Exponentiation order

Exponentiation order
You are encouraged to solve this task according to the task description, using any language you may know.

This task will demonstrate the order of exponentiation   (x^y)   when there are multiple exponents.

(Many programming languages,   especially those with extended─precision integer arithmetic,   usually support one of **, ^, ↑ or some such for exponentiation.)

Task requirements

Show the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).

If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.

Using whatever operator or syntax your language supports (if any), show the results in three lines (with identification):

•   5**3**2
•   (5**3)**2
•   5**(3**2)

If there are other methods (or formats) of multiple exponentiations, show them as well.

See also

• MathWorld entry:   exponentiation

Related tasks

•   exponentiation operator
•   arbitrary-precision integers (included)
•   Exponentiation with infix operators in (or operating on) the base

11l

print(5 ^ 3 ^ 2)
print((5 ^ 3) ^ 2)
print(5 ^ (3 ^ 2))

Output:

1.95313e+06
15625
1.95313e+06

Action!

There is no power operator in Action! Power function for REAL type is used. But the precision is insufficient.

Library: Action! Tool Kit

INCLUDE "D2:REAL.ACT" ;from the Action! Tool Kit

PROC Main()
  REAL r2,r3,r5,tmp1,tmp2

  Put(125) PutE() ;clear screen

  IntToReal(2,r2)
  IntToReal(3,r3)
  IntToReal(5,r5)

  PrintE("There is no power operator in Action!")
  PrintE("Power function for REAL type is used.")
  PrintE("But the precision is insufficient.")
  Power(r5,r3,tmp1)
  Power(tmp1,r2,tmp2)
  Print("(5^3)^2=")
  PrintRE(tmp2)

  Power(r3,r2,tmp1)
  Power(r5,tmp1,tmp2)
  Print("5^(3^2)=")
  PrintRE(tmp2)
RETURN

Output:

Screenshot from Atari 8-bit computer

There is no power operator in Action!
Power function for REAL type is used.
But the precision is insufficient.
(5^3)^2=15624.9977
5^(3^2)=1953124.17

Ada

5**3**2 is not a valid Ada expression. Parenthesis are mandatory.

with Ada.Text_IO;

procedure Exponentation_Order is
   use Ada.Text_IO;
begin
   --  Put_Line ("5**3**2   : " & Natural'(5**3**2)'Image);
   Put_Line ("(5**3)**2 : " & Natural'((5**3)**2)'Image);
   Put_Line ("5**(3**2) : " & Natural'(5**(3**2))'Image);
end Exponentation_Order;

Output:

(5**3)**2 :  15625
5**(3**2) :  1953125

ALGOL 68

Algol 68 provides various alternative symbols for the exponentiation operator generally, "**", "^" and "UP" can be used.

print( ( "5**3**2:   ", 5**3**2, newline ) );
print( ( "(5**3)**2: ", (5**3)**2, newline ) );
print( ( "5**(3**2): ", 5**(3**2), newline ) )

Output:

5**3**2:        +15625
(5**3)**2:      +15625
5**(3**2):    +1953125

ALGOL W

The Algol W exponentiation operator always produces a real result and requires an integer right operand, hence the round functions in the following.

begin
    write( "5**3**2:   ", round( 5 ** 3 ** 2 ) );
    write( "(5**3)**2: ", round( ( 5 ** 3 ) ** 2 ) );
    write( "5**(3**2): ", round( 5 ** round( 3 ** 2 ) ) )
end.

Output:

5**3**2:            15625
(5**3)**2:          15625
5**(3**2):        1953125

APL

APL has no order of precedence other than right-to-left operation. * is the APL exponentiation operator.

      5*3*2
1953125
      (5*3)*2
15625
      5*(3*2)
1953125

AppleScript

AppleScript's compiler inserts its own parentheses with 5 ^ 3 ^ 2.

set r1 to 5 ^ 3 ^ 2 -- Changes to 5 ^ (3 ^ 2) when compiled.
set r2 to (5 ^ 3) ^ 2
set r3 to 5 ^ (3 ^ 2)

return "5 ^ 3 ^ 2 = " & r1 & "
(5 ^ 3) ^ 2 = " & r2 & "
5 ^ (3 ^ 2) = " & r3

Output:

"5 ^ 3 ^ 2 = 1.953125E+6
(5 ^ 3) ^ 2 = 1.5625E+4
5 ^ (3 ^ 2) = 1.953125E+6"

Arturo

print 5^3^2
print (5^3)^2
print 5^(3^2)

Output:

1953125
15625
1953125

AWK

# syntax: GAWK -f EXPONENTIATION_ORDER.AWK
BEGIN {
    printf("5^3^2   = %d\n",5^3^2)
    printf("(5^3)^2 = %d\n",(5^3)^2)
    printf("5^(3^2) = %d\n",5^(3^2))
    exit(0)
}

output:

5^3^2   = 1953125
(5^3)^2 = 15625
5^(3^2) = 1953125

BASIC

Sinclair ZX81 BASIC

10 PRINT "5**3**2   = ";5**3**2
20 PRINT "(5**3)**2 = ";(5**3)**2
30 PRINT "5**(3**2) = ";5**(3**2)

Output:

5**3**2   = 15625
(5**3)**2 = 15625
5**(3**2) = 1953125

BBC BASIC

PRINT "5^3^2   = "; 5^3^2
PRINT "(5^3)^2 = "; (5^3)^2
PRINT "5^(3^2) = "; 5^(3^2)

Output:

5^3^2   = 15625
(5^3)^2 = 15625
5^(3^2) = 1953125

IS-BASIC

100 PRINT "5^3^2   =";5^3^2
110 PRINT "(5^3)^2 =";(5^3)^2
120 PRINT "5^(3^2) =";5^(3^2)

Output:

5^3^2   = 15625
(5^3)^2 = 15625
5^(3^2) = 1953125

QBasic

Works with: QBasic version 1.1

Works with: QuickBasic version 4.5

Works with: FreeBASIC

Works with: True BASIC

Works with: BASIC256

Works with: Run BASIC

PRINT "5^3^2   ="; 5^3^2
PRINT "(5^3)^2 ="; (5^3)^2
PRINT "5^(3^2) ="; 5^(3^2)
END

Output:

5^3^2   = 15625
(5^3)^2 = 15625
5^(3^2) = 1953125

BASIC256

Works with: QBasic

Works with: FreeBASIC

Works with: True BASIC

Works with: Run BASIC

print "5^3^2   = "; 5^3^2
print "(5^3)^2 = "; (5^3)^2
print "5^(3^2) = "; 5^(3^2)
end

Output:

5^3^2   = 15625
(5^3)^2 = 15625
5^(3^2) = 1953125

Run BASIC

Works with: QBasic

Works with: FreeBASIC

Works with: True BASIC

Works with: BASIC256

print "5^3^2   = "; 5^3^2
print "(5^3)^2 = "; (5^3)^2
print "5^(3^2) = "; 5^(3^2)
end

Output:

5^3^2   = 15625
(5^3)^2 = 15625
5^(3^2) = 1953125

True BASIC

Works with: QBasic

Works with: FreeBASIC

Works with: BASIC256

Works with: Run BASIC

PRINT "5^3^2   ="; 5^3^2
PRINT "(5^3)^2 ="; (5^3)^2
PRINT "5^(3^2) ="; 5^(3^2)
END

Output:

5^3^2   = 15625
(5^3)^2 = 15625
5^(3^2) = 1953125

Yabasic

Works with: QBasic

Works with: FreeBASIC

Works with: True BASIC

Works with: BASIC256

print "5^3^2   = ", 5^3^2
print "(5^3)^2 = ", (5^3)^2
print "5^(3^2) = ", 5^(3^2)
end

Output:

5^3^2   = 15625
(5^3)^2 = 15625
5^(3^2) = 1953125

Bracmat

put$str$("5^3^2: " 5^3^2 "\n(5^3)^2: " (5^3)^2 "\n5^(3^2): " 5^(3^2) \n)

Output:

5^3^2: 1953125
(5^3)^2: 15625
5^(3^2): 1953125

C

C does not have an exponentiation operator. The caret operator '^' performs xor bitwise operation in C. The function pow in the standard C Math library takes two arguments.

#include<stdio.h>
#include<math.h>

int main()
{
    printf("(5 ^ 3) ^ 2 = %.0f",pow(pow(5,3),2));
    printf("\n5 ^ (3 ^ 2) = %.0f",pow(5,pow(3,2)));
	
    return 0;
}

Output:

(5 ^ 3) ^ 2 = 15625
5 ^ (3 ^ 2) = 1953125

C++

#include <iostream>
#include <cmath>

int main() {
    std::cout << "(5 ^ 3) ^ 2 = " << (uint) pow(pow(5,3), 2) << std::endl;
    std::cout << "5 ^ (3 ^ 2) = "<< (uint) pow(5, (pow(3, 2)));
	
    return EXIT_SUCCESS;
}

With permissive flag:

#include <iostream>
#include <cmath>

enum my_int {};
inline my_int operator^(my_int a, my_int b) { return static_cast<my_int>(pow(a,b)); }

int main() {
    my_int x = 5, y = 3, z = 2;
    std::cout << "(5 ^ 3) ^ 2 = " << ((x^y)^z) << std::endl;
    std::cout << "5 ^ (3 ^ 2) = "<< (x^(y^z));
	
    return EXIT_SUCCESS;
}

Output:

(5 ^ 3) ^ 2 = 15625
5 ^ (3 ^ 2) = 1953125

C#

using System;

namespace exponents
{
    class Program
    {
        static void Main(string[] args)
        {
            /* 
             * Like C, C# does not have an exponent operator.
             * Exponentiation is done via Math.Pow, which
             * only takes two arguments 
             */
            Console.WriteLine(Math.Pow(Math.Pow(5, 3), 2));
            Console.WriteLine(Math.Pow(5, Math.Pow(3, 2)));
            Console.Read();
        }

    }
}

Output:

15625
1953125

Clojure

Clojure uses prefix notation and expt only takes 2 arguments for exponentiation, so "5**3**2" isn't represented.

(use 'clojure.math.numeric-tower)
;; (5**3)**2
(expt (expt 5 3) 2)      ; => 15625

;; 5**(3**2)
(expt 5 (expt 3 2))      ; => 1953125

;; (5**3)**2 alternative: use reduce 
(reduce expt [5 3 2])  ; => 15625

;; 5**(3**2) alternative: evaluating right-to-left with reduce requires a small modification
(defn rreduce [f coll] (reduce #(f %2 %) (reverse coll)))
(rreduce expt [5 3 2]) ; => 1953125

CLU

start_up = proc ()
    po: stream := stream$primary_output()
    
    stream$putl(po, "5**3**2   = " || int$unparse(5**3**2))
    stream$putl(po, "(5**3)**2 = " || int$unparse((5**3)**2))
    stream$putl(po, "5**(3**2) = " || int$unparse(5**(3**2)))
end start_up

Output:

5**3**2   = 1953125
(5**3)**2 = 15625
5**(3**2) = 1953125

Common Lisp

Because Common Lisp uses prefix notation and expt accepts only two arguments, it doesn't have an expression for 5**3**2. Just showing expressions for the latter two.

(expt (expt 5 3) 2)
(expt 5 (expt 3 2))

Output:

15625
1953125

D

void main() {
    import std.stdio, std.math, std.algorithm;

    writefln("5 ^^ 3 ^^ 2          = %7d", 5 ^^ 3 ^^ 2);
    writefln("(5 ^^ 3) ^^ 2        = %7d", (5 ^^ 3) ^^ 2);
    writefln("5 ^^ (3 ^^ 2)        = %7d", 5 ^^ (3 ^^ 2));
    writefln("[5, 3, 2].reduce!pow = %7d", [5, 3, 2].reduce!pow);
}

Output:

5 ^^ 3 ^^ 2          = 1953125
(5 ^^ 3) ^^ 2        =   15625
5 ^^ (3 ^^ 2)        = 1953125
[5, 3, 2].reduce!pow =   15625

EchoLisp

;; the standard and secure way is to use the (expt a b) function
(expt 5 (expt 3 2))  ;; 5 ** ( 3 ** 2)
    → 1953125
(expt (expt 5 3) 2) ;; (5 ** 3) ** 2
    → 15625

;; infix EchoLisp may use the ** operator, which right associates
(lib 'match)
(load 'infix.glisp)

(5 ** 3 ** 2)
    → 1953125
((5 ** 3) ** 2)
    → 15625
(5 ** (3 ** 2))
    → 1953125

Factor

Factor is a stack language where expressions take the form of reverse Polish notation, so there is no ambiguity here. It is up to you, the programmer, to perform operations in the order you intend.

USING: formatting math.functions ;

5 3 2 ^ ^
"5 3 2 ^ ^  %d\n" printf

5 3 ^ 2 ^
"5 3 ^ 2 ^  %d\n" printf

Output:

5 3 2 ^ ^  1953125
5 3 ^ 2 ^  15625

Factor also has syntax for infix arithmetic via the the infix vocabulary.

USING: formatting infix ;

[infix  5**3**2    infix]
"5**3**2   = %d\n" printf

[infix  (5**3)**2  infix]
"(5**3)**2 = %d\n" printf

[infix  5**(3**2)  infix]
"5**(3**2) = %d\n" printf

Output:

5**3**2   = 15625
(5**3)**2 = 15625
5**(3**2) = 1953125

Fortran

write(*, "(a, i0)") "5**3**2   = ", 5**3**2
write(*, "(a, i0)") "(5**3)**2 = ", (5**3)**2
write(*, "(a, i0)") "5**(3**2) = ", 5**(3**2)

Output:

5**3**2   = 1953125
(5**3)**2 = 15625
5**(3**2) = 1953125

FreeBASIC

' FB 1.05.0

' The exponentation operator in FB is ^ rather than **.
' In the absence of parenthesis this operator is
' left-associative. So the first example
' will have the same value as the second example.

Print "5^3^2   =>"; 5^3^2
Print "(5^3)^2 =>"; (5^3)^2
Print "5^(3^2) =>"; 5^(3^2)
Sleep

Output:

5^3^2   => 15625
(5^3)^2 => 15625
5^(3^2) => 1953125

Frink

Frink correctly follows standard mathematical notation that exponent towers are performed from "top to bottom" or "right to left."

println["5^3^2   = " + 5^3^2]
println["(5^3)^2 = " + (5^3)^2]
println["5^(3^2) = " + 5^(3^2)]

Output:

5^3^2   = 1953125
(5^3)^2 = 15625
5^(3^2) = 1953125

Go

package main

import "fmt"
import "math"

func main() {
    var a, b, c float64
    a = math.Pow(5, math.Pow(3, 2))
    b = math.Pow(math.Pow(5, 3), 2)
    c = math.Pow(5, math.Pow(3, 2))
    fmt.Printf("5^3^2   = %.0f\n", a)
    fmt.Printf("(5^3)^2 = %.0f\n", b)
    fmt.Printf("5^(3^2) = %.0f\n", c)
}

Output:

5^3^2   = 1953125
(5^3)^2 = 15625
5^(3^2) = 1953125

Groovy

Solution:

println(" 5 ** 3 ** 2 == " + 5**3**2)
println("(5 ** 3)** 2 == " + (5**3)**2)
println(" 5 **(3 ** 2)== " + 5**(3**2))

Output:

 5 ** 3 ** 2 == 15625
(5 ** 3)** 2 == 15625
 5 **(3 ** 2)== 1953125

Haskell

Haskell has three infix exponentiation operators dealing with different domains:

λ> :i (^)
(^) :: (Num a, Integral b) => a -> b -> a 	-- Defined in ‘GHC.Real’
infixr 8 ^
λ> :i (**)
class Fractional a => Floating a where
  ...
  (**) :: a -> a -> a
  ...
  	-- Defined in ‘GHC.Float’
infixr 8 **
λ> :i (^^)
(^^) :: (Fractional a, Integral b) => a -> b -> a  -- Defined in ‘GHC.Real’
infixr 8 ^^

All of them are right-associative.

λ> 5^3^2
1953125
λ> (5^3)^2
15625
λ> 5^(3^2)
1953125
λ> 5**3**2 == 5**(3**2)
True

However natural chaining of (^^) operator is impossible:

 5^^3^^2 = 5^^(3^^2)

but (3^^2) is not Integral any longer, so evaluation leads to the type error. Left-assiciative chain is Ok:

λ> (5^^3)^^2
15625.0
λ> ((5^^3)^^2)^^4
5.9604644775390624e16

Io

Io> 5**3**2
==> 15625
Io> (5**3)**2
==> 15625
Io> 5**(3**2)
==> 1953125
Io> 5 pow(3) pow(2)
==> 15625
Io> 5 **(3) **(2)
==> 15625
Io> Number getSlot("**") == Number getSlot("pow")
==> true
Io> 

Operators in Io are implemented as methods. Here the ** method is the same as the pow method. Syntax sugar converts "normal" mathematical expressions to messages.

J

J uses the same evaluation order for exponentiation as it does for assignment. That is to say: the bottom up view is right-to-left and the top-down view is left-to-right.

   5^3^2
1.95312e6
   (5^3)^2
15625
   5^(3^2)
1.95312e6

---

Java

Java has no exponentiation operator, but uses the static method java.lang.Math.pow(double a, double b). There are no associativity issues.

jq

Requires: jq 1.5 or higher

jq's built-in for exponentiation is an arity-two function and thus no ambiguity arising from infix-notation is possible. Here's an example:

jq -n 'pow(pow(5;3);2)'
15625

For chaining, one could use `reduce`:

    def pow: reduce .[1:] as $i (.[0]; pow(.;$i))

    [5,3,2] | pow

Result: 15625

Julia

Works with: Julia version 0.6

@show 5 ^ 3 ^ 2 # default: power operator is read right-to-left
@show (5 ^ 3) ^ 2
@show 5 ^ (3 ^ 2)
@show reduce(^, [5, 3, 2])
@show foldl(^, [5, 3, 2]) # guarantees left associativity
@show foldr(^, [5, 3, 2]) # guarantees right associativity

Output:

5 ^ (3 ^ 2) = 1953125
(5 ^ 3) ^ 2 = 15625
5 ^ (3 ^ 2) = 1953125
reduce(^, [5, 3, 2]) = 15625
foldl(^, [5, 3, 2]) = 15625
foldr(^, [5, 3, 2]) = 1953125

Kotlin

Kotlin does not have a dedicated exponentiation operator and we would normally use Java's Math.pow function instead. However, it's possible to define an infix function which would look like an operator and here we do so for integer base and exponent. For simplicity we disallow negative exponents altogether and consider 0 ** 0 == 1. Associativity would, of course, be the same as for a normal function call.

// version 1.0.5-2

infix fun Int.ipow(exp: Int): Int = when {
    exp < 0   -> throw IllegalArgumentException("negative exponents not allowed")
    exp == 0  -> 1
    else      -> {
        var ans = 1
        var base = this
        var e = exp
        while(e != 0) {
            if (e and 1 == 1) ans *= base
            e = e shr 1
            base *= base
        }
        ans
    }
} 

fun main(args: Array<String>) {
    println("5**3**2   = ${5 ipow 3 ipow 2}") 
    println("(5**3)**2 = ${(5 ipow 3) ipow 2}")
    println("5**(3**2) = ${5 ipow (3 ipow 2)}")
}

Output:

5**3**2   = 15625
(5**3)**2 = 15625
5**(3**2) = 1953125

Lambdatalk

Because lambdatalk uses prefix notation and {pow a b} accepts only two arguments, it doesn't have an expression for 5**3**2. Just showing expressions for the latter two.

'{pow {pow 5 3} 2}
-> {pow {pow 5 3} 2}
'{pow 5 {pow 3 2}}
-> {pow 5 {pow 3 2}}

Latitude

5 ^ 3 ^ 2.   ;; 1953125
(5 ^ 3) ^ 2. ;; 15625
5 ^ (3 ^ 2). ;; 1953125

Lua

print("5^3^2 = " .. 5^3^2)
print("(5^3)^2 = " .. (5^3)^2)
print("5^(3^2) = " .. 5^(3^2))

Output:

5^3^2 = 1953125
(5^3)^2 = 15625
5^(3^2) = 1953125

Lua also has math.pow(a, b), which is identical to pow(a, b) in C. Since function arguments are contained in brackets anyway, the associativity of nested uses of math.pow will be obvious.

Maple

5^3^2;
(5^3)^2;
5^(3^2);

Output:

Error, ambiguous use of `^`, please use parentheses
15625
1953125

Mathematica / Wolfram Language

a = "5^3^2";
Print[a <> " = " <> ToString[ToExpression[a]]]
b = "(5^3)^2";
Print[b <> " = " <> ToString[ToExpression[b]]]
c = "5^(3^2)";
Print[c <> " = " <> ToString[ToExpression[c]]]

Output:

5^3^2 = 1953125
(5^3)^2 = 15625
5^(3^2) = 1953125

min

As with other postfix languages, there is no ambiguity because all operators have the same precedence.

Works with: min version 0.19.6

5 3 2 pow pow
"5 3 2 ^ ^  " print! puts!

5 3 pow 2 pow
"5 3 ^ 2 ^  " print! puts!

Output:

5 3 2 ^ ^  1953125.0
5 3 ^ 2 ^  15625.0

Nanoquery

Nanoquery uses the '^' operator, which performs exponentiation in order like multiplication. Parenthesis are often needed to perform operations like 5^3^2 correctly.

% println 5^3^2
15625
% println (5^3)^2
15625
% println 5^(3^2)
1953125

Nim

<Lang Nim>import math, sequtils

echo "5^3^2 = ", 5^3^2 echo "(5^3)^2 = ", (5^3)^2 echo "5^(3^2) = ", 5^(3^2) echo "foldl([5, 3, 2], a^b) = ", foldl([5, 3, 2], a^b) echo "foldr([5, 3, 2], a^b) = ", foldr([5, 3, 2], a^b)</syntaxhighlight>

Output:

5^3^2 =   1953125
(5^3)^2 = 15625
5^(3^2) = 1953125
foldl([5, 3, 2], a^b) = 15625
foldr([5, 3, 2], a^b) = 1953125

OCaml

OCaml language has '**' as an exponentiation symbol for floating point integers <OCaml code>

1. 5. ** 3. ** 2. ;;
2. 5. **( 3. ** 2.) ;;
3. (5. ** 3. ) **2. ;;

Output:

- :   float = 1953125.
- :     float = 1953125. 
- :   float = 15625.

PARI/GP

Exponentiation is right-associative in GP.

f(s)=print(s" = "eval(s));
apply(f, ["5^3^2", "(5^3)^2", "5^(3^2)"]);

Output:

5^3^2 = 1953125
(5^3)^2 = 15625
5^(3^2) = 1953125

Perl

say "$_ = " . eval($_)  for  qw/5**3**2  (5**3)**2  5**(3**2)/;

Output:

5**3**2 = 1953125
(5**3)**2 = 15625
5**(3**2) = 1953125

Phix

Library: Phix/basics

Phix has a power function rather than an infix power operator, hence there is no possible confusion.

?power(power(5,3),2)
?power(5,power(3,2))

Output:

15625
1953125

PicoLisp

The PicoLisp '**' exponentiation function takes 2 arguments

: (** (** 5 3) 2)
-> 15625

: (** 5 (** 3 2))
-> 1953125

PL/I

exponentiation: procedure options(main);
    put skip edit('5**3**2   = ', 5**3**2)   (A,F(7));
    put skip edit('(5**3)**2 = ', (5**3)**2) (A,F(7));
    put skip edit('5**(3**2) = ', 5**(3**2)) (A,F(7));
end exponentiation;

Output:

5**3**2   =   15625
(5**3)**2 =   15625
5**(3**2) = 1953125

Python

>>> 5**3**2
1953125
>>> (5**3)**2
15625
>>> 5**(3**2)
1953125
>>> # The following is not normally done
>>> try: from functools import reduce # Py3K
except: pass

>>> reduce(pow, (5, 3, 2))
15625
>>>

Quackery

Quackery uses Reverse Polish Notation, so there is no ambiguity and no need for parenthesising.

As a dialogue in the Quackery Shell…

Welcome to Quackery.

Enter "leave" to leave the shell.

/O> $ "5 3 2 ** **" dup echo$ say " returns " quackery echo cr
... $ "5 3 ** 2 **" dup echo$ say " returns " quackery echo cr
... 
5 3 2 ** ** returns 1953125
5 3 ** 2 ** returns 15625

Stack empty.

R

The 'Operator Syntax and Precedence' documentation tells us that "^" is "exponentiation (right to left)". The 'Arithmetic Operators' documentation also tells us that the parser translates "**" to "^", but its depreciation status is complicated.

It turns out that the parser is so blind to "**" that we cannot even quote it. The following are identical:

print(quote(5**3))
print(quote(5^3))

Another method is to use "^" as if it is an ordinary function of two arguments. It appears that "**" does not support this. As there is no potential for ambiguity in the operator precedence, we will not print this result below. For example:

'^'('^'(5, 3), 2)

is clearly (5^3)^2 i.e. 15625, whereas

'^'(5, '^'(3, 2))

is clearly 5^(3^2) i.e. 1953125.

As for actually solving the task, the requirement that each output be on a new line causes us a surprising amount of difficulty. To avoid repeating ourselves, we must almost resort to metaprogramming:

inputs <- alist(5^3^2, (5^3)^2, 5^(3^2), 5**3**2, (5**3)**2, 5**(3**2))
invisible(sapply(inputs, function(x) cat(deparse(x), "returns: ", eval(x), "\n")))

Alternatively, we could print out a matrix or data frame:

print(matrix(sapply(inputs, eval), dimnames = list(inputs, "Outputs")))
print(data.frame(Inputs = sapply(inputs, deparse), Outputs = sapply(inputs, eval))))

Output:

> print(quote(5**3))
5^3
> print(quote(5^3))
5^3
> invisible(sapply(inputs, function(x) cat(deparse(x), "returns: ", eval(x), "\n")))
5^3^2 returns:  1953125 
(5^3)^2 returns:  15625 
5^(3^2) returns:  1953125 
5^3^2 returns:  1953125 
(5^3)^2 returns:  15625 
5^(3^2) returns:  1953125
> print(matrix(sapply(inputs, eval), dimnames = list(inputs, "Outputs")))
        Outputs
5^3^2   1953125
(5^3)^2   15625
5^(3^2) 1953125
5^3^2   1953125
(5^3)^2   15625
5^(3^2) 1953125
> print(data.frame(Inputs = sapply(inputs, deparse), Outputs = sapply(inputs, eval)))
   Inputs Outputs
1   5^3^2 1953125
2 (5^3)^2   15625
3 5^(3^2) 1953125
4   5^3^2 1953125
5 (5^3)^2   15625
6 5^(3^2) 1953125

Racket

#lang racket
;; 5**3**2 depends on associativity of ** : Racket's (scheme's) prefix function
;; calling syntax only allows for pairs of arguments for expt.

;; So no can do for 5**3**2
;; (5**3)**2
(displayln "prefix")
(expt (expt 5 3) 2)
;; (5**3)**2
(expt 5 (expt 3 2))

;; There is also a less-used infix operation (for all functions, not just expt)... which I suppose
;; might do with an airing. But fundamentally nothing changes.
(displayln "\"in\"fix")
((5 . expt . 3) . expt .  2)
(5  . expt . (3 . expt . 2))

;; everyone's doing a reduction, it seems
(displayln "reduction")
(require (only-in srfi/1 reduce reduce-right))
(reduce expt 1 '(5 3 2))
(reduce-right expt 1 '(5 3 2))

Output:

prefix
15625
1953125
"in"fix
15625
1953125
reduction
14134776518227074636666380005943348126619871175004951664972849610340958208
1953125

Raku

(formerly Perl 6)

Works with: rakudo version 2016.08

Note that the reduction forms automatically go right-to-left because the base operator is right-associative. Most other operators are left-associative and would automatically reduce left-to-right instead.

use MONKEY-SEE-NO-EVAL;
sub demo($x) { say "  $x\t───► ", EVAL $x }

demo '5**3**2';      # show ** is right associative
demo '(5**3)**2';
demo '5**(3**2)';

demo '[**] 5,3,2';   # reduction form, show only final result
demo '[\**] 5,3,2';  # triangle reduction, show growing results

# Unicode postfix exponents are supported as well:

demo '(5³)²';
demo '5³²';

Output:

  5**3**2	───► 1953125
  (5**3)**2	───► 15625
  5**(3**2)	───► 1953125
  [**] 5,3,2	───► 1953125
  [\**] 5,3,2	───► 2 9 1953125
  (5³)²	───► 15625
  5³²	───► 23283064365386962890625

The Unicode exponent form without parentheses ends up raising to the 32nd power. Nor are you even allowed to parenthesize it the other way: 5(³²) would be a syntax error. Despite all that, for programs that do a lot of squaring or cubing, the postfix forms can enhance both readability and concision.

Red

In Red, operators simply evaluate left to right. As this differs from mathematical order of operations, Red provides the math function which evaluates a block using math rules instead of Red's default evaluation. One could also use the power function, sidestepping the issue of evaluation order entirely. All three approaches are shown.

Red["Exponentiation order"]

exprs: [
    [5 ** 3 ** 2]
    [(5 ** 3) ** 2]
    [5 ** (3 ** 2)]
    [power power 5 3 2]   ;-- functions too
    [power 5 power 3 2]
]

foreach expr exprs [
    print [mold/only expr "=" do expr]
    if find expr '** [
        print [mold/only expr "=" math expr "using math"]
    ]
]

Output:

5 ** 3 ** 2 = 15625
5 ** 3 ** 2 = 1953125 using math
(5 ** 3) ** 2 = 15625
(5 ** 3) ** 2 = 15625 using math
5 ** (3 ** 2) = 1953125
5 ** (3 ** 2) = 1953125 using math
power power 5 3 2 = 15625
power 5 power 3 2 = 1953125

REXX

/*REXX program demonstrates various ways of multiple exponentiations.   */
/*┌────────────────────────────────────────────────────────────────────┐
  │ The REXX language uses      **      for exponentiation.            │
  │                   Also,    *  *     can be used.                   │
  |    and even                */*power of*/*                          |
  └────────────────────────────────────────────────────────────────────┘*/

say '   5**3**2   ───► '    5**3**2
say '   (5**3)**2 ───► '    (5**3)**2
say '   5**(3**2) ───► '    5**(3**2)
                                       /*stick a fork in it, we're done.*/

output

   5**3**2   ───►  15625
   (5**3)**2 ───►  15625
   5**(3**2) ───►  1953125

Ring

In the Ring it is impossible to show the result of: 5^3^2

see "(5^3)^2 =>" + pow(pow(5,3),2) + nl
see "5^(3^2) =>" + pow(5,pow(3,2)) + nl

Output:

(5^3)^2 =>15625
5^(3^2) =>1953125

Ruby

ar = ["5**3**2", "(5**3)**2", "5**(3**2)", "[5,3,2].inject(:**)"]
ar.each{|exp| puts "#{exp}:\t#{eval exp}"}

Output:

5**3**2:	1953125
(5**3)**2:	15625
5**(3**2):	1953125
[5,3,2].inject(:**):	15625

Rust

fn main() {
    println!("5**3**2   = {:7}", 5u32.pow(3).pow(2));
    println!("(5**3)**2 = {:7}", (5u32.pow(3)).pow(2));
    println!("5**(3**2) = {:7}", 5u32.pow(3u32.pow(2)));
}

Output:

5**3**2   =   15625
(5**3)**2 =   15625
5**(3**2) = 1953125

Scala

Scal has no exponentiation operator, but uses the function (scala.)math.pow(x: Double, y: Double): Double function in the Scala runtime library.
Integer exponentiation can be done with e.g. BigInt or BigInteger.pow(n: Int) method.
There are no associativity issues.

Seed7

$ include "seed7_05.s7i";

const proc: main is func
  begin
    writeln("5**3**2   = " <& 5**3**2);
    writeln("(5**3)**2 = " <& (5**3)**2);
    writeln("5**(3**2) = " <& 5**(3**2));
  end func;

Output:

5**3**2   = 1953125
(5**3)**2 = 15625
5**(3**2) = 1953125

Sidef

In Sidef, the whitespace between the operands and the operator controls the precedence of the operation.

var a = [
    '5**3**2',
    '(5**3)**2',
    '5**(3**2)',
    '5 ** 3 ** 2',
    '5 ** 3**2',
    '5**3 ** 2',
    '[5,3,2]«**»',
]

a.each {|e|
    "%-12s == %s\n".printf(e, eval(e))
}

Output:

5**3**2      == 1953125
(5**3)**2    == 15625
5**(3**2)    == 1953125
5 ** 3 ** 2  == 15625
5 ** 3**2    == 1953125
5**3 ** 2    == 15625
[5,3,2]«**»  == 15625

Simula

OutText("5** 3 **2: "); OutInt(5** 3 **2, 0); Outimage;
OutText("(5**3)**2: "); OutInt((5**3)**2, 0); Outimage;
OutText("5**(3**2): "); OutInt(5**(3**2), 0); Outimage

Output:

5** 3 **2: 15625
(5**3)**2: 15625
5**(3**2): 1953125

Smalltalk

Works in Smalltalk/X ¹

Smalltalk strictly evaluates left to right; operators are not known to the language/parser, but instead message sends to the receiver on the left side (aka: virtual function calls) .

Transcript show:'5**3**2 => '; showCR: 5**3**2.
Transcript show:'(5**3)**2 => '; showCR: (5**3)**2.
Transcript show:'5**(3**2) => '; showCR: 5**(3**2).

Output:

5**(3**2) => 1953125
5**3**2 => 15625
(5**3)**2 => 15625

Note ¹ other Smalltalk's may define ** to simply call "raisedTo:", which is standard.

Stata

. di (5^3^2)
15625

. di ((5^3)^2)
15625

. di (5^(3^2))
1953125

Likewise in Mata:

. mata (5^3^2)
  15625

. mata ((5^3)^2)
  15625

. mata (5^(3^2))
  1953125

Swift

Swift doesn't have an exponentiation operator, however it's possible to define one, including the precedence and associativity.

precedencegroup ExponentiationPrecedence {
  associativity: left
  higherThan: MultiplicationPrecedence
}

infix operator ** : ExponentiationPrecedence

@inlinable
public func ** <T: BinaryInteger>(lhs: T, rhs: T) -> T {
  guard lhs != 0 else {
    return 1
  }

  var x = lhs
  var n = rhs
  var y = T(1)

  while n > 1 {
    switch n & 1 {
    case 0:
      n /= 2
    case 1:
      y *= x
      n = (n - 1) / 2
    case _:
      fatalError()
    }

    x *= x
  }

  return x * y
}

print(5 ** 3 ** 2)
print((5 ** 3) ** 2)
print(5 ** (3 ** 2))

Output:

15625
15625
1953125

Tcl

foreach expression {5**3**2 (5**3)**2 5**(3**2)} {
    puts "${expression}:\t[expr $expression]"
}

Output:

5**3**2:	1953125
(5**3)**2:	15625
5**(3**2):	1953125

There's also a binary pow() expression function that always converts its arguments to floating point numbers and then applies the exponentiation operation; it's now largely obsolete because of the ** operator, but is retained for backward compatibility with older programs.

VBA

Public Sub exp()
    Debug.Print "5^3^2", 5 ^ 3 ^ 2
    Debug.Print "(5^3)^2", (5 ^ 3) ^ 2
    Debug.Print "5^(3^2)", 5 ^ (3 ^ 2)
End Sub

Output:

5^3^2          15625 
(5^3)^2        15625 
5^(3^2)        1953125 

VBScript

WScript.StdOut.WriteLine "5^3^2 => " & 5^3^2
WScript.StdOut.WriteLine "(5^3)^2 => " & (5^3)^2
WScript.StdOut.WriteLine "5^(3^2) => " & 5^(3^2)

Output:

5^3^2 => 15625
(5^3)^2 => 15625
5^(3^2) => 1953125

Verbexx

// Exponentiation order example:

@SAY "5**3**2   = " ( 5**3**2   ); 
@SAY "(5**3)**2 = " ( (5**3)**2 );
@SAY "5**(3**2) = " ( 5**(3**2) );

/] Output:
  
    5**3**2   =  1953125
    (5**3)**2 =  15625
    5**(3**2) =  1953125

Wren

Library: Wren-fmt

Wren doesn't have an exponentiation operator as such but the Num class has a pow method which does the same thing.

import "/fmt" for Fmt

var ops = [ "5**3**2", "(5**3)**2", "5**(3**2)" ]
var results = [ 5.pow(3).pow(2), (5.pow(3)).pow(2), 5.pow(3.pow(2)) ]
for (i in 0...ops.count) {
    System.print("%(Fmt.s(-9, ops[i])) -> %(results[i])")
}

Output:

5**3**2   -> 15625
(5**3)**2 -> 15625
5**(3**2) -> 1953125

zkl

Translation of: C

zkl does not have an exponentiation operator but floats have a pow method.

println("5 ^ 3 ^ 2   = %,d".fmt((5.0).pow((3.0).pow(2))));
println("(5 ^ 3) ^ 2 = %,d".fmt((5.0).pow(3).pow(2)));
println("5 ^ (3 ^ 2) = %,d".fmt((5.0).pow((3.0).pow(2))));

Output:

5 ^ 3 ^ 2   = 1,953,125
(5 ^ 3) ^ 2 = 15,625
5 ^ (3 ^ 2) = 1,953,125