Is O(n^(1/logn)) actually constant?

Assuming log is the natural log, then this is equivalent to e, not 2, but either way it's a constant.

First, let:

k = n^(1 / log n)

Then take the log of both sides:

log k = (1 / log n) * log n

So:

log k = 1

Now raise both sides to the power of e to get:

e^(log k) = e^(1)

And thus:

k = e.