Question

The value of $m$ which satisfies $(2{)}^{m+1}\times 2^5=4^{-2}$ is:

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Answer 1

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