d
(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.