say i have something like:(making up #'s here so they might not work out)16=log(.5x)+log(.1x)how do i solve for x?
11/19/2007 11:14:48 PM
flush 'em
11/19/2007 11:15:12 PM
I usually pinch them off in the toilet.
11/19/2007 11:15:19 PM
it's big, it's heavy, it's MATHEMATICS
11/19/2007 11:15:59 PM
eor inverse of log ?i done forgot sir
^ nah man, e is for natural logsbase 10 that motherfucker
11/19/2007 11:18:26 PM
you mean 10^x that bitch
11/19/2007 11:19:26 PM
aaahh yeah.base 10.true.
11/19/2007 11:20:22 PM
^^that is exactly what I meantthat motherfucker is base 10[Edited on November 19, 2007 at 11:21 PM. Reason : UNDO IT]
11/19/2007 11:20:44 PM
well base 10 would just be putting the number into base 10which it already should be
11/19/2007 11:21:24 PM
16=log(.5x)+log(.1x)10^(16) = 10^(log(.5x*.1x))10^(16) = (.5x*.1x)solve for x
11/19/2007 11:22:20 PM
pfffffffft
11/19/2007 11:24:39 PM
also, burn them
11/19/2007 11:25:28 PM
^burn them
11/19/2007 11:27:23 PM
11/19/2007 11:28:25 PM
clearcut and burn a forest
11/19/2007 11:29:11 PM
Walter is a moron
11/19/2007 11:30:41 PM
Get a calculator.
11/19/2007 11:52:40 PM
thanks StillFuchsia, now my brain hurts
11/20/2007 12:14:56 AM
you're welcome
11/20/2007 12:25:59 AM
someone didnt pay attention in 8th grade algebra.
11/20/2007 12:28:14 AM
now just solve this equation explicitlyxln(x)=17
11/20/2007 12:39:13 AM
^ hahahaha
11/20/2007 12:43:11 AM
^^gg
11/20/2007 12:52:24 AM
if people paid attention in class this all would never have happened...
11/20/2007 2:07:41 AM
^if noobs would STFU then that would never have happened...]
11/20/2007 2:08:20 AM
11/20/2007 2:08:40 AM
11/20/2007 2:16:12 AM