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: \[ 3^{x^2} = 3^3 \times (3^2)^x. \] Now simplify the right-hand side: \[ 3^{x^2} = 3^3 \times 3^{2x}. \] Using the property of exponents \(a^m \times a^n = a^{m+n}\), we combine the powers of 3: \[ 3^{x^2} = 3^{3 + 2x}. \] Since the bases are the same, we can equate the exponents: \[ x^2 = 3 + 2x. \] Rearranging the equation: \[ x^2 - 2x - 3 = 0. \] Factoring the quadratic equation: \[ (x - 3)(x + 1) = 0. \] Thus, \(x = 3\) or \(x = -1\). 
Step 2: Evaluate \(\frac{2^{x^2}}{(2^x)^2}\).
We now substitute \(x = 3\) and \(x = -1\) into the expression \(\frac{2^{x^2}}{(2^x)^2}\): When \(x = 3\): \[ \frac{2^{3^2}}{(2^3)^2} = \frac{2^9}{2^6} = 2^{9-6} = 2^3. \] When \(x = -1\): \[ \frac{2^{(-1)^2}}{(2^{-1})^2} = \frac{2^1}{2^{-2}} = 2^{1 - (-2)} = 2^3. \] Thus, the value is \(2^3\), which corresponds to Option (C).