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Simplifying \[\-e^{-\ln(x)}\]
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\[\ -e^{-ln(x)}\]
Lol oh @Jhannybean look at the message I sent :P \[\large -e^{-\ln(x)} = \frac{1}{-e^{\ln(x)}}\]
\[\LARGE -e^{-lnx}=-e^{\ln(1/x)}=-\frac{1}{x}\] alternatively we can write: \[\LARGE -e^{-lnx}=-(e^{lnx})^{-1}=-x^{-1}\]
@johnweldon1993 I understood how you solved this, but i'm just wondering if there was an alternative way by factoring out the negatives
Ah ok :) thanks you guys!
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