If ${{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}} \over {{{({2^{m + 1}})}^n}{2^{2m}}}} = 1,$ then $m =$
If ${{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}} \over {{{({2^{m + 1}})}^n}{2^{2m}}}} = 1,$ then $m =$
- A
$0$
- B
$1$
- C
$n$
- D
$2n$
Similar Questions
$\root 4 \of {(17 + 12\sqrt 2 )} = $
$\root 4 \of {(17 + 12\sqrt 2 )} = $
The rationalising factor of $2\sqrt 3 - \sqrt 7 $ is
The rationalising factor of $2\sqrt 3 - \sqrt 7 $ is
${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $
${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $
If $x = {2^{1/3}} - {2^{ - 1/3}},$ then $2{x^3} + 6x = $
If $x = {2^{1/3}} - {2^{ - 1/3}},$ then $2{x^3} + 6x = $
The rationalising factor of ${a^{1/3}} + {a^{ - 1/3}}$ is
The rationalising factor of ${a^{1/3}} + {a^{ - 1/3}}$ is