If x = logc (ab), y = loga(bc), z = logb(ca), where a, b, c > 0 are each…

Concept

For positive a, b (a ≠ 1), the change-of-base rule gives loga(b) = log(b)/log(a) for any common base. Also, the product rule states log(mn) = log(m) + log(n). (Assume a, b, c > 0, each different from 1, and abc ≠ 1, so every logarithm here and each of 1+x, 1+y, 1+z is defined and nonzero.)

Application

1. Using the change-of-base rule, x = logc(ab) = log(ab)/log(c) = [log(a) + log(b)]/log(c). Similarly, y = loga(bc) = [log(b) + log(c)]/log(a), and z = logb(ca) = [log(c) + log(a)]/log(b).

2. Add 1 to x: 1 + x = 1 + [log(a) + log(b)]/log(c) = [log(c) + log(a) + log(b)]/log(c) = log(abc)/log(c).

3. By the same steps, 1 + y = log(abc)/log(a) and 1 + z = log(abc)/log(b).

4. Sum the reciprocals: 1/(1+x) + 1/(1+y) + 1/(1+z) = log(c)/log(abc) + log(a)/log(abc) + log(b)/log(abc) = [log(a) + log(b) + log(c)]/log(abc) = log(abc)/log(abc) = 1.

Cross-check

Independent check: take a = b = c = 10. Then ab = bc = ca = 100, so x = y = z = log10(100) = 2, and 1 + x = 1 + y = 1 + z = 3. The sum of reciprocals is 1/3 + 1/3 + 1/3 = 1, matching the general result.

So (1+x)−1 + (1+y)−1 + (1+z)−1 = 1 holds identically, confirming this option.