Question

If `2 log  (x - y)/2 = log x + log y`, prove that x² + y² – 6xy = 0.

Answer 1

Given: `2 log ((x - y)/2) = log x + log y`

To Prove: x² + y² – 6xy = 0

Proof [Step-wise]:

1. Use the power rule for logarithms: 

`2 log ((x - y)/2) = log [((x - y)/2)^2]`

2. Use the product rule for logarithms on the right: 

log x + log y = log (xy)

3. Equate the two single logarithms logs defined ⇒ arguments positive: 

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

4. Multiply both sides by 4: 

(x – y)² = 4xy

5. Expand the left-hand side: 

x² – 2xy + y² = 4xy

6. Bring all terms to one side: 

x² + y² – 2xy – 4xy = 0

⇒ x² + y² – 6xy = 0

Also note domain conditions from the logs: 

x > 0, y > 0

And x – y > 0

i.e. x > y

From the given equation we obtain x² + y² – 6xy = 0, as required.