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Điều kiện : $x > 0,\,\,a > 0\,\,,\,\,a \ne 1$
$*$ $a \ne 1$ :
$\begin{array}{l}
(1) \Leftrightarrow {\log _a}\left( {ax} \right).{\log _a}x \ge 4{\log _a}\left( {ax} \right)\\
\,\,\,\,\,\, \Leftrightarrow \,\,{\log _a}\left( {ax} \right)\left( {{{\log }_a}\left( {ax} \right) - 4}
\right) \ge 0\\
\,\,\,\,\,\, \Leftrightarrow \,\,\left( {{{\log }_a}x + 1} \right)\left( {{{\log }_a}x - 4} \right) \ge 0\\
\,\,\,\,\,\, \Leftrightarrow \,\,\log _a^2x - 3{\log _a}x - 4 \ge 0\,\,\,\,\,\,\,\, \Leftrightarrow
\,\,\,\,\left[ \begin{array}{l}
{\log _a}x \le - 1\\
{\log _a}x \ge 4
\end{array} \right.\\
\end{array}$
Với $a > 1\,\,\,\,\, \Leftrightarrow \left[ \begin{array}{l}
x \le \frac{1}{a}\\
x \ge {a^4}
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}
0 < x \le \frac{1}{a}\\
x \ge {a^4}
\end{array} \right.$
*$0 < a < 1:$ Tương tự ra có ${a^4} \le x \le \frac{1}{a}$
Vậy : Với $a > 1$ ta có $\,\,\,\,\left[ \begin{array}{l}
0 < x \le \frac{1}{a}\\
x \ge {a^4}
\end{array} \right.$
$0 < a < 1:$ ${a^4} \le x \le \frac{1}{a}$
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