Solve the following equations:
$2^{2 x + 8} + 1 = 32 . 2^{x}$ .

- 2, - 2

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2, - 2

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- 4, - 4

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- 4, 4

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Solution

The correct option is B $- 4, - 4$
Given, $2^{2 x + 8} + 1 = 32. 2^{x}$
$\Rightarrow 2^{2 x} . 2^{8} + 1 = 32. 2^{x}$
$\Rightarrow 256 {(2^{x})}^{2} + 1 = 32. 2^{x}$
Put $2^{x} = t$
Thus $256 t^{2} + 1 = 32 t$
$\Rightarrow 256 t^{2} - 32 t + 1 = 0$
$\Rightarrow 256 t^{2} - 16 t - 16 t + 1 = 0$
$\Rightarrow 16 t (16 t - 1) - 1 (16 t - 1) = 0$
$\Rightarrow (16 t - 1) (16 t - 1) = 0$
$\Rightarrow (16 t - 1)^{2} = 0$
$\Rightarrow t = \frac{1}{16}$
Resubstituting value, we get
$2^{x} = t$
$\Rightarrow 2^{x} = \frac{1}{16} = \frac{1}{2^{4}}$
$\Rightarrow 2^{x} = 2^{- 4}$
$\Rightarrow x = - 4$
Therefore, values of $x$ are $- 4, - 4$ .

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