# How does the exponent raise to log works?

I have this simplification in my textbook, $$e^{-(x+3\log|x|)} = x^{-3}e^{-x}$$

I cant get how did we got that? i know this rules $e^{\log|x|}=x$ and $e^{x+y}= e^xe^y$

Are those things related? Can someone explain me this? Thanks

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That rule should say $e^{\log|x|}=|x|$, which is equal to $x$ if $x$ is positive. Also, recall that all this assumes that "$\log$" means natural logarithm, i.e. the base is $e$. – Michael Hardy Dec 28 '12 at 19:30

## 1 Answer

Well $$e^{-x-3\log x}=e^{-x}e^{-3\log x}=e^{-x}(e^{\log x})^{-3}=e^{-x}x^{-3}$$ and there you have it. Note that I used that $x^{yz}=(x^y)^z$

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