If frac{{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}}} | vedclass

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(D) Given the expression: $\frac{{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}}}{{{{({2^{m + 1}})}^n}{2^{2m}}}} = 1$Simplify the numerator: ${2^{m(n+1)}} \cdot

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$2n$

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(D) Given the expression: $\frac{{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}}}{{{{({2^{m + 1}})}^n}{2^{2m}}}} = 1$Simplify the numerator: ${2^{m(n+1)}} \cdot {2^{2n}} \cdot {2^n} = {2^{mn+m+3n}}$Simplify the denominator: ${2^{n(m+1)}} \cdot {2^{2m}} = {2^{mn+n+2m}}$Equating the numerator and denominator since the fraction equals $1$:${2^{mn+m+3n}} = {2^{mn+n+2m}}$Equating the exponents:$mn + m + 3n = mn + n + 2m$Subtract $mn$ from both sides:$m + 3n = n + 2m$Rearranging the terms:$3n - n = 2m - m$$2n = m$Therefore,$m = 2n$.

If $\frac{{{{({2^{n + 1}})}^m}({2^{2n}}){2^n}}}{{{{({2^{m + 1}})}^n}{2^{2m}}}} = 1$,then $m =$

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