Discussion

Explanation:

Given, ${\log }_{27}\left(8\right)$ ≤ ${\log }_3\left(x\right)$ < $9^{\frac{1}{{\log }_2\left(3\right)}}$ 

⇒ ${\log }_{3^3}\left(2^3\right)$ ≤ ${\log }_3\left(x\right)$ < $9^{{\log }_3\left(2\right)}$ 

⇒ $\frac{3}{3}{\log }_3\left(2\right)$ ≤ ${\log }_3\left(x\right)$ < $2^{{\log }_3\left(9\right)}$ 

⇒ ${\log }_3\left(2\right)$ ≤ ${\log }_3\left(x\right)$ < $2^2$ 

⇒ ${\log }_3\left(2\right)$ ≤ ${\log }_3\left(x\right)$ < 4

⇒ ${\log }_3\left(2\right)$ ≤ ${\log }_3\left(x\right)$ < ${\log }_3\left(3^4\right)$

⇒ 2 ≤ x < 81

∴ x ∈ [2, 81)

Hence, option (a).

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