Alg
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
a^{loga 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
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