If a = b^x, b = c^y and c = a^z for positive real numbers a, b and c not equal to 1, what is the value of the product x y z?

Introduction / Context:
This question involves exponential relationships between three positive real numbers a, b and c. Each number is expressed as a power of another, and we are asked to find the value of the product x y z. It tests the ability to work with logarithms and exponents in a system of equations and to recognise how cyclic relationships can force a simple condition on the exponents.

Given Data / Assumptions:
- a = b^x.
- b = c^y.
- c = a^z.
- The numbers a, b and c are positive and not equal to 1.
- We must find the numerical value of x y z that always holds for such a system.

Concept / Approach:
We use logarithms to convert the exponential equations into linear equations in terms of log a, log b and log c. This makes the relationships easier to handle algebraically. After expressing log a, log b and log c in terms of one another and the exponents x, y and z, we combine the three equations in a way that isolates the product x y z. The crucial step is multiplying the equations so that the logs cancel and we can solve for x y z as a constant.

Step-by-Step Solution:
Start with the three equations: a = b^x, b = c^y and c = a^z. Take natural logs (or any consistent log base) of all three equations. From a = b^x, we get ln a = x ln b. From b = c^y, we get ln b = y ln c. From c = a^z, we get ln c = z ln a. Now substitute ln b from the second equation into the first: ln a = x (y ln c) = x y ln c. Then substitute ln c from the third equation: ln a = x y (z ln a) = x y z ln a. So ln a = x y z ln a. Since a is positive and not equal to 1, ln a is not zero. Divide both sides by ln a to obtain 1 = x y z. Therefore x y z = 1.

Verification / Alternative check:
To verify, we can construct a simple example. Let a, b and c all be equal to 2. Then a = b^x becomes 2 = 2^x so x = 1. Similarly b = c^y becomes 2 = 2^y so y = 1, and c = a^z becomes 2 = 2^z so z = 1. Then x y z = 1 * 1 * 1 = 1, matching the derived condition. If we choose any other positive a, b, c not equal to 1 that satisfy the given cyclic relations, the same logarithmic argument will show that x y z must always be 1.

Why Other Options Are Wrong:
-1: This would imply that the product of the exponents is negative, which contradicts the derived equation ln a = x y z ln a for non zero ln a.
0: This would require at least one of x, y or z to be zero, which would break one of the original exponential relationships unless one of a, b or c were equal to 1, which is not allowed.
abc: This is a product of the bases, not of the exponents, and has no support from the logarithmic analysis of the system.

Common Pitfalls:
A common mistake is to attempt to substitute the expressions directly in their exponential form without using logarithms, which can become messy and confusing. Another error is to cancel terms incorrectly when manipulating exponents, especially when raising one equation to another power. Some students also forget the condition that a, b and c are not equal to 1, which ensures that ln a, ln b and ln c are non zero and division is valid. Using logarithms systematically and recalling that logs convert exponents into multipliers leads directly to the simple result x y z = 1.

Final Answer:
The value of the product x y z is 1.