1. **State the problem:** Solve the inequality $$\log_{\frac{1}{3}}(x^2 - 1) \leq \log_{\frac{1}{3}}(2x + 7)$$ for $x$. 2. **Recall the properties of logarithms:** - The base of the logarithm is $\frac{1}{3}$, which is between 0 and 1. - For logarithms with base $b$ where $0 < b < 1$, the logarithm function is decreasing. - Therefore, if $$\log_b(A) \leq \log_b(B)$$ with $0 < b < 1$, then $$A \geq B$$. 3. **Set the domain:** - Arguments of logarithms must be positive: $$x^2 - 1 > 0 \implies x < -1 \text{ or } x > 1$$ $$2x + 7 > 0 \implies x > -\frac{7}{2}$$ - Combine domain conditions: $$x < -1 \text{ or } x > 1$$ and $$x > -\frac{7}{2}$$ - Since $-\frac{7}{2} = -3.5$, the domain is: $$(-3.5, -1) \cup (1, \infty)$$ 4. **Rewrite the inequality using the decreasing property:** $$x^2 - 1 \geq 2x + 7$$ 5. **Bring all terms to one side:** $$x^2 - 1 - 2x - 7 \geq 0$$ $$x^2 - 2x - 8 \geq 0$$ 6. **Factor the quadratic:** $$x^2 - 2x - 8 = (x - 4)(x + 2)$$ 7. **Solve the inequality:** $$(x - 4)(x + 2) \geq 0$$ - The product is nonnegative when both factors are positive or both are negative. - Intervals: - $x \leq -2$ - $x \geq 4$ 8. **Combine with domain:** - Domain: $(-3.5, -1) \cup (1, \infty)$ - Inequality solution: $(-\infty, -2] \cup [4, \infty)$ - Intersection: - $(-3.5, -1) \cap (-\infty, -2] = (-3.5, -2]$ - $(1, \infty) \cap [4, \infty) = [4, \infty)$ 9. **Final solution:** $$x \in (-3.5, -2] \cup [4, \infty)$$