Sum of the roots of the equation $4^{x} - 3 (2^{x + 3}) + 128 = 0$ , is:

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Solution

The correct option is B 7
$4^{x} - 3 (2^{x + 3}) + 128 = 0$
$\Rightarrow (2^{x})^{2} - {24.2}^{x} + 128 = 0$
$\Rightarrow y^{2} - 24 y + 128 = 0$
$\Rightarrow y^{2} - 8 y - 16 y + 128 = 0$
$\Rightarrow y (y - 8) - 16 (y - 8) = 0$
$\Rightarrow (y - 8) (y - 16) = 0$
$y = 8$ or $y = 16$
$2^{x} = 2^{3}$ $2^{x} = 2^{4}$
$x = 3 x = 4$
sum of roots $= 3 + 4 = 7$

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4^{x} - 3 (2^{x + 3}) + 128 = 0

Sum of the roots of the equation is $4^{x} - 3 (2^{x + 3}) + 128 = 0$

4^{x} - 3 (2^{x + 3}) + 128 = 0

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4^{x} - 3 (2^{x + 3}) + 128 = 0

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