How do you calculate log_16 512log16512?
1 Answer
May 21, 2016
Explanation:
The change of base formula tells us that if
log_a b = (log_c b)/(log_c a)logab=logcblogca
Note that
16 = 2^416=24
512 = 2^9512=29
So we find:
log_(16) 512 = (log_2 512)/(log_2 16) = (log_2 2^9)/(log_2 2^4) = 9/4log16512=log2512log216=log229log224=94