How do you calculate #log_16 512#?

1 Answer
May 21, 2016

Answer:

#log_(16) 512 = 9/4#

Explanation:

The change of base formula tells us that if #a, b, c > 0# then:

#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#