Question #c7a86

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Answer 1 · 2016-11-28T17:55:27

The answer is $=2^(2x)$

$S(x)=x^2$

$P(x)=2^x$

$SoP(x)=S(P(x))=S(2^x)=(2^x)^2=2^(2x)$

$PoS(x)=P(S(x))=P(x^2)=2^(x^2)$

$SoP(x)!=PoS(x)$

Answer 2 · 2016-11-28T17:55:51

$:. (S@P)(y) = 4^y$

$(S@P)(y)$ means the composite $S(p(y))$

$:. (S@P)(y) = S(P(y))$
$:. (S@P)(y) = S(2^y)$
$:. (S@P)(y) = (2^y)^2$
$:. (S@P)(y) = 2^(2y)$
$:. (S@P)(y) = 4^y$