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

Difficulty: Easy

Correct Answer: 1

Explanation:


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:

Let x = log_a b, y = log_b c, z = log_c a.Then xyz = (log_a b)(log_b c)(log_c a).Using change of base: log_b c = (log_a c)/(log_a b), and log_c a = 1/(log_a c).Thus xyz = (log_a b) * [(log_a c)/(log_a b)] * [1/(log_a c)] = 1.


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

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