If a^x = b, b^y = c, and c^z = a (all positive, a, b, c ≠ 1), find the value of xyz.

Introduction / Context:This is a classic log–exponent chain identity. Converting each equation to logarithmic form and multiplying yields a telescoping product that evaluates neatly to 1.

Given Data / Assumptions:

• a, b, c > 0 and not equal to 1 (to avoid degenerate logs).
• a^x = b, b^y = c, c^z = a.

Concept / Approach:

• Take logarithms to a common base (any base is fine): log_a b, log_b c, log_c a.
• Use the change-of-base property to relate the factors.

Step-by-Step Solution:

Verification / Alternative check:Pick a = 2, b = 4, c = 16: then x = 2, y = 2, z = 1/4 gives xyz = 1, confirming the identity.

Why Other Options Are Wrong:

• 0, 2, 4 contradict the invariant telescoping product.

Common Pitfalls:

• Mistaking log multiplication rules; the telescoping only occurs with change-of-base handled correctly.

Final Answer:1