Upload
Login or register
Anonymous comments allowed.
#10 - anon
Reply 0
(07/04/2013) [-]
e^-ln(x)...

So... -(x)...
#15 to #10 - anonymoose
Reply +3
(07/04/2013) [-]
Nope, 1/(x)
#14 to #10 - ilgattozaiga
Reply +1
(07/04/2013) [-]
e^-lnx
=e^ln(x^-1)
=e^ln(1/x)
=1/x
#12 to #10 - anon
Reply 0
(07/04/2013) [-]
no, actually that would be 1/x. Since e^-Ln(x) equals ( e^Ln(x) )^-1 = (x)^-1 = 1/x