If ${x^{{3 \over 4}{{({{\log }_3}x)}^2} + {{\log }_3}x - {5 \over 4}}} = \sqrt 3 $ then $x$ has
One positive integral value
One irrational value
Two positive rational values
All of These
Logarithm of $32\root 5 \of 4 $ to the base $2\sqrt 2 $ is
The set of real values of $x$ satisfying ${\log _{1/2}}({x^2} - 6x + 12) \ge - 2$ is
If ${x_n} > {x_{n - 1}} > ... > {x_2} > {x_1} > 1$ then the value of ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}.....{\log _{{x_n}}}{x_n}^{x_{n - 1}^{{ {\mathinner{\mkern2mu\raise1pt\hbox{.}\mkern2mu \raise4pt\hbox{.}\mkern2mu\raise7pt\hbox{.}\mkern1mu}} ^{{x_1}}}}}$ is equal to
The value of $\sqrt {(\log _{0.5}^24)} $ is
The solution of the equation ${\log _7}{\log _5}$ $(\sqrt {{x^2} + 5 + x} ) = 0$