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 Posted: Mon, 17th May 2010 20:28 Post subject: e^(2x) = k (root x) |
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fuck I don't remember calc. Question is find k so that the equation has one solution. I know the answer is k = 2e^(1/2), but how do I arrive at that?
First one to answer gets a e-blowjob. Cheers,
-t
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| Posted: Mon, 17th May 2010 20:49 Post subject: |
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I assume you mean the equation is e^(2x) = x^(1/k).
1)e^(2x) = x^(1/k)
2)e^(2x)-x^(1/k) = 0
3)-x^(1/k) = -e^(2x)
4)x^(1/k) = e^(2x)
5)1/k = (log(e^(2x)))/(log(x))
6)log(x) = k*log(e^(2x))
7)-k*log(e^(2x)) = -log(x)
8 ) k=(log(x))/(log(e^(2x)))
Preliminary precautions: x can't be zero, log(x) (base is e) can't be zero, e^(2x) can't be zero, log(e^(2x))+2*i*pi*n can't be zero
I really can't fathom how the answer can be as the one you claim it to be. Well, that can be the case if the equation is not the same as the one I presume it to be.
Last edited by human_steel on Mon, 17th May 2010 20:55; edited 1 time in total
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| Posted: Mon, 17th May 2010 20:55 Post subject: |
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| Quote: | | I assume you mean the equation is e^(2x) = x^(1/k). |
Sorry, wrong assumption. Equation is:
e^(2x) = k * x^1/2.
Wrote it in a hurry, sorry =/
Sense Amid Madness, Wit Amidst Folly
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| Posted: Mon, 17th May 2010 20:59 Post subject: |
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Heh, at that point the equation is a piece of cake. You just have to divide both sides by sqrt(x), provided you have beforehand reserved that sqrt(x) is not equal to zero.
Edit: Hmm, you have to have the equation be compliant to only one x in order to calculate k... Let me think.
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TSR69
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| Posted: Mon, 17th May 2010 23:13 Post subject: |
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calculate the derivative of e^(-1/2 ln x +2 x) = 0
insert x and you're done.
the graf of e^(-1/2 ln x + 2 x) has 2 values except at (e^(-1/2 ln x + 2 x))' = 0
edit:
(e^(-1/2 ln x + 2 x))' = 0 => 2-1/(2x) = 0 => x=1/4
e^(-1/2 ln (1/4) + 2 *1/4) = 2 e^(1/2) = k
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| Posted: Tue, 18th May 2010 00:34 Post subject: |
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Why is everyone speaking alien here all of a sudden?
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HubU
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| Posted: Tue, 18th May 2010 16:46 Post subject: |
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Tout par l'État, rien hors de l'État, rien contre l'État!
"Music washes away from the soul the dust of everyday life." ~Berthold Auerbach
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HubU
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| Posted: Tue, 18th May 2010 16:58 Post subject: |
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Tous sur le cul de ta mère, merdeuuuu
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ixigia
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| Posted: Tue, 18th May 2010 19:28 Post subject: |
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ixigia
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Frant
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| Posted: Tue, 18th May 2010 19:56 Post subject: |
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2 ^ x + 3 ^ x = 50
Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn!
"The sky was the color of a TV tuned to a dead station" - Neuromancer
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Epsilon
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| Posted: Wed, 19th May 2010 01:09 Post subject: |
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LeoNatan
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| Posted: Wed, 19th May 2010 10:03 Post subject: |
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I think the question means "Find k so that the equation e^(2x) - k*x^(1/2) = 0 has one solution."
This is what is called an implicit function, i.e. you cannot make split it into an independent and dependent parts like y = f(x). In Hebrew, we call these functions "dumb functions" ("פונקציות סתומות"). 
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deelix
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| Posted: Wed, 19th May 2010 11:40 Post subject: |
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Fuck off I solved the shit out of this already!
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| Posted: Wed, 19th May 2010 13:12 Post subject: |
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I just noticed that I couldn't find any analytical solution, or numerical, when taking k into consideration... But it seems that Atropa is right...
"Quantum mechanics is actually, contrary to it's reputation, unbeliveably simple, once you take the physics out."
Scott Aaronson | chiv wrote: | | thats true you know. newton didnt discover gravity. the apple told him about it, and then he killed it. the core was never found. |
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| Posted: Wed, 19th May 2010 14:02 Post subject: |
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spankie
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| Posted: Wed, 19th May 2010 14:09 Post subject: |
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Frant
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| Posted: Wed, 19th May 2010 19:20 Post subject: |
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Hah, wolfram couldn't compute a^n + b^n = c^n.
It did however correctly find X in 2 ^ x + 3 ^ x = 50. It just gives the answer though, not the breakdown.
Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn!
"The sky was the color of a TV tuned to a dead station" - Neuromancer
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TSR69
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| Posted: Wed, 19th May 2010 19:23 Post subject: |
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Fermat lol
Formerly known as iconized
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Frant
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| Posted: Thu, 20th May 2010 21:07 Post subject: |
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| Posted: Thu, 20th May 2010 21:24 Post subject: Re: e^(2x) = k (root x) |
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| tainted4ever wrote: |
First one to answer gets a e-blowjob. Cheers,
-t |
Ahem..
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| Posted: Thu, 20th May 2010 21:27 Post subject: Re: e^(2x) = k (root x) |
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| Atropa wrote: | | tainted4ever wrote: |
First one to answer gets a e-blowjob. Cheers,
-t |
Ahem.. |
what do you think a favour (e-favour) means here ?  |
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