Question

If `x = log  2/3, y = log  3/7, z = log  7/2`, find the value of `2^(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 (a) + log (b) + log (c) = log (abc): 

x + y + z 

= `log [(2/3) xx (3/7) xx (7/2)]` 

= log (1)

= 0

3. Therefore

`2^(x  +  y  +  z)`

= 2⁰

= 1