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The inequality you provided is:

\[
3^x - 3^{x-2} \geq 24
\]

### Step-by-step solution:

1. **Rewrite the second term**:
   \[
   3^{x-2} = \frac{3^x}{3^2} = \frac{3^x}{9}
   \]

2. **Substitute this into the inequality**:
   \[
   3^x - \frac{3^x}{9} \geq 24
   \]

3. **Factor out \(3^x\) from the left-hand side**:
   \[
   3^x \left( 1 - \frac{1}{9} \right) \geq 24
   \]

   Simplifying inside the parentheses:
   \[
   1 - \frac{1}{9} = \frac{8}{9}
   \]

   So, the inequality becomes:
   \[
   3^x \cdot \frac{8}{9} \geq 24
   \]

4. **Multiply both sides by \(\frac{9}{8}\)** to isolate \(3^x\):
   \[
   3^x \geq 24 \cdot \frac{9}{8} = 27
   \]

5. **Solve for \(x\)**:
   \[
   3^x \geq 27
   \]
   Since \(27 = 3^3\), we can rewrite the inequality as:
   \[
   3^x \geq 3^3
   \]

   Therefore:
   \[
   x \geq 3
   \]

### Final Answer:
\[
x \geq 3
\]

Let me know if you need any further clarifications or details!

Here are 5 related questions you can explore:
1. What happens if the base of the exponent is different from 3?
2. How would the solution change if the inequality were strict, i.e., \(>\) instead of \(\geq\)?
3. What is the general method to solve inequalities involving exponents?
4. How would you solve a similar inequality with base 2, such as \(2^x - 2^{x-1} \geq 10\)?
5. How does the logarithmic function help solve inequalities involving exponents?

**Tip**: Whenever you have exponents with different powers, try to rewrite them with a common base or factor them out for easier simplification.

Solve the inequality: 3^x - 3^(x-2) ≥ 24

Learn how to solve the exponential inequality 3^x - 3^(x-2) ≥ 24 using properties of exponents and factorization. This step-by-step solution is perfect for students in Grades 9-12, helping them understand key algebraic techniques for solving inequalities with exponential terms.

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