Đáp án:
\(
\left[ {\begin{array}{*{20}c}
{x = \log _3 10} \\
{x = \log _3 \frac{{28}}{{27}}} \\
\end{array}} \right.
\)
Giải thích các bước giải:
\(
\begin{array}{l}
\log _3 (3^x - 1)\log _3 (3^{x + 1} - 3) = 6 \\
Đk:x > 0 \\
\Leftrightarrow \log _3 (3^x - 1).\log _3 3(3^x - 1) = 6 \\
\Leftrightarrow \log _3 (3^x - 1){\rm{[1 + log}}_{\rm{3}} (3^x - 1){\rm{] = 6}} \\
\Leftrightarrow {\rm{log}}_{\rm{3}} ^2 (3^x - 1) + \log _3 (3^x - 1) - 6 = 0 \\
\Leftrightarrow \left[ {\begin{array}{*{20}c}
{\log _3 (3^x - 1) = 2} \\
{\log _3 (3^x - 1) = - 3} \\
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}c}
{3^x - 1 = 3^2 = 9} \\
{3^x - 1 = 3^{ - 3} = \frac{1}{{27}}} \\
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}c}
{x = \log _3 10} \\
{x = \log _3 \frac{{28}}{{27}}} \\
\end{array}} \right.(tm) \\
\end{array}
\)
