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The given equation is:

\[
\log_x 25 = -2
\]

We are asked to solve for \( x \).

### Step-by-step solution:

1. Recall that a logarithmic equation \( \log_b a = c \) can be rewritten in exponential form as:
   \[
   b^c = a
   \]
   Here, \( b = x \), \( a = 25 \), and \( c = -2 \).

   So, the equation becomes:
   \[
   x^{-2} = 25
   \]

2. To solve for \( x \), first rewrite \( x^{-2} = 25 \) as:
   \[
   \frac{1}{x^2} = 25
   \]

3. Now, take the reciprocal of both sides to get rid of the fraction:
   \[
   x^2 = \frac{1}{25}
   \]

4. Finally, take the square root of both sides:
   \[
   x = \pm \frac{1}{5}
   \]

### Solution:

The solution set is:
\[
x = \pm \frac{1}{5}
\]

Would you like further details or clarification?

Here are five related questions to expand your understanding:

1. How do you convert a logarithmic equation into its exponential form?
2. What is the meaning of a negative logarithmic value?
3. How would the solution change if the base and the argument in the logarithm were different?
4. What happens if we raise a negative base to a negative exponent?
5. Can you apply the same method to solve equations with natural logarithms?

**Tip**: When solving logarithmic equations, always ensure that the base \( x \) must be positive and not equal to 1 for the logarithmic function to be valid.

Learn how to solve the logarithmic equation log_x 25 = -2 by converting it into exponential form and applying the laws of exponents. This problem involves key concepts in logarithms and exponents, suitable for high school students in Grades 9-12.

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