Question

How do you calculate `log_(1/6) 216`?

Answer 1

Use the change of base rule `log_an = logn/loga`

Explanation:

`= log216/log(1/6)`

`= log(6^3)/log(6^-1)`

Use the rule `loga^n = nloga`

`= (3log6)/(-1log6)`

`= -3`

Hopefully this helps!