Giải phương trình: $log_{3}x$.$log_{9}x$.$log_{27}x$.$log_{81}x$ $=$ $\frac{2}{3}$

Đáp án:

\(S = \left\{\dfrac19;9\right\}\) 

Giải thích các bước giải:

\(\begin{array}{l}
\log_3x.\log_9x.\log_{27}x.\log_{81}x = \dfrac23\qquad (ĐK: x >0)\\
\Leftrightarrow \log_3x\cdot \left(\dfrac12\log_3x\right)\cdot \left(\dfrac13\log_3x\right)\cdot \left(\dfrac14\log_3x\right) = \dfrac23\\
\Leftrightarrow \dfrac{1}{24}\log_3^4x = \dfrac23\\
\Leftrightarrow \log_3^4x = 16\\
\Leftrightarrow \left[\begin{array}{l}\log_3^2x = 4\\\log_3^2x = -4\quad (vn)\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}\log_3x = 2\\\log_3x = -2\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = 9\\x = \dfrac19\end{array}\right.\quad (nhận)\\
\text{Vậy}\ S = \left\{\dfrac19;9\right\}
\end{array}\)