If {x^{{3 over 4}{{({{log }_3}x)}^2} + {{log | vedclass

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Question

(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

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(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, $LHS =$ ${3^{{3 \over 4}{{.1}^2} + 1 - \left( {{5 \over 4}} ight)}} = {3^{{2 \over 4}}} = {3^{{1 \over 2}}} = {m{RHS}}$.

$	herefore x = 3$ is a solution, which is a $ + ve$ integer.

Next, $\left[ {{3 \over 4}{{({{\log }_3}x)}^2} + {{\log }_3}x - {5 \over 4}} ight]\,{\log _3}x = {1 \over 2}$

$ \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}}$; $x = 3,\,{1 \over 9},\,{1 \over {oot 3 \of 3 }}$

Thus, there is  $one +ve$ integral value, one irrational value, two positive rational values.

If ${x^{{3 \over 4}{{({{\log }_3}x)}^2} + {{\log }_3}x - {5 \over 4}}} = \sqrt 3 $ then $x$ has

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