If 2\(^{x + y}\) = 16 and 4\(^{x - y} = \frac{1}{32}\), find the values of x and y.
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Correct Answer: Option B
Explanation:
2\(^{x + y}\) = 16 ; 4\(^{x - y}\) = \(\frac{1}{32}\).
\(\implies 2^{x + y} = 2^4\)
\(x + y = 4 ... (1)\)
\(2^{2(x - y)} = 2^{-5} \)
\(2^{2x - 2y} = 2^{-5}\)
\(\implies 2x - 2y = -5 ... (2)\)
Solving the equations (1) and (2) simultaneously, we have
x = \(\frac{3}{4}\) and y = \(\frac{13}{4}\)
2\(^{x + y}\) = 16 ; 4\(^{x - y}\) = \(\frac{1}{32}\).
\(\implies 2^{x + y} = 2^4\)
\(x + y = 4 ... (1)\)
\(2^{2(x - y)} = 2^{-5} \)
\(2^{2x - 2y} = 2^{-5}\)
\(\implies 2x - 2y = -5 ... (2)\)
Solving the equations (1) and (2) simultaneously, we have
x = \(\frac{3}{4}\) and y = \(\frac{13}{4}\)