -2
\]

Would you like more details or have any questions?

Here are some related questions:
1. How would you simplify \(\frac{\log 8}{\log \frac{1}{2}}\)?
2. What is the value of \(\log_2 \frac{1}{4}\)?
3. How does the logarithm property \(\log a^b = b \log a\) apply to other bases?
4. What is the result of \(\frac{\log 16}{\log \frac{1}{4}}\)?
5. How do you simplify \(\frac{\log 9}{\log \frac{1}{3}}\)?

**Tip:** Remember that \(\log \left(\frac{1}{a}\right) = -\log a\), which can simplify many logarithmic expressions.

Learn how to simplify the logarithmic expression \(\frac{\log 4}{\log \frac{1}{2}}\) with a detailed step-by-step solution. Understand the logarithm properties and apply them to solve similar problems. Suitable for high school students and anyone preparing for logarithm-related questions.

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