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
${{\sqrt 2 } \over {\sqrt {(2 + \sqrt 3 )} - \sqrt {(2 - \sqrt 3 } )}} = $
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 $x = \sqrt 7 + \sqrt 3 $ and $xy = 4,$then ${x^4} + {y^4}=$
If $a = \sqrt {(21)} - \sqrt {(20)} $ and $b = \sqrt {(18)} - \sqrt {(17),} $ then
For $x \ne 0,{\left( {{{{x^l}} \over {{x^m}}}} \right)^{({l^2} + lm + {m^2})}}$${\left( {{{{x^m}} \over {{x^n}}}} \right)^{({m^2} + nm + {n^2})}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{({n^2} + nl + {l^2})}}=$