${a^{m{{\log }_a}n}} = $
${a^{m{{\log }_a}n}} = $
- A
${a^{mn}}$
- B
${m^n}$
- C
${n^m}$
- D
None of these
Similar Questions
The number of integers $q , 1 \leq q \leq 2021$, such that $\sqrt{ q }$ is rational, and $\frac{1}{ q }$ has a terminating decimal expansion, is
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- [KVPY 2021]
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 $x = {{\sqrt 5 + \sqrt 2 } \over {\sqrt 5 - \sqrt 2 }},y = {{\sqrt 5 - \sqrt 2 } \over {\sqrt 5 + \sqrt 2 }},$ then $3{x^2} + 4xy - 3{y^2} = $
$\sqrt {(3 + \sqrt 5 )} $ is equal to
$\sqrt {(3 + \sqrt 5 )} $ is equal to
If ${a^x} = {(x + y + z)^y},{a^y} = {(x + y + z)^z}$, ${a^z} = {(x + y + z)^x},$ then
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${{\sqrt {(5/2)} + \sqrt {(7 - 3\sqrt 5 )} } \over {\sqrt {(7/2)} + \sqrt {(16 - 5\sqrt 7 )} }}=$
${{\sqrt {(5/2)} + \sqrt {(7 - 3\sqrt 5 )} } \over {\sqrt {(7/2)} + \sqrt {(16 - 5\sqrt 7 )} }}=$