Nth root

func nthroot(n, a, precision=1e-5) {
  var x = 1.float
  var prev = 0.float
  while (abs(prev-x) > precision) {
    prev = x
    x = (((n-1)*prev + a/(prev**(n-1))) / n)
  }
  return x
}
 
say nthroot(5, 34)  # => 2.024397458501034082599817835297912829678314204

A minor optimization would be to calculate the successive int(n-1) square roots of a number, then raise the result to the power of 2**(int(n-1) / n).

func nthroot_fast(n, a, precision=1e-5) {
  { a = nthroot(2, a, precision) } * int(n-1)
  a ** (2**int(n-1) / n)
}
 
say nthroot_fast(5, 34, 1e-64)  # => 2.02439745849988504251081724554193741911462170107