Question #80447

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Question #80447

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Answer 1 · 2017-08-24T00:36:21

${(- e^{ln x} x^{3} e^{1 - x})}^{2} = x^{8} e^{2 - 2 x}$ $( - e^(lnx) x^3 e^(1-x) )^2 = x^8 e^(2-2x)$

Explanation:

${(- e^{ln x} x^{3} e^{1 - x})}^{2}$ $( - e^(lnx) x^3 e^(1-x) )^2$

First simplify what is inside the parentheses, note that $e^{ln x} = x$ $e^lnx=x$

$= {(- x \cdot x^{3} \cdot e^{1 - x})}^{2}$ $= ( -x*x^3*e^(1-x) )^2$

$= {(- x^{4} e^{1 - x})}^{2}$ $= ( -x^4e^(1-x) )^2$

Distribute the exponent among what is inside the base of the exponent:
$= {(- 1)}^{2} \cdot {(x^{4})}^{2} \cdot {(e^{1 - x})}^{2}$ $= (-1)^2 * (x^4)^2 * (e^(1-x))^2$

Simplify:

$= x^{8} e^{2 - 2 x}$ $= x^8 e^(2-2x)$

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Question #80447

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