Question

How do you condense `2log_3 x -3log_3 y +log_3 8`?

Answer 1

`2log_(3)x-3log_(3)y+log_(3)8=log_3((8x^2)/y^3)`

Explanation:

Using `plog_am=log_am^p` and `loga+logb-logc=log((ab)/c)`

`2log_(3)x-3log_(3)y+log_(3)8`

= `log_(3)x^2-log_(3)y^3+log_(3)8`

= `log_3((8x^2)/y^3)`