Discussion
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
The value is \(2^{3}\).
Write both sides as powers of 3: \(27 \times 9^{x} = 3^{3} \times 3^{2x} = 3^{3 + 2x}\). Equating exponents:
\[ \begin{aligned} x^{2} &= 3 + 2x \\ x^{2} - 2x &= 3 \end{aligned} \]
The expression simplifies to the same combination:
\[ \frac{2^{x^{2}}}{(2^{x})^{2}} = 2^{x^{2} - 2x} = 2^{3} \]
There is no need to solve for x (it is 3 or −1, and both give \(2^{3}\)). \(2^{15}\) comes from substituting x = 3 into \(2^{x^{2}} \times 2^{2x}\), multiplying instead of dividing, and \(2^{0}\) and \(2^{-1}\) do not follow from the exponent equation.
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