Có bao nhiêu số nguyên \(x\) thỏa mãn \({\log _2}\frac{{{x^2} - 1}}{{81}} < {\log _3}\frac{{{x^2} - 1}}{{16}}\)?

  

Điều kiện: \({x^2} - 1 > 0 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x > 1}\\{x < - 1}\end{array}} \right..\)

Ta có \({\log _2}\frac{{{x^2} - 1}}{{81}} < {\log _3}\frac{{{x^2} - 1}}{{16}}\)

\( \Leftrightarrow {\log _2}\left( {{x^2} - 1} \right) - {\log _2}81 < {\log _3}\left( {{x^2} - 1} \right) - {\log _3}16\)\( \Leftrightarrow {\log _2}3 \cdot {\log _3}\left( {{x^2} - 1} \right) - 4{\log _2}3 < {\log _3}\left( {{x^2} - 1} \right) - 4{\log _3}2\)

\( \Leftrightarrow \left( {{{\log }_2}3 - 1} \right){\log _3}\left( {{x^2} - 1} \right) < 4\left( {{{\log }_2}3 - {{\log }_3}2} \right)\)

\( \Leftrightarrow {\log _3}\left( {{x^2} - 1} \right) < \frac{{4\left( {{{\log }_2}3 - {{\log }_3}2} \right)}}{{{{\log }_2}3 - 1}}\)\( \Leftrightarrow {\log _3}\left( {{x^2} - 1} \right) < \frac{{4\left( {\frac{1}{{{{\log }_3}2}} - {{\log }_3}2} \right)}}{{\frac{1}{{{{\log }_3}2}} - 1}}\)

\( \Leftrightarrow {\log _3}\left( {{x^2} - 1} \right) < 4\left( {1 + {{\log }_3}2} \right) \Leftrightarrow {\log _3}\left( {{x^2} - 1} \right) < {\log _3}{6^4}\)

\( \Leftrightarrow 0 < {x^2} - 1 < {6^4} \Leftrightarrow \left[ \begin{array}{l}1 < x < \sqrt {1297} \\ - \sqrt {1297} < x < - 1\end{array} \right.\).

Vì \(x\) là số nguyên nên \(x \in \left\{ { - 36\,;\,\, - 35\,;\,\, \ldots ;\,\, - 2\,;\,\,2\,;\,\, \ldots ;\,\,35\,;\,\,36} \right\}.\)

Vậy có 70 số nguyên \(x\) thỏa mãn yêu cầu bài toán. Chọn C.