1. **Stating the problems:** a) Solve the inequality $$\lg^2(2x + 3) - 12\lg(2x + 3) + 20 \leq 0$$. b) Solve the inequality $$2\lg \frac{1}{5} x - 5\log_{\frac{1}{5}}(6x + 7) > 0$$. c) Solve the inequality $$\log_{0.3}\left(\frac{x^2 - 1}{x + 3}\right) \geq 0$$. d) Solve the inequality $$\log_2(x - 1) - \log_2(x + 1) + \log_{\frac{x + 2}{x - 1}} \leq 0$$. 2. **Important rules and formulas:** - The domain of logarithmic functions must be positive inside the log. - Change of base formula: $$\log_a b = \frac{\log_c b}{\log_c a}$$. - For inequalities involving logarithms, consider the base: - If base $>1$, inequality direction is preserved. - If base $\in (0,1)$, inequality direction is reversed. - Quadratic inequalities can be solved by factoring or using the quadratic formula. 3. **Problem 6a:** Let $$y = \lg(2x + 3)$$. The inequality becomes: $$y^2 - 12y + 20 \leq 0$$. Solve quadratic inequality: $$\Delta = (-12)^2 - 4 \cdot 1 \cdot 20 = 144 - 80 = 64$$. Roots: $$y_1 = \frac{12 - \sqrt{64}}{2} = \frac{12 - 8}{2} = 2$$ $$y_2 = \frac{12 + \sqrt{64}}{2} = \frac{12 + 8}{2} = 10$$ Since coefficient of $y^2$ is positive, inequality holds between roots: $$2 \leq y \leq 10$$ Substitute back: $$2 \leq \lg(2x + 3) \leq 10$$ Convert from logarithm base 10: $$10^2 \leq 2x + 3 \leq 10^{10}$$ $$100 \leq 2x + 3 \leq 10^{10}$$ Solve for $x$: $$100 - 3 \leq 2x \leq 10^{10} - 3$$ $$97 \leq 2x \leq 10^{10} - 3$$ $$\frac{97}{2} \leq x \leq \frac{10^{10} - 3}{2}$$ Domain condition: $$2x + 3 > 0 \Rightarrow x > -\frac{3}{2}$$ This is satisfied by the solution interval. **Final solution for 6a:** $$x \in \left[\frac{97}{2}, \frac{10^{10} - 3}{2}\right]$$ 4. **Problem 6b:** Given: $$2\lg \frac{1}{5} x - 5\log_{\frac{1}{5}}(6x + 7) > 0$$ Change all logs to base 10: $$\lg_{\frac{1}{5}} t = \frac{\lg t}{\lg \frac{1}{5}}$$ Note that $$\lg \frac{1}{5} = \lg 1 - \lg 5 = 0 - \lg 5 = -\lg 5 < 0$$ Rewrite inequality: $$2 \cdot \frac{\lg x}{\lg \frac{1}{5}} - 5 \cdot \frac{\lg(6x + 7)}{\lg \frac{1}{5}} > 0$$ Multiply both sides by $$\lg \frac{1}{5} < 0$$, inequality reverses: $$2 \lg x - 5 \lg(6x + 7) < 0$$ Rewrite: $$2 \lg x < 5 \lg(6x + 7)$$ Use power property: $$\lg x^2 < \lg (6x + 7)^5$$ Since $$\lg$$ is increasing, inequality is: $$x^2 < (6x + 7)^5$$ Domain: $$x > 0$$ and $$6x + 7 > 0 \Rightarrow x > -\frac{7}{6}$$, so domain is $$x > 0$$. Since $(6x + 7)^5$ grows faster than $x^2$ for $x > 0$, inequality holds for all $x > 0$ except possibly near zero. Check at $x=0.1$: $$0.1^2 = 0.01$$ $$6(0.1) + 7 = 0.6 + 7 = 7.6$$ $$7.6^5$$ is very large, so inequality holds. **Final solution for 6b:** $$x > 0$$ 5. **Problem 7a:** $$\log_{0.3} \left( \frac{x^2 - 1}{x + 3} \right) \geq 0$$ Domain: $$x + 3 > 0 \Rightarrow x > -3$$ $$x^2 - 1 > 0 \Rightarrow x < -1 \text{ or } x > 1$$ Also, the argument of the log must be positive: $$\frac{x^2 - 1}{x + 3} > 0$$ Analyze sign: - Numerator zero at $x = \pm 1$ - Denominator zero at $x = -3$ Intervals: 1) $(-\infty, -3)$: denominator negative, numerator positive for $x < -1$, so fraction negative. 2) $(-3, -1)$: denominator positive, numerator negative, fraction negative. 3) $(-1, 1)$: numerator negative, denominator positive, fraction negative. 4) $(1, \infty)$: numerator positive, denominator positive, fraction positive. So domain is $(1, \infty)$. Since base $0.3 \in (0,1)$, logarithm is decreasing. Inequality: $$\log_{0.3} (\text{expr}) \geq 0 \Rightarrow \text{expr} \leq 1$$ So: $$\frac{x^2 - 1}{x + 3} \leq 1$$ Multiply both sides by $x + 3 > 0$ (since $x > 1 > -3$): $$x^2 - 1 \leq x + 3$$ Rearrange: $$x^2 - x - 4 \leq 0$$ Solve quadratic: $$\Delta = (-1)^2 - 4 \cdot 1 \cdot (-4) = 1 + 16 = 17$$ Roots: $$x = \frac{1 \pm \sqrt{17}}{2}$$ Approximate roots: $$x_1 \approx \frac{1 - 4.123}{2} = -1.5615$$ $$x_2 \approx \frac{1 + 4.123}{2} = 2.5615$$ Since coefficient of $x^2$ is positive, inequality holds between roots: $$-1.5615 \leq x \leq 2.5615$$ Intersect with domain $(1, \infty)$: $$x \in [1, 2.5615]$$ **Final solution for 7a:** $$x \in [1, \frac{1 + \sqrt{17}}{2}]$$ 6. **Problem 7b:** $$\log_2(x - 1) - \log_2(x + 1) + \log_{\frac{x + 2}{x - 1}} \leq 0$$ Domain: $$x - 1 > 0 \Rightarrow x > 1$$ $$x + 1 > 0 \Rightarrow x > -1$$ (already satisfied by $x > 1$) Base of last log: $$a = \frac{x + 2}{x - 1}$$ For log base: $$a > 0, a \neq 1$$ Since $x > 1$, denominator positive. Numerator $x + 2 > 0$ for all $x > 1$. So $a > 0$. Check $a \neq 1$: $$\frac{x + 2}{x - 1} = 1 \Rightarrow x + 2 = x - 1 \Rightarrow 2 = -1$$ no solution. So $a \neq 1$ always. Rewrite inequality: $$\log_2 \frac{x - 1}{x + 1} + \log_a \leq 0$$ The last term is incomplete in the problem statement; assuming it is $$\log_a (x + 2)$$ or similar, but since problem is incomplete, we consider only given expression. Since problem 7b is incomplete, we cannot solve it fully. **Summary of solutions:** - 6a: $$x \in \left[\frac{97}{2}, \frac{10^{10} - 3}{2}\right]$$ - 6b: $$x > 0$$ - 7a: $$x \in \left[1, \frac{1 + \sqrt{17}}{2}\right]$$ - 7b: Problem incomplete, cannot solve.