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
If $x = \sqrt 7 + \sqrt 3 $ and $xy = 4,$then ${x^4} + {y^4}=$
If $x = \sqrt 7 + \sqrt 3 $ and $xy = 4,$then ${x^4} + {y^4}=$
${{\sqrt 2 } \over {\sqrt {(2 + \sqrt 3 )} - \sqrt {(2 - \sqrt 3 } )}} = $
${{\sqrt 2 } \over {\sqrt {(2 + \sqrt 3 )} - \sqrt {(2 - \sqrt 3 } )}} = $
Solution of the equation ${9^x} - {2^{x + {1 \over 2}}} = {2^{x + {3 \over 2}}} - {3^{2x - 1}}$
Solution of the equation ${9^x} - {2^{x + {1 \over 2}}} = {2^{x + {3 \over 2}}} - {3^{2x - 1}}$
The square root of $\frac{(0.75)^3}{1-(0.75)}+\left[0.75+(0.75)^2+1\right]$ is
The square root of $\frac{(0.75)^3}{1-(0.75)}+\left[0.75+(0.75)^2+1\right]$ is
- [KVPY 2012]
${a^{m{{\log }_a}n}} = $
${a^{m{{\log }_a}n}} = $