# How do you evaluate #log_27 (1/3)#?

##### 1 Answer

Aug 4, 2016

#log_27 (1/3) = -1/3#

#### Explanation:

By definition

#log_b (b^x) = x# for all Real values of#x#

#b^(log_b x) = x# for all#x > 0#

The change of base formula tells us that for any

#log_a b = (log_c b) / (log_c a)#

So we find:

#log_27 (1/3) = (log_3 (1/3)) / (log_3 27) = (log_3 3^(-1)) / (log_3 3^3) = -1/3#