How do you calculate log_16 512log16512?

1 Answer
May 21, 2016

log_(16) 512 = 9/4log16512=94

Explanation:

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

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