Question

Show that: `1/(log_36 12) + 1/(log_6 12) + 1/(log_8 12) = 3`

Answer 1

Given: The identity to prove is `1/(log_36 12) + 1/(log_6 12) + 1/(log_8 12) = 3`

To Prove: Show that `1/(log_36 12) + 1/(log_6 12) + 1/(log_8 12) = 3`

Proof [Step-wise]:

1. Use the change-of-base reciprocal identity:

`1/(log_a b) = log_b a`

2. Apply this to each term:

`1/(log_36 12) = log_12 36`

`1/(log_6 12) = log_12 6`

`1/(log_8 12) = log_12 8`

3. Sum the three:

log₁₂ 36 + log₁₂ 6 + log₁₂ 8 

= log₁₂ (36 × 6 × 8)

Since log_b x + log_b y = log_b (xy)

4. Compute the product:

36 × 6 × 8

= 36 × 48

= 1728

= 12³

5. Therefore, the sum

= log₁₂ (12³)

= 3

Hence `1/(log_36 12) + 1/(log_6 12) + 1/(log_8 12) = 3`, as required.