Question

How do you evaluate `log_(1/3) (1/27)`?

Answer 1

`3`

Explanation:

`log_(1/3)(1/27)`
`=log_(1/3)(1/3)^3`
`=3log_(1/3)(1/3)`
`=3`

Alternatively, you can do it this way :

Recall that `log_ab=c, =>a^c=b`

let `log_(1/3)(1/27)=x`
`=> (1/3)^x=1/27=(1/3)^3`
Hence, `x=3`