Question 214824
{{{s(x) = 2^x}}}
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{{{h(x) = x^2}}}
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you replace x in h(x) with (2^x) to get:
{{{h(s(x)) = h(2^x) = (2^x)^2}}}
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To prove this is true, take any value for x.
let x = 5
{{{s(x) = 2^x = 2^5 = 32}}}
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{{{h(s(x)) = (2^x)^2 = (2^5)^2 = 2^(10) = 1024}}}
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if x = 32
then {{{h(x) = h(32) = 32^2 = 1024}}}
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You got the same result if you solved for {{{h(s(x))}}} directly, or if you solved for {{{s(x)}}} first and then placed that value into {{{h(x)}}}.  This shows that the two equations are equivalent.
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Your answer is:
{{{h(s(x)) = (2^x)^2}}}