Question

Prove that: `(log x)^2 - (log y)^2 = log (x/y) · log (xy)`.

Answer 1

Given:

x > 0, y > 0 and log denotes logarithm to the same base for both x and y so the laws of logs apply.

To Prove:

`(log x)^2 - (log y)^2 = log(x/y) xx log(xy)`

Proof [Step-wise]:

1. Start with the left-hand side:

(log x)² – (log y)².

2. Recognize this as a difference of squares and factor:

(log x – log y)(log x + log y)

3. Use logarithm laws:

`log x - log y = log (x/y)`

log x + log y = log (xy)

4. Substitute these into the factorization to obtain:

(log x – log y)(log x + log y) 

= `log (x/y) xx log (xy)`

Therefore `(log x)^2 - (log y)^2 = log(x/y) xx log(xy)`, as required.