${a^{m{{\log }_a}n}} = $
${a^{mn}}$
${m^n}$
${n^m}$
None of these
(c) ${a^{m{{\log }_a}n}} = {a^{{{\log }_a}{n^m}}} = {n^m}$.
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} = $
${{3\sqrt 2 } \over {\sqrt 6 + \sqrt 3 }} – {{4\sqrt 3 } \over {\sqrt 6 + \sqrt 2 }} + {{\sqrt 6 } \over {\sqrt 3 + \sqrt 2 }} = $
The value of ${{15} \over {\sqrt {10} + \sqrt {20} + \sqrt {40} – \sqrt 5 – \sqrt {80} }}$ is
If ${a^x} = bc,{b^y} = ca,\,{c^z} = ab,$ then $xyz$=
If ${({a^m})^n} = {a^{{m^n}}}$, then the value of $'m'$ in terms of $'n'$ is
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