Có bao nhiêu số nguyên \(x\) thỏa mãn \(\left[ {{{\log }_2}\left( {{x^2} + 1} \right) - {{\log }_2}\left( {x + 31} \right)} \right]\left( {32 - {2^{x - 1}}} \right) \ge 0\)?

Điều kiện: \(x > 0\).

Ta có \({\log _2}\left( {2x} \right) \cdot \log \left( {\frac{{100}}{x}} \right) > 2\)\( \Leftrightarrow \left( {1 + {{\log }_2}x} \right)\left( {2 - \log x} \right) > 2\)

\( \Leftrightarrow 2 - \log x + 2{\log _2}x - \log x \cdot {\log _2}x > 2\)\( \Leftrightarrow 2{\log _2}x - \log 2 \cdot {\log _2}x - \log x \cdot {\log _2}x > 0\)

\[ \Leftrightarrow {\log _2}x\left( {2 - \log 2 - \log x} \right) > 0 \Leftrightarrow {\log _2}x\left( {\log 50 - \log x} \right) > 0\]

\( \Leftrightarrow \left[ {\left\{ {\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{{{\log }_2}x > 0}\\{\log 50 - \log x > 0}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{l}}{{{\log }_2}x < 0}\\{\log 50 - \log x < 0}\end{array}} \right.}\end{array}} \right.} \right.\)\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\left\{ {\begin{array}{*{20}{l}}{x > 1}\\{x < 50}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{l}}{x < 1}\\{x > 50}\end{array}} \right.}\end{array} \Leftrightarrow 1 < x < 50} \right.\).

Vậy có 48 số nguyên thỏa mãn. Chọn B.