Đáp án:
\(x = {\log _3}\dfrac{{26}}{{17}}\)
Giải thích các bước giải:
\(\begin{array}{l}
b{.3^{x + 1}} - {3^{x - 2}} - {3^x} = \dfrac{{26}}{9}\\
\to {3.3^x} - \dfrac{1}{{{3^2}}}{.3^x} - {3^x} = \dfrac{{26}}{9}\\
\to {3^x}\left( {3 - \dfrac{1}{9} - 1} \right) = \dfrac{{26}}{9}\\
\to {3^x}.\dfrac{{17}}{9} = \dfrac{{26}}{9}\\
\to {3^x} = \dfrac{{26}}{{17}}\\
\to x = {\log _3}\dfrac{{26}}{{17}}
\end{array}\)
