Question: User: $5^{x+2} + 5^{x+1} < 5^{2x} + 125$
1. **State the problem:** Solve the inequality $$5^{x+2} + 5^{x+1} < 5^{2x} + 125$$.
2. **Rewrite the terms using properties of exponents:**
$$5^{x+2} = 5^x \cdot 5^2 = 25 \cdot 5^x$$
$$5^{x+1} = 5^x \cdot 5 = 5 \cdot 5^x$$
$$5^{2x} = (5^x)^2$$
3. **Substitute to simplify:**
Let $$y = 5^x$$ (note that $$y > 0$$ since $$5^x$$ is always positive).
The inequality becomes:
$$25y + 5y < y^2 + 125$$
4. **Combine like terms:**
$$30y < y^2 + 125$$
5. **Bring all terms to one side:**
$$0 < y^2 - 30y + 125$$
6. **Rewrite as:**
$$y^2 - 30y + 125 > 0$$
7. **Find the roots of the quadratic:**
Use the quadratic formula:
$$y = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 1 \cdot 125}}{2} = \frac{30 \pm \sqrt{900 - 500}}{2} = \frac{30 \pm \sqrt{400}}{2} = \frac{30 \pm 20}{2}$$
So the roots are:
$$y_1 = \frac{30 - 20}{2} = 5$$
$$y_2 = \frac{30 + 20}{2} = 25$$
8. **Analyze the inequality:**
Since the quadratic opens upward (coefficient of $$y^2$$ is positive), the inequality $$y^2 - 30y + 125 > 0$$ holds when $$y < 5$$ or $$y > 25$$.
9. **Recall that $$y = 5^x$$ and $$y > 0$$:**
- For $$y < 5$$, since $$5^x$$ is increasing, this means $$x < 1$$.
- For $$y > 25$$, this means $$x > 2$$.
10. **Final solution:**
$$x < 1 \quad \text{or} \quad x > 2$$.
**Answer:** The solution set is $$(-\infty, 1) \cup (2, \infty)$$.