How do you evaluate #log_81 27#?
1 Answer
Aug 3, 2016
#log_81 27 = 3/4#
Explanation:
Using the change of base formula:
#log_a b = (log_c b)/(log_c a)#
and the definition of logarithm as the inverse of exponentiation:
#log_a a^b = b#
we find:
#log_81 27 = (log_3 27)/(log_3 81) = (log_3 3^3)/(log_3 3^4) = 3/4#
