Question

How do I solve `log_(2/3) (8/27)`?

Answer 1

3

Explanation:

Let , `m =log_"(2/3)" (8/27)`

`rArr (2/3)^m = 8/27`

`rArr (2/3)^m = (2^3)/(3^3)`

`rArr (2/3)^m = (2/3)^3`

`rArr m = 3`