posted by Anonymous .
How would you evaluate this n^(log(n lowered)3)?
I don't know how to make the n lower than the log.
What would the answer be?
one of the fundamental properties of logs is that
aloga k = k
(hope this came out ok)
e.g. (one we can actually evaluate)
2log2 8 = 8
so for your problem the answer would be 3