Correct answer: 5^(-16)
Setting t = log₅(x), we get 16t⁴ - 68t² + 16 = 0. Solving gives t = ±1/2, ±2. Since x > 0, all t values are valid. The product of all positive x values equals 5 raised to the sum of all t values that give x > 0, but we must only take positive x (all t values give positive x since 5^t > 0 always). Product = 5^(t1+t2+t3+t4) = 5^(1/2 - 1/2 + 2 - 2) = 5⁰ = 1... Re-examining: 16t⁴ - 68t² + 16 = 0 → 4t⁴ - 17t² + 4 = 0 → t² = (17 ± sqrt(289-64))/8 = (17 ± 15)/8. So t² = 4 or t² = 1/4. Thus t = ±2, ±1/2. Sum of all t = 0. Product of all positive x = 5⁰ = 1.