How do you evaluate #log_(1/4) (1/4)#?
1 Answer
Jul 27, 2016
Explanation:
For any base
#log_b b = 1#
If you like, use the change of base formula:
#log_a b = (log_c b)/(log_c a)#
So we find:
#log_(1/4) (1/4) = (log (1/4))/(log (1/4)) = 1#