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.