Solve the following equations:
$2^{2 x + 3} - 57 = 65 (2^{x} - 1)$ .

- 3, - 3

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\pm 3

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\pm 4

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2, - 3

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Solution

The correct option is B $\pm 3$
Given, $2^{2 x + 3} - 57 = 65 (2^{x} - 1)$
$\Rightarrow 2^{2 x} . 2^{3} - 57 = 65 (2^{x} - 1)$
$\Rightarrow 8 {(2^{x})}^{2} - 57 = 65 (2^{x} - 1)$
Put $2^{x} = t$
Therefore, $8 t^{2} - 57 = 65 (t - 1)$
$\Rightarrow 8 t^{2} - 57 - 65 t + 65 = 0$
$\Rightarrow 8 t^{2} - 65 t + 8 = 0$
$\Rightarrow 8 t^{2} - 64 t - t + 8 = 0$
$\Rightarrow (8 t - 1) (t - 8) = 0$
$\Rightarrow t = \frac{1}{8}, 8$
Resubstituting values, we get
$2^{x} = t$
$\Rightarrow 2^{x} = \frac{1}{8} = {(\frac{1}{2})}^{3}$
$\Rightarrow 2^{x} = {(2^{- 1})}^{3} = 2^{- 3}$
$\Rightarrow x = - 3$
and $2^{x} = 8$
$\Rightarrow 2^{x} = 2^{3}$
$\Rightarrow x = 3$
So, complete solution is $x = \pm 3$ .

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