Question

Show that: log_b a · log_c d = log_c a · log_b d.

Answer 1

Given: a, b, c, d > 0 with b ≠ 1, c ≠ 1 so all logarithms below are defined.

To Prove: log_b a · log_c d = log_c a · log_b d

Proof [Step-wise]:

1. Use the change-of-base formula:

For any positive x, y with base ≠ 1.

`log_x y = (ln y)/(ln x)` or any common logarithm.

2. Write each logarithm with natural logs:

`log_b a = (ln a)/(ln b)` 

`log_c d = (ln d)/(ln c)` 

`log_c a = (ln a)/(ln c)` 

`log_b d = (ln d)/(ln b)`

3. Compute the left-hand side:

log_b a · log_c d

= `(ln a/ln b) · (ln d/ln c)`

= `(ln a · ln d) / (ln b · ln c)`

4. Compute the right-hand side:

log_c a · log_b d

= `(ln a / ln c) · (ln d / ln b)`

= `(ln a · ln d) / (ln c · ln b)`

5. Since multiplication is commutative in the denominator

(ln b · ln c) = (ln c · ln b)

The two expressions are equal.

Therefore the LHS = RHS.

log_b a · log_c d = log_c a · log_b d, as required.