Question

Let `j(x)=x^2` and `k(x)=x^3`, does `k(j(x))=j(k(x))`?

Answer 1

Yes

Explanation:

`k(j(x)) = k(x^2) = (x^2)^3 = x^6 = (x^3)^2 = j(x^3) = j(k(x))`

More generally...

If `m, n` are positive integers and `x` is any number, then we find:

`(x^m)^n = overbrace(x^m xx x^m xx ... xx x^m)^"n times"`

`color(white)((x^m)^n) = overbrace(overbrace((x xx x xx ... xx x))^"m times" xx overbrace((x xx x xx ... xx x))^"m times" xx ... xx overbrace((x xx x xx ... xx x))^"m times")^"n times"`

`color(white)((x^m)^n) = overbrace(x xx x xx ... xx x)^"mn times"`

`color(white)((x^m)^n) = x^(mn)`

So in particular:

`(x^m)^n = x^(mn) = x^(nm) = (x^n)^m`

`color(white)()`
Footnote

`(x^m)^n = x^(mn)` holds in any of the following circumstances:

• Any number `x` (Real or complex), with `m, n` positive integers.

• Any non-zero number `x` (Real or complex), with `m, n` any integers.

• Any positive (Real) number `x`, with `m, n` any numbers.

It does not hold in all cases.

For example:

`-1 = (-1)^1 = (-1)^(3/2*2/3) != ((-1)^(3/2))^(2/3) = (-i)^(2/3) = 1/2-sqrt(3)/2i`