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

$6$

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$5$

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$7$

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$8$

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Solution

The correct option is C
$7$

Put $2^{x} = y$ , then the equation becomes:

$y^{2} - 3 (8 y) + 128 = 0 \Rightarrow y^{2} - 24 y + 128 = 0 \Rightarrow (y - 8) (y - 16) = 0 \Rightarrow y = 16, 8 \Rightarrow 2^{x} = 16, 8 \Rightarrow x = 4, 3$

$∴$ Sum of the roots is $7$ .

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

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

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