If a real variable \(x\) satisfies \(3^{x^{2}} = 27 \times 9^{x}\), then the value of \(\dfrac{2^{x^{2}}}{(2^{x})^{2}}\) is:
Answer
(C)
Explanation
Write the right side as a power of 3: \(27 \times 9^{x} = 3^{3} \times 3^{2x} = 3^{2x + 3}\). Equating exponents gives \(x^{2} = 2x + 3\), that is, \(x^{2} - 2x = 3\).
The required value is \(\dfrac{2^{x^{2}}}{(2^{x})^{2}} = \dfrac{2^{x^{2}}}{2^{2x}} = 2^{x^{2} - 2x} = 2^{3}\).
There is no need to find \(x\): both roots, \(x = 3\) and \(x = -1\), satisfy \(x^{2} - 2x = 3\). Adding the exponents instead of subtracting them (with \(x = 3\)) would give the wrong option \(2^{15}\).
Answer: C