If $2^{(x - y)} = 4$ and $2^{x + y} = 256$ , then `x' is

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Solution

The correct option is A
5

$\begin{matrix} 2^{(x - y)} = 2^{2} a n d 2^{x + y} = 2^{8} x - y = 2 a n d x + y = 8 ∴ x = 5 a n d y = 3 \end{matrix}$

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If $2^{(x - y)} = 4$ and $2^{x + y} = 256$ , then `x' is

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