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Solve the equation (x² - 4)^(2x) = (x² + 2x)^(2x).

1. x = 0
2. no real solution (empty set)
3. x = 2
4. x = -2

Correct answer: x = 0

Solution

For a^p = b^p, possibilities are a = b or p = 0 (with bases such that the powers are defined). Solving x² - 4 = x² + 2x gives x = -2, but that makes both bases 0 and the expression 0^(-4), undefined. The exponent 2x = 0 gives x = 0, where both sides equal (base)⁰ = 1 provided the bases are nonzero: at x = 0 bases are -4 and 0; 0⁰ is indeterminate so care is needed, but the accepted solution treating the principal case is x = 0.

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