Bài 2 (Trang 90 SGK Toán Giải Tích 12):

Giải các bất phương trình lôgarit:

a) ${\log }_8(4 - 2x) \ge 2$;

b) ${\log }_{\frac{1}{5}}(3x - 5) > l{\mathrm{og}}_{\frac{1}{5}}(x + 1)$;

c) ${\log }_{0,2}x - {\log }_5(x - 2 ) <{\log }_{0,2}3$;

d) $\log \frac{2}{3}x - 5{\log }_{3}x + 6\le 0$.

Hướng dẫn giải:

$a) {\log }_8(4 -2x ) \ge 2 \iff 4 - 2x \ge 8^2 \iff 2x \le -60 \iff x\le -30. Vậy S=(-\infty ;-30]$.

$$\begin{gathered} b) {\log }_{\frac{1}{5}}(3x - 5) > {\log }_{\frac{1}{5}}(x + 1 ) \iff 0 < 3x - 5 < x + 1 \\ \iff \left\{\begin{array}{l}3x - 5 > 0\\ 3x - 5 < x + 1\end{array}\right. \iff \left\{\begin{array}{l}x>\frac{5}{3} \\ x<3\end{array}\right.\iff \frac{5}{3}<x<3 \\ Vậy S = \left(\frac{5}{3};3\right) \end{gathered}$$

$$\begin{gathered} \mathrm{c}) {\log }_{\frac{1}{5}}\mathrm{x} - {\log }_5(\mathrm{x} -2) < {\log }_{\frac{1}{5}}3 \iff -{\log }_5\mathrm{x} - {\log }_5(\mathrm{x} -2) <-{\log }_{5}3 \\ \iff {\log }_5\mathrm{x} + {\log }_5(\mathrm{x} -2) >{\log }_{5}3 \iff \left\{\begin{array}{l}\mathrm{x}>2\\ \mathrm{x}(\mathrm{x}-2) > 3\end{array}\right. \\ \iff \left\{\begin{array}{l}\mathrm{x}>2\\ {\mathrm{x}}^2 - 2\mathrm{x} -3 >0\end{array}\right. \iff \left\{\begin{array}{l}\mathrm{x}>2\\ \mathrm{x}<-1 \mathrm{hoặc} \mathrm{x}>3 \end{array}\right.\iff \mathrm{x}>3 \\ \mathrm{Vậy} \mathrm{S}=(3;+\infty ). \end{gathered}$$

d) Điều kiện: x > 0.

$$\begin{gathered} \mathrm{Đặt} \mathrm{t}={\log }_3\mathrm{x}, \mathrm{ta} \mathrm{có}: {\mathrm{t}}^2 -5\mathrm{t} +6 \le 0 \iff 2 \le \mathrm{t} \le 3 \\ \iff 2 \le {\log }_3\mathrm{x} \le 3 \iff 9 \le \mathrm{x} \le 27. \mathrm{Vậy} \mathrm{S} = [9;27]. \end{gathered}$$