Let $j (x) = x^{2}$ $j(x)=x^2$ and $k (x) = x^{3}$ $k(x)=x^3$ , does $k (j (x)) = j (k (x))$ $k(j(x))=j(k(x))$ ?

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Let $j (x) = x^{2}$ $j(x)=x^2$ and $k (x) = x^{3}$ $k(x)=x^3$ , does $k (j (x)) = j (k (x))$ $k(j(x))=j(k(x))$ ?

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Answer 1 · 2017-01-08T14:08:04

Yes

Explanation:

$k (j (x)) = k (x^{2}) = {(x^{2})}^{3} = x^{6} = {(x^{3})}^{2} = j (x^{3}) = j (k (x))$ $k(j(x)) = k(x^2) = (x^2)^3 = x^6 = (x^3)^2 = j(x^3) = j(k(x))$

More generally...

If $m, n$ $m, n$ are positive integers and $x$ $x$ 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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Let $j (x) = x^{2}$ $j(x)=x^2$ and $k (x) = x^{3}$ $k(x)=x^3$ , does $k (j (x)) = j (k (x))$ $k(j(x))=j(k(x))$ ?

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Explanation:

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Let j(x)=x^2j(x)=x2j(x)=x^2 and k(x)=x^3k(x)=x3k(x)=x^3, does k(j(x))=j(k(x))k(j(x))=j(k(x))k(j(x))=j(k(x))?

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Explanation:

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Let $j (x) = x^{2}$ $j(x)=x^2$ and $k (x) = x^{3}$ $k(x)=x^3$ , does $k (j (x)) = j (k (x))$ $k(j(x))=j(k(x))$ ?