How do you calculate #log_16 512#?
1 Answer
May 21, 2016
Explanation:
The change of base formula tells us that if
#log_a b = (log_c b)/(log_c a)#
Note that
#16 = 2^4#
#512 = 2^9#
So we find:
#log_(16) 512 = (log_2 512)/(log_2 16) = (log_2 2^9)/(log_2 2^4) = 9/4#