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

Question

I came across this time complexity function and according to me, it is actually constant. Please correct me if I am wrong.

n^(1/logn) => (2^m)^(1/log(2^m)) => (2^m)^(1/m) => 2 

Since any n can be written as a power of 2, I can do the above simplification and prove that it is constant, right?

Answer 1

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.