# How do you evaluate #log_(1/3) (1/81)#?

##### 1 Answer

Dec 16, 2016

#### Explanation:

#= log_(1/3) (81^-1)#

Use the change of base rule

#= log81^-1/log(1/3)#

#=log(3^4)^-1/log(3^-1)#

#= log(3^-4)/log3^-1#

Use the exponent rule that

#=(-4log3)/(-1log3)#

#=4#

Hopefully this helps!