Đề kiểm tra Phương trình, bất phương trình mũ và lôgarit (có lời giải) - Đề 2

Ta có: \({9^{x + 1}} = {27^{2x + 1}}\)\( \Leftrightarrow {3^{2x + 2}} = {3^{6x + 3}}\)\( \Leftrightarrow 2x + 2 = 6x + 3\)\( \Leftrightarrow x =  - \frac{1}{4}\).

\({\log _2}\left( {x - 1} \right) < 3\)\( \Leftrightarrow \left\{ \begin{array}{l}x - 1 > 0\\x - 1 < {2^3}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x > 1\\x < 9\end{array} \right. \Leftrightarrow 1 < x < 9\).

Điều kiện \(x - 3 > 0 \Leftrightarrow x > 3\).

\({\log _{\frac{1}{2}}}\left( {x - 3} \right) \ge 2\)\( \Leftrightarrow x - 3 \le {\left( {\frac{1}{2}} \right)^2}\)\( \Leftrightarrow x \le \frac{{13}}{4}\).

Điều kiện \(\left\{ \begin{array}{l}{x^2} + 4x > 0\\2x + 3 > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x > 0\\x <  - 4\end{array} \right.\\x >  - \frac{3}{2}\end{array} \right. \Leftrightarrow x > 0\).

Phương trình \( \Leftrightarrow {\log _3}\left( {{x^2} + 4x} \right) = {\log _3}\left( {2x + 3} \right)\)\( \Leftrightarrow {x^2} + 4x = 2x + 3\)\( \Leftrightarrow {x^2} + 2{\rm{x}} - 3 = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}x = 1\\x =  - 3\end{array} \right.\).

\[{\log _{\frac{2}{5}}}\left( {x - 4} \right) + 1 > 0\]\[ \Leftrightarrow \]\[{\log _{\frac{2}{5}}}\left( {x - 4} \right) >  - 1\]\[ \Leftrightarrow \]\[0 < x - 4 < \frac{5}{2}\]\[ \Leftrightarrow \]\[4 < x < \frac{{13}}{2}\].

Ta có: \[{\log _{\frac{1}{{\sqrt 5 }}}}\left( {{6^{x + 1}} - {{36}^x}} \right) =  - 2 \Leftrightarrow {6^{x + 1}} - {36^x} = 5 \Leftrightarrow \left[ \begin{array}{l}{6^x} = 1\\{6^x} = 5\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = {\log _6}5\end{array} \right.\].