1. Problem: Solve the inequality $$\log_{\frac{1}{5}}(x - 5) > -2$$. 2. Formula and rules: For logarithmic inequalities with base $a$ where $0 < a < 1$, the inequality direction reverses when removing the log. Also, the argument of the log must be positive: $$x - 5 > 0 \Rightarrow x > 5$$. 3. Solve the inequality: $$\log_{\frac{1}{5}}(x - 5) > -2 \Rightarrow x - 5 < \left(\frac{1}{5}\right)^{-2}$$ because base $\frac{1}{5} < 1$ reverses inequality. 4. Calculate the right side: $$\left(\frac{1}{5}\right)^{-2} = 5^2 = 25$$. 5. So, $$x - 5 < 25 \Rightarrow x < 30$$. 6. Combine domain and inequality: $$5 < x < 30$$. 7. Final answer: Interval notation is $$]5; 30[\,$$ which corresponds to option C. --- 1. Problem: Solve $$\log_{\frac{1}{9}}(x + 3) > -0.5$$. 2. Domain: $$x + 3 > 0 \Rightarrow x > -3$$. 3. Since base $\frac{1}{9} < 1$, inequality reverses: $$x + 3 < \left(\frac{1}{9}\right)^{-0.5}$$. 4. Calculate right side: $$\left(\frac{1}{9}\right)^{-0.5} = 9^{0.5} = 3$$. 5. So, $$x + 3 < 3 \Rightarrow x < 0$$. 6. Combine domain and inequality: $$-3 < x < 0$$. 7. Final answer: $$]-3; 0[\,$$ which is option C. --- 1. Problem: Solve $$\log_{\frac{1}{4}}(x - 3) > 1.5$$. 2. Domain: $$x - 3 > 0 \Rightarrow x > 3$$. 3. Base $\frac{1}{4} < 1$ reverses inequality: $$x - 3 < \left(\frac{1}{4}\right)^{1.5}$$. 4. Calculate right side: $$\left(\frac{1}{4}\right)^{1.5} = 4^{-1.5} = \frac{1}{4^{1.5}} = \frac{1}{8} = 0.125$$. 5. So, $$x - 3 < 0.125 \Rightarrow x < 3.125$$. 6. Combine domain and inequality: $$3 < x < 3.125$$. 7. Final answer: $$]3; 3.125[\,$$ which is option E. --- 1. Problem: Solve $$\log_{\frac{1}{3}}(x - 5) > -3$$. 2. Domain: $$x - 5 > 0 \Rightarrow x > 5$$. 3. Base $\frac{1}{3} < 1$ reverses inequality: $$x - 5 < \left(\frac{1}{3}\right)^{-3}$$. 4. Calculate right side: $$\left(\frac{1}{3}\right)^{-3} = 3^3 = 27$$. 5. So, $$x - 5 < 27 \Rightarrow x < 32$$. 6. Combine domain and inequality: $$5 < x < 32$$. 7. Final answer: $$]5; 32[\,$$ which is option D. --- 1. Problem: Solve $$\log_{\frac{1}{2}}(x + 3) > -2$$. 2. Domain: $$x + 3 > 0 \Rightarrow x > -3$$. 3. Base $\frac{1}{2} < 1$ reverses inequality: $$x + 3 < \left(\frac{1}{2}\right)^{-2}$$. 4. Calculate right side: $$\left(\frac{1}{2}\right)^{-2} = 2^2 = 4$$. 5. So, $$x + 3 < 4 \Rightarrow x < 1$$. 6. Combine domain and inequality: $$-3 < x < 1$$. 7. Final answer: $$]-3; 1[\,$$ which is option C. --- 1. Problem: Solve $$\log_{\frac{1}{3}}(x - 2) > -1$$. 2. Domain: $$x - 2 > 0 \Rightarrow x > 2$$. 3. Base $\frac{1}{3} < 1$ reverses inequality: $$x - 2 < \left(\frac{1}{3}\right)^{-1}$$. 4. Calculate right side: $$\left(\frac{1}{3}\right)^{-1} = 3$$. 5. So, $$x - 2 < 3 \Rightarrow x < 5$$. 6. Combine domain and inequality: $$2 < x < 5$$. 7. Final answer: $$]2; 5[\,$$ which is option B. --- 1. Problem: Solve $$\log_{\frac{1}{2}}(x - 5) > -2$$. 2. Domain: $$x - 5 > 0 \Rightarrow x > 5$$. 3. Base $\frac{1}{2} < 1$ reverses inequality: $$x - 5 < \left(\frac{1}{2}\right)^{-2}$$. 4. Calculate right side: $$\left(\frac{1}{2}\right)^{-2} = 2^2 = 4$$. 5. So, $$x - 5 < 4 \Rightarrow x < 9$$. 6. Combine domain and inequality: $$5 < x < 9$$. 7. Final answer: $$]5; 9[\,$$ which is option D. --- 1. Problem: Solve $$\log_{3}(x + 20) < 3$$. 2. Domain: $$x + 20 > 0 \Rightarrow x > -20$$. 3. Base $3 > 1$ keeps inequality direction: $$x + 20 < 3^3$$. 4. Calculate right side: $$3^3 = 27$$. 5. So, $$x < 7$$. 6. Combine domain and inequality: $$-20 < x < 7$$. 7. Final answer: $$]-20; 7[\,$$ which is option D. --- 1. Problem: Solve $$\log_{5}(x + 13) < 2$$. 2. Domain: $$x + 13 > 0 \Rightarrow x > -13$$. 3. Base $5 > 1$ keeps inequality direction: $$x + 13 < 5^2$$. 4. Calculate right side: $$5^2 = 25$$. 5. So, $$x < 12$$. 6. Combine domain and inequality: $$-13 < x < 12$$. 7. Final answer: $$]-13; 12[\,$$ which is option E. --- 1. Problem: Solve $$\log_{\frac{1}{8}}(x - 7) > -\frac{2}{3}$$. 2. Domain: $$x - 7 > 0 \Rightarrow x > 7$$. 3. Base $\frac{1}{8} < 1$ reverses inequality: $$x - 7 < \left(\frac{1}{8}\right)^{-\frac{2}{3}}$$. 4. Calculate right side: $$\left(\frac{1}{8}\right)^{-\frac{2}{3}} = 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4$$. 5. So, $$x - 7 < 4 \Rightarrow x < 11$$. 6. Combine domain and inequality: $$7 < x < 11$$. 7. Final answer: $$]7; 11[\,$$ which is option C.