Question

If `x = log  2/3, y = log  3/7, z = log  7/2`, find the value of x + y + z.

Answer 1

Given: `x = log (2/3), y = log (3/7), z = log (7/2)`

Step-wise calculation:

1. `x + y + z = log(2/3) + log(3/7) + log(7/2)`

2. Use log addition:

`log [(2/3) xx (3/7) xx (7/2)]`

3. Multiply and cancel:

`(2/3) xx (3/7) xx (7/2) = 1`

4. Therefore x + y + z = log (1) = 0.