(D) ${x^{{3 \over 4}{{({{\log }_3}x)}^2} + {{\log }_3}x – {5 \over 4}}}$=$\sqrt 3 = {3^{{1 \over 2}}}$.

There is a possibility of a solution $x = 3$

For this value, $LHS =$ ${3^{{3 \over 4}{{.1}^2} + 1 – \left( {{5 \over 4}} \right)}} = {3^{{2 \over 4}}} = {3^{{1 \over 2}}} = {\rm{RHS}}$.

$\therefore x = 3$ is a solution, which is a $ + ve$ integer.

Next, $\left[ {{3 \over 4}{{({{\log }_3}x)}^2} + {{\log }_3}x – {5 \over 4}} \right]\,{\log _3}x = {1 \over 2}$

$ \Rightarrow $$[3\,{({\log _3}x)^2} + 4{\log _3}x – 5]\,{\log _3}x – 2 = 0$

$ \Rightarrow $$3{t^3} + 4{t^2} – 5t – 2 = 0$, [$t = {\log _3}x$]

$ \Rightarrow $$3{t^3} – 3{t^2} + 7{t^2} – 7t + 2t – 2 = 0$

$ \Rightarrow $$(3{t^2} + 7t + 2)\,(t – 1) = 0$$ \Rightarrow $ $(3t + 1)\,(t + 2)\,(t – 1) = 0$

$ \Rightarrow $$t = 1,\, – 2,\, – {1 \over 3}$$ \Rightarrow $${\log _3}x = 1,\, – 2,\, – {1 \over 3}$

$ \Rightarrow $$x = {3^1},\,{3^{ – 2}},{3^{ – 1/3}}$; $x = 3,\,{1 \over 9},\,{1 \over {\root 3 \of 3 }}$

Thus, there is  $one +ve$ integral value, one irrational value, two positive rational values.