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
The equation $\sqrt {(x + 1)} - \sqrt {(x - 1)} = \sqrt {(4x - 1)} $, $x \in R$ has
If ${a^{x - 1}} = bc,{b^{y - 1}} = ca,{c^{z - 1}} = ab,$then $\sum {(1/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 ${{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}} \over {{{({2^{m + 1}})}^n}{2^{2m}}}} = 1,$ then $m =$
${{{{[4 + \sqrt {(15)} ]}^{3/2}} + {{[4 - \sqrt {(15)} ]}^{3/2}}} \over {{{[6 + \sqrt {(35)} ]}^{3/2}} - {{[6 - \sqrt {(35)} ]}^{3/2}}}} = $