If $2^{x + 7} = 2048$ and $4^{x} . 3^{y} = 768$ , then the value of $(x^{2} + y^{3})$ is

31

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17

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28

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9

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Solution

The correct option is B $17$
$2^{x + 7} = 2048, 4^{x} 3^{y} = 768$

$2^{x + 7} = 2^{11}, 2^{2 x} 3^{y} = 2^{8} \times 3$

Equating the powers of same bases,

$\Rightarrow x + 7 = 11, 2 x = 8, y = 1$

$\Rightarrow x = 11 - 7, 2 x = 8, y = 1$

$\Rightarrow x = 4, x = 4, y = 1$

$∴ x = 4, y = 1$

$x^{2} + y^{3} = 4^{2} + 1^{3} = 16 + 1 = 17$

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