SOLUTION: One more function I can't seem to get right. s(x)=2^x h(x)=x^2 find h(s(x)) My answer was 4^x but was wrong of course.

Algebra ->  Functions -> SOLUTION: One more function I can't seem to get right. s(x)=2^x h(x)=x^2 find h(s(x)) My answer was 4^x but was wrong of course.      Log On


   

Question 214824: One more function I can't seem to get right.
s(x)=2^x h(x)=x^2 find h(s(x))
My answer was 4^x but was wrong of course.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
s%28x%29+=+2%5Ex
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h%28x%29+=+x%5E2
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you replace x in h(x) with (2^x) to get:
h%28s%28x%29%29+=+h%282%5Ex%29+=+%282%5Ex%29%5E2
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To prove this is true, take any value for x.
let x = 5
s%28x%29+=+2%5Ex+=+2%5E5+=+32
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h%28s%28x%29%29+=+%282%5Ex%29%5E2+=+%282%5E5%29%5E2+=+2%5E%2810%29+=+1024
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if x = 32
then h%28x%29+=+h%2832%29+=+32%5E2+=+1024
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You got the same result if you solved for h%28s%28x%29%29 directly, or if you solved for s%28x%29 first and then placed that value into h%28x%29. This shows that the two equations are equivalent.
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Your answer is:
h%28s%28x%29%29+=+%282%5Ex%29%5E2