log4(3^x-1).log1/4(3^x-1/16)<=3/4 tìm nghiệm bpt

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1. Đáp án: $\,\left[ \begin{array}{l}
   x \ge 2\\
   0 < x \le 1
   \end{array} \right.$

    

   Giải thích các bước giải:

   $\begin{array}{l}
   Đkxđ:{3^x} – 1 > 0 \Rightarrow x > 0\\
   {\log _4}\left( {{3^x} – 1} \right).{\log _{\frac{1}{4}}}\left( {\frac{{{3^x} – 1}}{{16}}} \right) \le \frac{3}{4}\\
    \Rightarrow {\log _4}\left( {{3^x} – 1} \right).\left[ { – {{\log }_4}\frac{{\left( {{3^x} – 1} \right)}}{{16}}} \right] \le \frac{3}{4}\\
    \Rightarrow  – {\log _4}\left( {{3^x} – 1} \right).\left( {{{\log }_4}\left( {{3^x} – 1} \right) – {{\log }_4}16} \right) \le \frac{3}{4}\\
    \Rightarrow  – {\log _4}\left( {{3^x} – 1} \right).\left[ {{{\log }_4}\left( {{3^x} – 1} \right) – 2} \right] \le \frac{3}{4}\\
   Dat:{\log _4}\left( {{3^x} – 1} \right) = t\\
    \Rightarrow  – t\left( {t – 2} \right) \le \frac{3}{4}\\
    \Rightarrow {t^2} – 2t + \frac{3}{4} \ge 0\\
    \Rightarrow \left[ \begin{array}{l}
   t \ge \frac{3}{2}\\
   t \le \frac{1}{2}
   \end{array} \right. \Rightarrow \left[ \begin{array}{l}
   {\log _4}\left( {{3^x} – 1} \right) \ge \frac{3}{2}\\
   {\log _4}\left( {{3^x} – 1} \right) \le \frac{1}{2}
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   {3^x} – 1 \ge {4^{\frac{3}{2}}}\\
   {3^x} – 1 \le {4^{\frac{1}{2}}}
   \end{array} \right. \Rightarrow \left[ \begin{array}{l}
   {3^x} – 1 \ge 8\\
   {3^x} – 1 \le 2
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x \ge 2\\
   x \le 1
   \end{array} \right.\\
   Vậy\,\left[ \begin{array}{l}
   x \ge 2\\
   0 < x \le 1
   \end{array} \right.
   \end{array}$

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