Solve: \(4(2^{x^2}) = 8^{x}\)
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Correct Answer: Option A
Explanation:
\(4(2^{x^2}) = 8^{x} \equiv (2^{2})(2^{x^2}) = (2^{3})^{x}\)
\(\implies 2^{2 + x^{2}} = 2^{3x}\)
Comparing bases, we have
\(2 + x^{2} = 3x \implies x^{2} - 3x + 2 = 0\)
\(x^{2} - 2x - x + 2 = 0 \)
\(x(x - 2) - 1(x - 2) = 0\)
\((x - 1) = 0\) or \((x - 2) = 0\)
\(x = \text{1 or 2}\)
\(4(2^{x^2}) = 8^{x} \equiv (2^{2})(2^{x^2}) = (2^{3})^{x}\)
\(\implies 2^{2 + x^{2}} = 2^{3x}\)
Comparing bases, we have
\(2 + x^{2} = 3x \implies x^{2} - 3x + 2 = 0\)
\(x^{2} - 2x - x + 2 = 0 \)
\(x(x - 2) - 1(x - 2) = 0\)
\((x - 1) = 0\) or \((x - 2) = 0\)
\(x = \text{1 or 2}\)