If {{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}} over {{{({2^{m + 1}})}^n}{2^{2m}}}} = 1, then m =

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Basic of Logarithms

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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

$2n$

Solution

(d) ${2^{m(n + 1) + 2n + n}} = {2^{(m + 1)n + 2m}}$

$ \Rightarrow $$mn + m + 3n = mn + 2m + n$$ \Rightarrow $$m = 2n$.

Standard 11

Mathematics

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