Question

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

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• A. -7
• B. 7
• C. 10
• D. -10

Answer 1

The correct option is D -10

Given, $(2{)}^{m+1}\times 2^5=4^{-2}$
$\Rightarrow$ $(2{)}^{m+1}\times 2^5={\left(2^2\right)}^{-2}$

We know that​​​​​​, for any non-zero integer 'a' and any integers 'm' and 'n',
$a^m\times a^n=(a{)}^{m+n}$
$\Rightarrow (2{)}^{m+1}\times 2^5=(2{)}^{m+1+5}$
and $(a^m{)}^n=a^{mn}$
$\Rightarrow {\left(2^2\right)}^{-2}=(2{)}^{2\times -2}=(2{)}^{-4}$

$\therefore (2{)}^{m+1+5}=(2{)}^{-4}$

Since the powers on the LHS and the RHS are equal and the bases are the same, the exponents must also be equal.

This means, $m+1+5=-4$
$\therefore m=-10$