Number of integers less than or equal to 10 satisfying the $\text{ inequality }2{\log }_{1/2}(x-1)\le \frac{1}{3}-\frac{1}{{\log }_{x^2-x}8}\text{ is }$

$\begin{array}{l}\text{ We must have }x>1\\ 2{\log }_{1/2}(x-1)\le \frac{1}{3}-\frac{1}{{\log }_{x^2-x}8}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{3}-\frac{{\log }_2\left(x^2-x\right)}{3}+2{\log }_2(x-1)\ge 0\\ \Rightarrow {\log }_{2}2-{\log }_2\left(x^2-x\right)+6{\log }_2(x-1)\ge 0\\ \Rightarrow {\log }_2\frac{2(x-1{)}^6}{x(x-1)}\ge 0\\ \Rightarrow \frac{2(x-1{)}^5}{x}\ge 1\\ \text{ Putting }x-1=y\Rightarrow y>0\\ \Rightarrow \frac{2y^5}{y+1}-1\ge 0\\ \Rightarrow \frac{2y^5-y-1}{y+1}\ge 0\\ \Rightarrow \frac{2y^5-2y+y-1}{y+1}\ge 0\\ \Rightarrow \frac{2y\left(y^4-1\right)+y-1}{y+1}\ge 0\end{array}$

$\begin{array}{l}\Rightarrow \frac{(y-1)\left[2y(y+1)\left(y^2+1\right)+1\right]}{y+1}\ge 0\\ \Rightarrow \frac{y-1}{y+1}\ge 0\\ \Rightarrow y\ge 1\\ \Rightarrow x\ge 2\end{array}$