Question

If $(2{)}^{m+1}\times 2^5=4^{-2}$ the value of $m$ is

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Answer 1

Given, $(2{)}^{m+1}\times 2^5=4^{-2}$
$\Rightarrow$ $(2{)}^{m+1}\times 2^5={\left(\right(2{)}^2)}^{-2}(\because (a^m{)}^n=a^{mn})$
$\Rightarrow$ $(2{)}^{m+1+5}=(2{)}^{-4}(\because a^m\times a^n=a^{m+n})$
Since the bases on both the sides are equal, therefore the exponents must also be equal. This means, $m+1+5=-4⟹m=-10$.