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#10 - StarvedSouthKorean (07/04/2013) [-]
e^-ln(x)...

So... -(x)...
User avatar #15 to #10 - anonymoose (07/04/2013) [-]
Nope, 1/(x)
#14 to #10 - ilgattozaiga (07/04/2013) [-]
e^-lnx
=e^ln(x^-1)
=e^ln(1/x)
=1/x
#12 to #10 - StarvedSouthKorean (07/04/2013) [-]
no, actually that would be 1/x. Since e^-Ln(x) equals ( e^Ln(x) )^-1 = (x)^-1 = 1/x
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