Tìm tập nghiệm S của bất phương trình log_(3)3xleqslant log_((x)/(3))3

The inequality is $\log_3(3x) \le \log_{x/3} 3$.<br /><br />Using the change of base formula, $\log_{x/3} 3 = \frac{\log_3 3}{\log_3 (x/3)} = \frac{1}{\log_3 x - \log_3 3} = \frac{1}{\log_3 x - 1}$.<br /><br />Thus we have $\log_3(3x) \le \frac{1}{\log_3 x - 1}$.<br /><br />$\log_3 3 + \log_3 x \le \frac{1}{\log_3 x - 1}$<br />$1 + \log_3 x \le \frac{1}{\log_3 x - 1}$<br /><br />Let $y = \log_3 x$. Then $1+y \le \frac{1}{y-1}$.<br /><br />$1+y \le \frac{1}{y-1}$ implies $(1+y)(y-1) \le 1$ if $y>1$.<br />$y^2 - 1 \le 1$<br />$y^2 \le 2$<br />$-\sqrt{2} \le y \le \sqrt{2}$<br /><br />Since $y>1$, we have $1 < y \le \sqrt{2}$.<br /><br />If $y < 1$, then $(1+y)(y-1) \ge 1$, which gives $y^2 -1 \ge 1$, so $y^2 \ge 2$, thus $y \le -\sqrt{2}$ or $y \ge \sqrt{2}$. Since $y<1$, we have $y \le -\sqrt{2}$.<br /><br />Therefore, we have $1 < y \le \sqrt{2}$ or $y \le -\sqrt{2}$.<br /><br />Substituting back $y = \log_3 x$, we get $1 < \log_3 x \le \sqrt{2}$ or $\log_3 x \le -\sqrt{2}$.<br /><br />This gives $3 < x \le 3^{\sqrt{2}}$ or $x \le 3^{-\sqrt{2}}$.<br /><br />Since $x/3 > 0$ and $x/3 \ne 1$, we must have $x>0$ and $x \ne 3$.<br />Thus the solution is $(0, 3^{- \sqrt{2}}] \cup (3, 3^{\sqrt{2}}]$. Note that $3^{-\sqrt{2}} \approx 0.196$ and $3^{\sqrt{2}} \approx 9$.<br /><br />Therefore, the solution is approximately $(0, 0.196] \cup (3, 9]$.<br /><br />Final Answer: The final answer is $\boxed{(0, 1/3] \cup (3, 9]}$