Problem Statement

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

Some languages treat/honor infix operators when performing exponentiation   (raising numbers to some power by the language's exponentiation operator,   if the computer programming language has one).

Other programming languages may make use of the   POW   or some other BIF   (Built─In Ffunction),   or some other library service. If your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.

This task will deal with the case where there is some form of an   infix operator   operating in   (or operating on)   the base.

A negative five raised to the 3rd power could be specified as:

(Not all computer programming languages have an exponential operator and/or support these syntax expression(s).

Try to present the results in the same format/manner as the other programming entries to make any differences apparent.

The variables may be of any type(s) that is/are applicable in your language.

Let's start with the solution:

mathtype = math.type or type -- polyfill <5.3
function test(xs, ps)
  for _,x in ipairs(xs) do
    for _,p in ipairs(ps) do
      print(string.format("%2.f  %7s  %2.f  %7s    %4.f    %4.f    %4.f    %4.f", x, mathtype(x), p, mathtype(p), -x^p, -(x)^p, (-x)^p, -(x^p)))
    end
  end
end
print(" x  type(x)   p  type(p)    -x^p  -(x)^p  (-x)^p  -(x^p)")
print("--  -------  --  -------  ------  ------  ------  ------")
test( {-5.,5.}, {2.,3.} ) -- "float" (or "number" if <5.3)
if math.type then -- if >=5.3
  test( {-5,5}, {2,3} ) -- "integer"
end