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Weekly Cal Thread
Good [+1]Toggle ReplyLink» No_Comply replied on Wed Oct 22, 2008 @ 1:02pm |
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 replied on Wed Oct 22, 2008 @ 1:08pm |
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 replied on Wed Oct 22, 2008 @ 4:52pm |
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. | |
I'm feeling phd powa !!! right now.. |
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 |
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 |
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 |
pie and log. you mathematicians are perverts | |
I'm feeling 4.5kw 240vrms 45a right now.. |
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 |
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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