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

1 Answer
George C.
Aug 4, 2016

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

Explanation:

By definition #log# is the inverse of exponentiation and vice versa for any base #b#. That is:

#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 #a, b, c > 0# with #a != 1# and #c != 1#:

#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#

Related questions
See all questions in Logarithm-- Inverse of an Exponential Function
Impact of this question
6260 views around the world
You can reuse this answer
Creative Commons License