If ${({a^m})^n} = {a^{{m^n}}}$, then the value of $'m'$ in terms of $'n'$ is
$n$
${n^{1/m}}$
${n^{1/(n - 1)}}$
None of these
Solution of the equation ${4.9^{x - 1}} = 3\sqrt {({2^{2x + 1}})} $ has the solution
If ${x^y} = {y^x},$then ${(x/y)^{(x/y)}} = {x^{(x/y) - k}},$ where $k = $
${{\sqrt {6 + 2\sqrt 3 + 2\sqrt 2 + 2\sqrt 6 } - 1} \over {\sqrt {5 + 2\sqrt 6 } }}$
If ${a^{x - 1}} = bc,{b^{y - 1}} = ca,{c^{z - 1}} = ab,$then $\sum {(1/x) = } $
${{\sqrt 2 } \over {\sqrt {(2 + \sqrt 3 )} - \sqrt {(2 - \sqrt 3 } )}} = $