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Comments:
<0> 1x-.2x=.8x
<0> is that not right
<1> peachfuzz: correct. so you have: y = 0.8x
<1> peachfuzz: now rearrange so you have x =
<0> x =.8y
<1> Vegitto: log both sides
<1> peachfuzz: not quite
<1> peachfuzz: you want to divide both sides by 0.8
<0> opps y/.8=x
<0> when you type fast you make mistakes
<1> right. in fact what is 1/0.8 ?
<0> its like rushing on a exam
<1> indeed
<0> 1.25
<1> peachfuzz: so you answer is: 1.25 * original price
<1> peachfuzz: so you answer is: 1.25 * final_price
<1> i meant
<1> or final_price / 0.8
<0> 15737.5
<0> if my math is correct
<1> peachfuzz: well check it - subtract 20% and see if you get the right answer ;)
<0> yes I do
<0> have another silly question
<0> If Leah is 6 years older than her sister, Sue, and John is 5 years older than Leah, and the total of their ages is 41. Then how old is Sue?... My equations are Leah = sue +6, John = 5+(sue+6)=11+sue but then I get stuck
<2> damn, my pc just broke down
<1> Vegitto: :/
<1> Vegitto: how did you get on?
<2> did someone post anything revelant?
<2> dunno.. somehow
<1> Vegitto: log both sides
<2> JohnFlux.. like?
<3> peachfuzz solve for x+x+6+x +6 +5 =41
<2> I get log_x 4 = log_2 (x/2)
<4> What has lilo said about the netsplit?
<2> but what's next?
<1> you mean log_2 4x ?
<1> hmm
<1> no, what step did you do?
<2> okay.. look:
<1> what was the original equation again :)
<2> x^(log_2 (x)) = 4x
<2> x^(log_2 (x)) = 4 * x^1 | : x^1
<2> x^(log_2 (x) - 1) = 4
<2> x^(log_2 (x/2)) = 4
<2> it that okay so far?
<5> http://soerusabsolutus.so.funpic.de/formula3.jpeg <-- can someone please explain a bit how someone can get this result (a*b= a1b1+a2b2+a3b3)?
<1> Vegitto: what does that notation mean :)
<1> | :
<2> to divide both sides from x
<2> *by*
<1> hmm
<1> interesting way
<1> how about instead:
<1> Vegitto: log_2 (x ^ (log_2 x) = log_2(4x)
<1> -> log_2(x) * log_2(x) = log_2(4x)
<1> which is looking better, no?
<2> indeed
<2> thanks ;)
<1> you know what to do now?
<2> well, it's kinda easy then, huh?
<1> not so much
<2> log_2(4x) * x = log_2(4x)
<2> or am i missing something here?
<1> how did you get the LHS ?
<2> well, from your equation..
<6> hmmmm
<2> but oh well, i've probably screwed it ;)
<6> where does the log_2(x) * log_2(x) = log_2(4x)
<2> please get on with your equation then
<6> <1> which i
<6> ouch
<6> where does the log_2(x) * log_2(x) = log_2(4x) come from
<2> with your solution i mean
<1> kmh: i don't know
<1> kmh: oh
<1> kmh: sorry that
<2> log_something (x^(log_something) = log_something * log_something(x)
<2> or is it?
<1> right
<6> yes but that is log_2(4x) * x = 4x
<6> oh ic
<6> ignore that line
<7> log(x^y) = log x * log y
<7> hm
<7> no
<1> log(x^y) = y log x
<6> so you want to solve x^(log_2 (x)) = 4x ?
<7> yes
<6> ok
<1> I'm not sure where to go from: log_2(x) * log_2(x) = log_2(4x)
<1> it's looks simple, but i can't see what I'm missing...
<1> heh
<6> SoerenW : e1*e2=0, e2*e3=0, e1*e3=0
<6> log_2(4x)=log_2(4)+log_2(x)
<6> so you need to solve z^2=2+z for z
<6> and then compute 2^z
<1> oh yeah
<6> z=log_2(x)
<1> yeah course
<2> that's cool ;)
<2> thanks
<1> log_2(x) * log_2(x) = log_2(4x) -> log_2(x)^2 = log_2(x) + log_2(4) -> log_2(x)^2 - log_2(x) - log_2(4) = 0
<1> and a quadratic
<1> beautiful ;-)
<8> can someone tell me how I would evaluate a pi notation expression in octave or some other free program?
<0> If r = 5 z then 15 z = 3 y, then r =
<0> what am I not seeing
<5> kmh, but then i have products with e1^2, e2^2...
<1> peachfuzz: you have 15 z = 3 y so rearrange so it's just z=
<1> peachfuzz: or 5z = if you want
<8> its 5z = y, so r = 5z = y
<0> 15 z = 3y , z = 3y/15
<1> peachfuzz: okay now put that z into the r= 5z equation
<2> but it has no solution.. hmm
<6> SoerenW : yes but e1^2=1
<3> it works
<1> Vegitto: hmm? it has two solutions
<0> r= (5)(3y/15)
<2> wait.. no
<3> .
<2> it's okay
<0> so r = y
<1> Vegitto: :)
<2> :)
<0> ???
<6> JohnFlux : how's your japanese ?:)
<5> kmh, but e1*e2, too, or not? (i think they're oneness-vectors (is that the right word for a thing with a value=1?))
<1> kmh: i know "slower" "that hurts" and "dangerous"
<2> ha, anata wa nihon-jin desu ka?
<2> ah
<2> seems not ;)
<6> SoerenW : noe e1*e2=0
<8> so noone knows how to evaluate pi notation expressions?
<3> kmh: japanese? need something translated?
<2> maybe you do by chance listen to japanesepod101.com ? ;)
<6> SoerenW : e_i^2=1 and e_i*e_j=0 for i!=j
<5> ok
<6> JohnFlux : lol
<1> Vegitto: nah, nihongo sukoshi hanashimasu
<6> JohnFlux : i guess i better not ask about the context or inquire any further :)
<1> Vegitto: nah, nihongo au sukoshi hanashimasu
<8> can someone just reply so i know im not hidden or something?
<5> kmh, thx
<6> knoppix_ : what exactly ?
<1> knoppix_: i can't see you
<1> i don't know what pi notation is
<8> its sigma notation but for multiplication
<8> i need to multiply something 4.8 * 10 ^24 times
<8> but each thing is different from every other one
<6> SoerenW : think of e1 = (1,0,0), e2=(0,1,0) , e3=(0,0,1)
<1> knoppix_: what is each thing?
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