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

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

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$2n$

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(D) Given the equation: $\frac{{({2^{n + 1}})^m}({2^{2n}}){2^n}}{{({2^{m + 1}})^n}{2^{2m}}} = 1$ Simplify the numerator: ${2^{m(n+1)}} \cdot {2^{2n}} \cdot {2^n} = {2^{mn + m + 2n + n}} = {2^{mn + m + 3n}}$ Simplify the denominator: ${2^{n(m+1)}} \cdot {2^{2m}} = {2^{nm + n + 2m}}$ Equating the numerator and denominator: ${2^{mn + m + 3n}} = {2^{nm + n + 2m}}$ Since the bases are the same,equate the exponents: $mn + m + 3n = nm + n + 2m$ Subtract $mn$ from both sides: $m + 3n = n + 2m$ Rearrange the terms: $3n - n = 2m - m$ Therefore: $2n = m$ Thus,$m = 2n$.

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

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