[Question]
  [
Given a positive integer *n* and a number *a*, the *n*-th [tetration](https://en.wikipedia.org/wiki/Tetration) of *a* is defined as *a*^(*a*^(*a*^(...^*a*))), where *^* denotes exponentiation (or power) and the expression contains the number *a* exactly *n* times.
In other words, tetration is right-associative iterated exponentiation. For *n*=4 and *a*=1.6 the tetration is 1.6^(1.6^(1.6^1.6)) ≈ 3.5743.
The inverse function of tetration with respect to *n* is the [super-logarithm](https://en.wikipedia.org/wiki/Super-logarithm). In the previous example, 4 is the super-logarithm of 3.5743 with "super-base" 1.6.
# The challenge
Given a positive integer *n*, find *x* such that *n* is the super-logarithm of itself in super-base *x*. That is, find *x* such that *x*^(*x*^(*x*^(...^*x*))) (with *x* appearing *n* times) equals *n*.
# Rules
Program or function allowed.
Input and output formats are flexible as usual.
The algorithm should theoretically work for all positive integers. In practice, input may be limited to a maximum value owing to memory, time or data-type restrictions. However, the code must work for inputs up to `100` at least in less than a minute.
The algorithm should theoretically give the result with `0.001` precision. In practice, the output precision may be worse because of accumulated errors in numerical computations. However, the output must be accurate up to `0.001` for the indicated test cases.
Shortest code wins.
# Test cases
```
1    ->  1
3    ->  1.635078
6    ->  1.568644
10   ->  1.508498
25   ->  1.458582
50   ->  1.448504
100  ->  1.445673
```
# Reference implementation
Here's a **reference implementation** in Matlab / Octave (try it at [Ideone](http://ideone.com/6yuUaZ)).
```
N = 10; % input
t = .0001:.0001:2; % range of possible values: [.0001 .0002 ... 2]
r = t;
for k = 2:N
    r = t.^r; % repeated exponentiation, element-wise
end
[~, ind] = min(abs(r-N)); % index of entry of r that is closest to N
result = t(ind);
disp(result)
```
For `N = 10` this gives `result = 1.5085`.
The following code is a **check** of the output precision, using variable-precision arithmetic:
```
N = 10;
x = 1.5085; % result to be tested for that N. Add or subtract 1e-3 to see that
            % the obtained y is farther from N
s = num2str(x); % string representation
se = s;
for n = 2:N;
    se = [s '^(' se ')']; % build string that evaluates to iterated exponentiation
end
y = vpa(se, 1000) % evaluate with variable-precision arithmetic
```
This gives:
* For `x = 1.5085`: `y = 10.00173...`
* For `x = 1.5085 + .001`: `y = 10.9075`
* For `x = 1.5085 - .001` it gives `y = 9.23248`.
so `1.5085` is a valid solution with `.001` precision.
  
[Answer]
# Haskell, ~~55~~ ~~54~~ 52 bytes
```
s n=[x|x<-[2,1.9999..],n>iterate(x**)1!!floor n]!!0
```
Usage:
```
> s 100
1.445600000000061
```
Thanks to @nimi for 1 byte!
  
Thanks to @xnor for 2!
[Answer]
## Javascript, ES6: 77 bytes / ES7: ~~57~~ 53 bytes
### ES6
```
n=>eval("for(x=n,s='x';--x;s=`Math.pow(x,${s})`);for(x=2;eval(s)>n;)x-=.001")
```
### ES7
Using `**` as suggested by DanTheMan:
```
n=>eval("for(x=2;eval('x**'.repeat(n)+1)>n;)x-=.001")
```
### Example
```
let f =
n=>eval("for(x=n,s='x';--x;s=`Math.pow(x,${s})`);for(x=2;eval(s)>n;)x-=.001")
console.log(f(25));
```
[Answer]
# Mathematica, ~~40~~ 33 bytes
*Thanks to murphy for an almost 20% savings!*
```
1//.x_:>x+.001/;Nest[x^#&,1,#]<#&
```
`Nest[x^#&,1,n]` produces the nth tetration of x. So `Nest[x^#&,1,#]<#` tests whether the (input)th tetration of x is less than (input). We simply start at x=1 and add 0.001 repeatedly until the tetration is too large, then output the last x value (so the answer is guaranteed to be larger than the exact value, but within 0.001).
As I'm slowly learning: `//.x_:>y/;z` or `//.x_/;z:>y` means "look for anything that matches the template x, but only things for which the test z returns true; and then operate on x by the rule y; repeatedly until nothing changes". Here the template `x_` is just "any number I see", although in other contexts it could be further constrained.
When the input is at least 45, the tetration increases so rapidly that that last step causes an overflow error; but the value of x is still updated and output correctly. Decreasing the step-size from 0.001 to 0.0001 fixes this problem for inputs up to 112, and gives a more precise answer to boot (and still runs quickly, in about a quarter second). But that's one extra byte, so forget that!
Original version:
```
x=1;(While[Nest[x^#&,1,#]<#,x+=.001];x)&
```
[Answer]
# J, ~~39~~ ~~31~~ 28 bytes
```
(>./@(]*[>^/@#"0)1+i.%])&1e4
```
Based on the reference implementation. It is only accurate to three decimal places.
Saved 8 bytes using the method from @Adám's [solution](https://codegolf.stackexchange.com/a/93208/6710).
## Usage
Extra commands used to format multiple input/output.
```
   f =: (>./@(]*[>^/@#"0)1+i.%])&1e4
   (,.f"0) 1 3 6 10 25 50 100
  1      0
  3  1.635
  6 1.5686
 10 1.5084
 25 1.4585
 50 1.4485
100 1.4456
   f 1000
1.4446
```
## Explanation
```
(>./@(]*[>^/@#"0)1+i.%])&1e4  Input: n
                         1e4  The constant 10000
(                      )      Operate on n (LHS) and 10000 (RHS)
                   i.           The range [0, 10000)
                      ]         Get (RHS) 10000
                     %          Divide each in the range by 10000
                 1+             Add 1 to each
     (          )               Operate on n (LHS) and the range (RHS)
             #"0                  For each in the range, create a list of n copies
          ^/@                     Reduce each list of copies using exponentation
                                  J parses from right-to-left which makes this
                                  equivalent to the tetration
        [                         Get n
         >                        Test if each value is less than n
      ]                           Get the initial range
       *                          Multiply elementwise
 >./@                           Reduce using max and return
```
[Answer]
# Python, 184 bytes
```
def s(n):
 def j(b,i):
  if i<0.1**12:
   return b
  m = b+i
  try:
   v = reduce(lambda a,b:b**a,[m]*n)
  except:
   v = n+1
  return j(m,i/2) if v<n else j(b,i/2)
 return j(1.0,0.5)
```
Test output (skipping the actual print statements):
```
   s(1) 1.0
   s(3) 1.63507847464
   s(6) 1.5686440646
  s(10) 1.50849792026
  s(25) 1.45858186605
  s(50) 1.44850389566
 s(100) 1.44567285047
```
[Answer]
# [Dyalog APL](http://goo.gl/9KrKoM), ~~33~~ 25 [bytes](http://meta.codegolf.stackexchange.com/a/9429/43319)
Needs `⎕IO←0` which is default on many systems.
```
⌈/⎕(⊢×⊣≥(*/⍴)¨)(1+⍳÷⊢)1E4
```
Theoretically calculates for all integers, but practically limited to very small one only.
[TryAPL online!](http://tryapl.org/?a=%u2206%u21901%20%u22C4%20%u2308/%u2206%28%u22A2%D7%u22A3%u2265%28*/%u2374%29%A8%29%281+%u2373%F7%u22A2%291E4%20%u22C4%20%u2206%u21902%20%u22C4%20%u2308/%u2206%28%u22A2%D7%u22A3%u2265%28*/%u2374%29%A8%29%281+%u2373%F7%u22A2%291E4%20%u22C4%20%u2206%u21903%20%u22C4%20%u2308/%u2206%28%u22A2%D7%u22A3%u2265%28*/%u2374%29%A8%29%281+%u2373%F7%u22A2%291E4%20%u22C4%20%u2206%u21904%20%u22C4%20%u2308/%u2206%28%u22A2%D7%u22A3%u2265%28*/%u2374%29%A8%29%281+%u2373%F7%u22A2%291E4&run)
[Answer]
## Racket 187 bytes
```
(define(f x n)(define o 1)(for((i n))(set! o(expt x o)))o)
(define(ur y(l 0.1)(u 10))(define t(/(+ l u)2))(define o(f t y))
(cond[(<(abs(- o y)) 0.1)t][(> o y)(ur y l t)][else(ur y t u)]))
```
Testing: 
```
(ur 1)
(ur 3)
(ur 6)
(ur 10)
(ur 25)
(ur 50)
(ur 100)
```
Output:
```
1.028125
1.6275390625
1.5695312499999998
1.5085021972656247
1.4585809230804445
1.4485038772225378
1.4456728475168346
```
Detailed version: 
```
(define (f x n)
  (define out 1)
  (for((i n))
    (set! out(expt x out)))
  out)
(define (uniroot y (lower 0.1) (upper 10))
  (define trying (/ (+ lower upper) 2))
  (define out (f trying y))
  (cond
    [(<(abs(- out y)) 0.1)
     trying]
    [(> out y)
     (uniroot y lower trying)]
    [else
      (uniroot y trying upper)]))
```
[Answer]
# [Perl 6](https://perl6.org), 42 bytes
```
{(0,.0001…2).min:{abs $_-[**] $^r xx$_}}
```
( Translation of example Matlab code )
## Test:
```
#! /usr/bin/env perl6
use v6.c;
use Test;
my &code = {(0,.0001…2).min:{abs $_-[**] $^r xx$_}}
my @tests = (
  1   => 1,
  3   => 1.635078,
  6   => 1.568644,
  10  => 1.508498,
  25  => 1.458582,
  50  => 1.448504,
  100 => 1.445673,
);
plan +@tests + 1;
my $start-time = now;
for @tests -> $_ ( :key($input), :value($expected) ) {
  my $result = code $input;
  is-approx $result, $expected, "$result ≅ $expected", :abs-tol(0.001)
}
my $finish-time = now;
my $total-time = $finish-time - $start-time;
cmp-ok $total-time, &[<], 60, "$total-time.fmt('%.3f') is less than a minute";
```
```
1..8
ok 1 - 1 ≅ 1
ok 2 - 1.6351 ≅ 1.635078
ok 3 - 1.5686 ≅ 1.568644
ok 4 - 1.5085 ≅ 1.508498
ok 5 - 1.4586 ≅ 1.458582
ok 6 - 1.4485 ≅ 1.448504
ok 7 - 1.4456 ≅ 1.445673
ok 8 - 53.302 seconds is less than a minute
```
[Answer]
# PHP , 103 Bytes
```
$z=2;while($z-$a>9**-9){for($r=$s=($a+$z)/2,$i=0;++$i<$n=$argv[1];)$r=$s**$r;$r<$n?$a=$s:$z=$s;}echo$s;
```
[Answer]
**Axiom 587 bytes**
```
l(a,b)==(local i;i:=1;r:=a;repeat(if i>=b then break;r:=a^r;i:=i+1);r);g(q,n)==(local r,y,y1,y2,t,v,e,d,i;n<=0 or n>1000 or q>1000 or q<0 => 0;e:=1/(10**(digits()-3));v:=0.01; d:=0.01;repeat(if l(v,n)>=q then break;v:=v+d;if v>=1 and n>25 then d:=0.001;if v>=1.4 and n>40 then d:=0.0001;if v>=1.44 and n>80 then d:=0.00001;if v>=1.445 and n>85 then d:=0.000001);if l(v-d,n)>q then y1:=0.0 else y1:=v-d;y2:=v;y:=l(v,n);i:=1;if abs(y-q)>e then repeat(t:=(y2-y1)/2.0;v:=y1+t;y:=l(v,n);i:=i+1;if i>100 then break;if t<=e then break;if y<q then y1:=v else y2:=v);if i>100 then output "#i#";v)
```
less golfed + numbers
```
l(a,b)==
  local i
  i:=1;r:=a;repeat(if i>=b then break;r:=a^r;i:=i+1)
  r
g(q,n)==
 local r, y, y1,y2,t,v,e,d, i
 n<=0 or n>1000 or q>1000 or q<0 => 0  
 e:=1/(10**(digits()-3))
 v:=0.01; d:=0.01  
 repeat  --cerco dove vi e' il punto di cambiamento di segno di l(v,n)-q
    if l(v,n)>=q then break
    v:=v+d 
    if v>=1     and n>25 then d:=0.001
    if v>=1.4   and n>40 then d:=0.0001
    if v>=1.44  and n>80 then d:=0.00001
    if v>=1.445 and n>85 then d:=0.000001
 if l(v-d,n)>q then y1:=0.0
 else               y1:=v-d 
 y2:=v; y:=l(v,n); i:=1  -- applico il metodo della ricerca binaria
 if abs(y-q)>e then      -- con la variabile i di sicurezza
    repeat 
       t:=(y2-y1)/2.0; v:=y1+t; y:=l(v,n)
       i:=i+1
       if i>100 then break
       if t<=e  then break 
       if  y<q  then y1:=v
       else          y2:=v
 if i>100 then output "#i#"
 v
(3) -> [g(1,1), g(3,3), g(6,6), g(10,10), g(25,25), g(50,50), g(100,100)]
   Compiling function l with type (Float,PositiveInteger) -> Float
   Compiling function g with type (PositiveInteger,PositiveInteger) ->
      Float
   (3)
   [1.0000000000 000000001, 1.6350784746 363752387, 1.5686440646 047324687,
    1.5084979202 595960768, 1.4585818660 492876919, 1.4485038956 661040907,
    1.4456728504 738144738]
                                                             Type: List Float
```
[Answer]
# Common Lisp, 207 bytes
```
(defun superlog(n)(let((a 1d0)(i 0.5))(loop until(< i 1d-12)do(let((v(or(ignore-errors(reduce #'expt(loop for q below n collect(+ a i)):from-end t))(1+ n))))(when(< v n)(setq a (+ a i)))(setq i(/ i 2)))) a))
```
Using `reduce` with `:from-end t` avoids the need of doing a "reversing exponentiation" intermediate lambda (basically `(lambda (x y) (expt y x))`, saving 14 bytes (12, if you remove removable spaces).
We still need to handle float overflow, but an `ignore-errors` form returns `nil` if an error happened, so we can use `or` to provide a default value.
]