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

1 Answer
George C.
Jan 8, 2017

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#

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