Weekly Cal Thread - Page 1

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Weekly Cal Thread

Good [+1]Toggle ReplyLink» No_Comply replied on Wed Oct 22, 2008 @ 1:02pm


Coolness: 85015	Ok, so would LOGa(x^2/yz^3) = 2LOGaX - 3LOGaY + 3LOGaZ?

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Good [+1]Toggle ReplyLink» Kishmay_Pinas replied on Wed Oct 22, 2008 @ 1:08pm


Coolness: 103335	What rolls down stairs alone or in pairs, and over your neighbor's dog? What's great for a snack, And fits on your back? It's log, log, log It's log, it's log, It's big, it's heavy, it's wood. It's log, it's log, it's better than bad, it's good." Everyone wants a log You're gonna love it, log Come on and get your log Everyone needs a log log log log

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Good [+1]Toggle ReplyLink» flo replied on Wed Oct 22, 2008 @ 4:52pm


Coolness: 146405	Originally Posted By NO_COMPLY Ok, so would LOGa(x^2/yz^3) = 2LOGaX - 3LOGaY + 3LOGaZ? lol, you said GaY seriously : it depends whether it's (yz)^3, ie. (y^3 times z^3), or (y times z^3) ; i'll assume it's the first one (ie. the exponent is applied to both y and z) note that you don't care about the base "a" of the log, since it's always the same. then you just apply the logarithmic identities : log(x^2 / (yz)^3) = log(x^2) - log((yz)^3) = 2log(x) - log((yz)^3) = 2log(x) - log(y^3 times z^3) = 2log(x) - (log(y^3) + log(z^3)) = 2log(x) - 3log(y) - 3log(z) Thus you get the initial equality rewritten as : 2log(x) - 3log(y) - 3log(z) = 2log(x) - 3log(y) + 3log(z) which you easily simplify as : -3log(z) = 3log(z) which you can write : -6log(z) = 0, ie. log(z)=0 So the answer is yes if and only if log(z)=0 : this can never happen, since logarithms (in any base) have an asymptote whose equation is "x=0", ie. the vertical axis. The asymptote means that it's a limit value that's never actually reached. Update » flo wrote on Wed Oct 22, 2008 @ 5:15pm if it's not clear enough : the answer is NO

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Good [+1]Toggle ReplyLink» Mico replied on Wed Oct 22, 2008 @ 5:28pm

Good [+1]Toggle ReplyLink» flo replied on Wed Oct 22, 2008 @ 5:30pm

Good [+1]Toggle ReplyLink» No_Comply replied on Thu Oct 23, 2008 @ 10:24am


Coolness: 85015	Originally Posted By FLO Thus you get the initial equality rewritten as : 2log(x) - 3log(y) - 3log(z) = 2log(x) - 3log(y) + 3log(z) I thought that's what i said =p As for not needing the to write the base, i was told to whenever it's given, but i understand why its not important in this case.

	I'm feeling gettin more ink soon right now..

Good [+1]Toggle ReplyLink» flo replied on Fri Oct 24, 2008 @ 4:41am


Coolness: 146405	i think there were several words missing in your post but i guess i got what you meant after reading it twice :P ok so the second half of the proof i wrote is useless for you here, hence the answer to your initial question is yes.

	I'm feeling phd powa !!! right now..

Good [+1]Toggle ReplyLink» cutterhead replied on Fri Oct 24, 2008 @ 6:14am


Coolness: 131675	pie and log. you mathematicians are perverts

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Good [+1]Toggle ReplyLink» flo replied on Fri Oct 24, 2008 @ 8:34am

Good [+1]Toggle ReplyLink» No_Comply replied on Fri Oct 24, 2008 @ 10:18am


Coolness: 85015	Ahhh k, thanks Flo, was confoosed for a sec =] you mathematicians are perverts Oh fuck ya, my cocks^9, if ya know what i mean.

	I'm feeling gettin more ink soon right now..

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