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
${{\sqrt {(5/2)} + \sqrt {(7 - 3\sqrt 5 )} } \over {\sqrt {(7/2)} + \sqrt {(16 - 5\sqrt 7 )} }}=$
For $x \ne 0,{\left( {{{{x^l}} \over {{x^m}}}} \right)^{({l^2} + lm + {m^2})}}$${\left( {{{{x^m}} \over {{x^n}}}} \right)^{({m^2} + nm + {n^2})}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{({n^2} + nl + {l^2})}}=$
${{\sqrt {6 + 2\sqrt 3 + 2\sqrt 2 + 2\sqrt 6 } - 1} \over {\sqrt {5 + 2\sqrt 6 } }}$
Let ${7 \over {{2^{1/2}} + {2^{1/4}} + 1}}$$ = A + B{.2^{1/4}} + C{.2^{1/2}} + D{.2^{3/4}}$, then $A+B+C+D= . . .$
$\sqrt {(3 + \sqrt 5 )} - \sqrt {(2 + \sqrt 3 )} = $