I have to prove that
a^log n = n^log a
an example would be
2^log n = n^log2 = n
That isn't quite a true statement. It only works for a specific base:
So your statement needs to read:
Prove:
This isn't so much a proof as a definition: the logarithm and exponentiation operations are functional inverses.
What you need to do is show that each of these are equivalent to the same number.
We know that
So what is ? Well, call it x.
Then
Since log is a 1 to 1 function, we must have x = n. Thus
-Dan