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 

因为任何n都可以写为2的幂，所以我可以做上述简化并证明它是常数，对吗？

Answer 1

假设log是自然对数，则它等效于e，而不是2，但是无论哪种方式它都是常量。

首先，让：

k = n^(1 / log n)

然后取双方的日志：

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

所以：

log k = 1

因此

k = e.