Question

How do you evaluate `log_(1/11) (1/121)`?

Answer 1

`color(blue)(2)`

Explanation:

Using change of base formula:

`log_(1/11)(1/121)=(ln(1/121))/ln(1/11)`

`log(a/b)=loga-logb`

`(ln(1/121))/ln(1/11)=(ln(1)-ln(121))/(ln(1)-ln(11))=(0-ln(121))/(0-ln(11))=ln(121)/ln(11)`

`loga^b=bloga`

`=(ln(11^2))/ln(11)=(2ln(11))/ln(11)=2`