Solution of the equation ${9^x} - {2^{x + {1 \over 2}}} = {2^{x + {3 \over 2}}} - {3^{2x - 1}}$
${\log _9}(9/\sqrt 8 )$
${\log _{\left( {9/2} \right)}}(9/\sqrt 8 )$
${\log _e}(9/\sqrt 8 )$
None of these
The value of $\sqrt {[12 - \sqrt {(68 + 48\sqrt 2 )} ]} = $
If ${\left( {{2 \over 3}} \right)^{x + 2}} = {\left( {{3 \over 2}} \right)^{2 - 2x}},$then $x =$
If $x = {{\sqrt 5 + \sqrt 2 } \over {\sqrt 5 - \sqrt 2 }},y = {{\sqrt 5 - \sqrt 2 } \over {\sqrt 5 + \sqrt 2 }},$ then $3{x^2} + 4xy - 3{y^2} = $
If ${2^x} = {4^y} = {8^z}$ and $xyz = 288,$ then ${1 \over {2x}} + {1 \over {4y}} + {1 \over {8z}} = $
The rationalising factor of ${a^{1/3}} + {a^{ - 1/3}}$ is