To solve the equation \(\log(x^2) = -2\), we'll proceed step by step.

### Step 1: Convert the logarithmic equation to its exponential form.
Given that \(\log_b(y) = x\) implies \(b^x = y\), and assuming the base of the logarithm is 10 (common logarithm):

\[
\log(x^2) = -2 \implies x^2 = 10^{-2}
\]

### Step 2: Simplify the exponential expression.
\[
10^{-2} = \frac{1}{10^2} = \frac{1}{100} = 0.01
\]

So, we have:
\[
x^2 = 0.01
\]

### Step 3: Solve for \(x\).
To find \(x\), take the square root of both sides:

\[
x = \pm \sqrt{0.01}
\]

\[
x = \pm 0.1
\]

### Final Answer:
The solutions are \(x = 0.1\) and \(x = -0.1\).

Would you like more details on any of these steps, or do you have any questions?

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Here are 5 related questions to expand your understanding:

1. How would the solution change if the logarithm base was different?
2. What if the equation was \(\log(x^2) = 2\) instead of \(\log(x^2) = -2\)?
3. How do you solve logarithmic equations with variables in both the base and the argument?
4. What are the properties of logarithms that simplify solving equations like this?
5. How does the logarithmic function behave for negative arguments?

**Tip:** When dealing with logarithmic equations, always check the domain of the logarithmic function, as it requires the argument to be positive.

Explore how to solve the logarithmic equation log x^2 = -2 with a step-by-step solution. This problem highlights key concepts in logarithms and exponential equations, and it's perfect for students in Grades 9-12.

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