Question

If x² + y² = 34 xy, prove that `log ((x + y)/6) = 1/2 (log x + log y)`.

Answer 1

Given: x² + y² = 34 xy

To Prove: `log((x + y)/6) = 1/2 (log x + log y)` logs taken in any fixed base > 0, ≠ 1; assume x > 0, y > 0 so logs exist.

Proof [Step-wise]:

1. Start from the given and form (x + y)²:

(x + y)² = x² + 2xy + y²

2. Substitute x² + y² = 34xy into the expression:

(x + y)²

= (x² + y²) + 2xy 

= 34xy + 2xy

= 36xy

3. Therefore (x + y)² = 36xy.

So `x + y = ±6sqrt(xy)`.

4. Since x > 0 and y > 0 for logs to be defined, x + y > 0 and `sqrt(xy) > 0`. 

Hence, we take the positive root:

`x + y = 6sqrt(xy)` 

⇒ `(x + y)/6 = sqrt(xy)`

5. Apply the logarithm to both sides and use

`log(sqrt(xy)) = 1/2 xx log(xy)`

= `1/2 xx (log x + log y)` 

`log((x + y)/6)`

= `log(sqrt(xy))` 

= `1/2 (log x + log y)`

`log((x + y)/6) = 1/2 (log x + log y)`, as required.