Solve a basic exponential equation (repaired minimal typo): If a^(2x) = 1 with a > 0 and a ≠ 1, determine x.

Introduction / Context:
The equation a^(2x) = 1 asks for exponents that yield 1 for a positive base a ≠ 1. In real numbers, this happens exactly when the exponent is 0. (The originally provided “+ 2 = 1” appears to be a minor transcription error; applying the recovery-first policy, we correct to the standard solvable form.)

Given Data / Assumptions:

• a > 0, a ≠ 1.
• a^(2x) = 1.

Concept / Approach:
For a ≠ 1 and a > 0, a^t = 1 ⇔ t = 0. Therefore set 2x = 0 to get x. (We work over reals; complex solutions with periodic logarithms are beyond scope.)

Step-by-Step Solution:

Verification / Alternative check:
Pick a concrete base, say a = 2: 2^(2x) = 1 ⇒ 2x = 0 ⇒ x = 0; indeed 2^0 = 1.

Why Other Options Are Wrong:

• −2, −1, 1: Each gives a nonzero exponent, leading to values different from 1 for a ≠ 1.

Common Pitfalls:
Confusing special cases like a = 1 (any exponent gives 1) or a = 0 (undefined for negative exponents). The constraints a > 0, a ≠ 1 avoid these edge cases.

Final Answer: