Weekly Cal Thread - Page 1

Page: 1

Rating: Unrated [0]

Weekly Cal Thread

Good [+1]Toggle ReplyLink» No_Comply a répondu le Wed 22 Oct, 2008 @ 1:02pm


Coolness: 85155	Ok, so would LOGa(x^2/yz^3) = 2LOGaX - 3LOGaY + 3LOGaZ?

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

Good [+1]Toggle ReplyLink» Kishmay_Pinas a répondu le Wed 22 Oct, 2008 @ 1:08pm


Coolness: 103475	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

	I'm feeling in a meeting @ barin right now..

Good [+1]Toggle ReplyLink» flo a répondu le Wed 22 Oct, 2008 @ 4:52pm


Coolness: 146545	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. Mise À Jour » flo a écrit sur Wed 22 Oct, 2008 @ 5:15pm if it's not clear enough : the answer is NO

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

Good [+1]Toggle ReplyLink» Mico a répondu le Wed 22 Oct, 2008 @ 5:28pm

Good [+1]Toggle ReplyLink» flo a répondu le Wed 22 Oct, 2008 @ 5:30pm

Good [+1]Toggle ReplyLink» No_Comply a répondu le Thu 23 Oct, 2008 @ 10:24am


Coolness: 85155	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 a répondu le Fri 24 Oct, 2008 @ 4:41am


Coolness: 146545	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 a répondu le Fri 24 Oct, 2008 @ 6:14am


Coolness: 131815	pie and log. you mathematicians are perverts

	I'm feeling 4.5kw 240vrms 45a right now..

Good [+1]Toggle ReplyLink» flo a répondu le Fri 24 Oct, 2008 @ 8:34am

Good [+1]Toggle ReplyLink» No_Comply a répondu le Fri 24 Oct, 2008 @ 10:18am


Coolness: 85155	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..

Weekly Cal Thread

Page: 1

Poster Une Réponse

Vous devez être connecté pour soumettre une réponse.