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

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Solution

The correct option is D 2n
$\frac{(2^{n + 1})^{m} (2^{2})^{n} 2^{n}}{(2^{m + 1})^{n} 2^{2 m}} = 1$

$\Rightarrow$ $\frac{(2^{m n + m}) . (2^{2})^{n} {.2}^{n}}{(2^{m n + n}) {.2}^{2 m}} = 1$

$\Rightarrow$ $\frac{2^{m n} {.2}^{m} {.2}^{2 n} {.2}^{n}}{2^{m n} {.2}^{n} {.2}^{2 m}} = 1$

$\Rightarrow$ $\frac{2^{2 n}}{2^{m}} = 1$

$\Rightarrow$ $2^{2 n} = 2^{m}$

$∴$ $m = 2 n$

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(2n+1)m(22)n2n(2m+1)n22m=1 then m= ..........

The correct option is D 2n(2n+1)m(22)n2n(2m+1)n22m=1⇒ (2mn+m).(22)n.2n(2mn+n).22m=1⇒ 2mn.2m.22n.2n2mn.2n.22m=1⇒ 22n2m=1⇒ 22n=2m∴ m=2n

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