A) \[8\]
B) \[{{8}^{2}}\]
C) \[{{8}^{-2}}\]
D) \[{{8}^{3}}\]
Correct Answer: B
Solution :
\[{{T}_{4}}={{T}_{3+1}}=\left( \begin{matrix} 6 \\ 3 \\ \end{matrix} \right){{\left( \frac{2}{x} \right)}^{3}}.{{\left( {{x}^{{{\log }_{8}}x}} \right)}^{3}}\] \[20\times {{8}^{7}}=\frac{160}{{{x}^{3}}}.{{x}^{3{{\log }_{8}}x}}\] \[{{8}^{6}}={{x}^{{{\log }_{2}}x}}-3\] \[{{2}^{18}}={{x}^{{{\log }_{2}}x-3}}\] \[\Rightarrow 18=(lo{{g}_{2}}x-3)(lo{{g}_{2}}x)\] Let\[{{\log }_{2}}x=t\] \[\Rightarrow {{t}^{2}}-3t-18=0\]\[\Rightarrow (t-6)(t+3)=0\] \[\Rightarrow t=6,-3\] \[{{\log }_{2}}x=6\Rightarrow x={{2}^{6}}={{8}^{2}}\] \[{{\log }_{2}}x=-3\Rightarrow x={{2}^{-3}}={{8}^{-1}}\]
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