Question

If a real variable \( x \) satisfies \( 3^{x^2} = 27 \times 9^x \), then the value of \( \frac{2^{x^2}}{(2^x)^2} \) is:

• A. \(2^{-1}\)
• B. \(2^0\)
• C. \(2^3\)
• D. \(2^{15}\)

Hint:

When solving equations with exponents, simplify both sides of the equation, equate the exponents, and solve the resulting algebraic equation. Afterward, substitute the values into the given expression.

Answer 1

The Correct Option is C

Step 1: Solve the given equation \(3^{x^2} = 27 \times 9^x\).

We start with the equation:
\[ 3^{x^2} = 27 \times 9^x \]
We can rewrite 27 and 9 as powers of 3:
\[ 27 = 3^3 \quad \text{and} \quad 9 = 3^2 \]
Thus, the equation becomes:
\[ 3^{x^2} = 3^3 \times (3^2)^x = 3^3 \times 3^{2x} \]
Combine the exponents:
\[ 3^{x^2} = 3^{3 + 2x} \]
Since the bases are the same, equate the exponents:
\[ x^2 = 3 + 2x \]
Rearranging terms:
\[ x^2 - 2x - 3 = 0 \]
Factoring:
\[ (x - 3)(x + 1) = 0 \]
So the possible values of \( x \) are: \( x = 3 \) or \( x = -1 \)

Step 2: Evaluate \( \frac{2^{x^2}}{(2^x)^2} \)

For \( x = 3 \):
\[ \frac{2^{3^2}}{(2^3)^2} = \frac{2^9}{2^6} = 2^{9 - 6} = 2^3 \]
For \( x = -1 \):
\[ \frac{2^{(-1)^2}}{(2^{-1})^2} = \frac{2^1}{2^{-2}} = 2^{1 - (-2)} = 2^3 \]

Therefore, the value is \( 2^3 = 8 \), which corresponds to Option (C).