Question

If `1/(log_a x) + 1/(log_c x) = 2/(log_b x)`, then prove that b² = ac.

Answer 1

Given: `1/(log_a x) + 1/(log_c x) = 2/(log_b x)`, with a, b, c, x > 0 and a, b, c, x ≠ 1

To Prove: b² = ac

Proof (Step-wise):

1. Use the reciprocal identity for logarithms:

`1/(log_a x) = log_x a` and similarly for the others.

2. Replace each reciprocal in the given equation:

log_x a + log_x c = 2 log_x b

3. Combine the left-hand logs using the product rule:

log_x (ac) = log_x (b²)

4. Since log_x is one-to-one for x > 0, x ≠ 1, the arguments must be equal: 

ac = b²

Therefore b² = ac.