If 2n-1m22n2n2m-1n22m - 1- then m -

The correct option is B 2n (2^n+1)^m(2^2)^n2^n(2^m+1)^n2^2m=1 ⇒2^mn+m+2n+n2^mn+n+2m=2 ⇒ 2^mn+m+2n+n mn n 2m =2^0 ⇒ 2n=m ...

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Byju's Answer

Standard VIII

Mathematics

Multiplication of Monomials

If 2n+1m22n...

Question

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

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Solution

The correct option is B 2n
$\frac{(2^{n + 1})^{m} (2^{2})^{n} 2^{n}}{(2^{m + 1})^{n} 2^{2 m}} = 1$
$\Rightarrow \frac{2^{m n + m + 2 n + n}}{2^{m n + n + 2 m}} = 2$
$\Rightarrow 2^{m n + m + 2 n + n - m n - n - 2 m} = 2^{0}$
$\Rightarrow 2 n = m$

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Similar questions

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

then

m =

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If $\frac{2^{m + n}}{2^{n - m}} = 16, \frac{3^{p}}{3^{n}} = 81 a n d a = 2^{\frac{1}{10}}, t h e n \frac{a^{2 m + n - p}}{(a^{m - 2 n + 2 p})^{- 1}} =$

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If (2n+1)m(22)n2n(2m+1)n22m=1, then m = ____

The correct option is B 2n(2n+1)m(22)n2n(2m+1)n22m=1⇒2mn+m+2n+n2mn+n+2m=2⇒2mn+m+2n+n−mn−n−2m=20⇒2n=m

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

The correct option is B 2n
$\frac{(2^{n + 1})^{m} (2^{2})^{n} 2^{n}}{(2^{m + 1})^{n} 2^{2 m}} = 1$
$\Rightarrow \frac{2^{m n + m + 2 n + n}}{2^{m n + n + 2 m}} = 2$
$\Rightarrow 2^{m n + m + 2 n + n - m n - n - 2 m} = 2^{0}$
$\Rightarrow 2 n = m$