How would you evaluate this n^(log(n lowered)3)?

Question

I don't know how to make the n lower than the log.

What would the answer be?

Answer 1

one of the fundamental properties of logs is that

a^{log_a k} = k

(hope this came out ok)

e.g. (one we can actually evaluate)

2^{log₂ 8} = 8

so for your problem the answer would be 3