If the fourth term in the binomial expansion of

$\mathrm{T}_{4}=\mathrm{T}_{3+1}=\left(\begin{array}{l}6 \\ 3\end{array}\right)\left(\frac{2}{\mathrm{x}}\right)^{3} \cdot\left(\mathrm{x}^{\log _{\mathrm{s}} \mathrm{x}}\right)^{3}$

$20 \times 8^{7}=\frac{160}{x^{3}} \cdot x^{3 \log _{8} x}$

$8^{6}=x^{\log _{2} x}-3$

$2^{18}=x^{\log _{2} x-3}$

$\Rightarrow 18=\left(\log _{2} x-3\right)\left(\log _{2} x\right)$

Let $\log _{2} x=t$

$\Rightarrow t^{2}-3 t-18=0$

$\Rightarrow(t-6)(t+3)=0$

$\Rightarrow t=6,-3$

$\log _{2} x=6 \Rightarrow x=2^{6}=8^{2}$

$\log _{2} x=-3 \Rightarrow x=2^{-3}=8^{-1}$