If ${({a^m})^n} = {a^{{m^n}}}$, then the value of $'m'$ in terms of $'n'$ is
If ${({a^m})^n} = {a^{{m^n}}}$, then the value of $'m'$ in terms of $'n'$ is
- A
$n$
- B
${n^{1/m}}$
- C
${n^{1/(n - 1)}}$
- D
None of these
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The value of ${{15} \over {\sqrt {10} + \sqrt {20} + \sqrt {40} - \sqrt 5 - \sqrt {80} }}$ is
The value of ${{15} \over {\sqrt {10} + \sqrt {20} + \sqrt {40} - \sqrt 5 - \sqrt {80} }}$ is
${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $
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${a^{m{{\log }_a}n}} = $
${a^{m{{\log }_a}n}} = $
If ${2^x} = {4^y} = {8^z}$ and $xyz = 288,$ then ${1 \over {2x}} + {1 \over {4y}} + {1 \over {8z}} = $
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If ${a^x} = {(x + y + z)^y},{a^y} = {(x + y + z)^z}$, ${a^z} = {(x + y + z)^x},$ then
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