[MATH] a little harder one

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jasaiyajin

jasaiyajin

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x = e^((4+3i)x)
y = 12e^((-5+4i)x)

What is xy?

Can anyone solve is step by step?
i = imaginary number ^_^

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  • Sep 24, 2005
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one question, x isn't a whole # or a real #, right? if yes, then I'm not gonna bother to do it, I have no clue.

edit: it can't be, I can't think of a way to cancel out the i. dude, you're making this too difficult, ...no...maybe I'm just no smart enough to figure this out. x.x

  • Sep 24, 2005

jasaiyajin

jasaiyajin

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lol, its not an easy one...

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  • Sep 24, 2005

j0n0

j0n0

Increadibly Cute

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This isn't that hard. The answer is xy = jjasdkfjljvviruskjhvisrus.jkdlfuckjfdsljdf

An eye for an eye brings justice, but it is compassion that changes a man.

Another point of view doesn't necessarily make yours more or less right.

  • Sep 24, 2005

tevi

tevi

Misa's evil twin

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xy =
12e^[((4+3i)x) + ((-5+4i)x)]
12e^[x(4+3i-5+4i)]
12e^[x(-1+7i)]

*blank* Tell me if I'm right?
Is the answer further than this? ~_~

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- same person; no difference at all, just a different sex

  • Sep 25, 2005
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^ can you leave a X on the left side of the equation???

I thought xy has to equal certain #... if not, then is xy = 12e^[2x^2(-32+i)]??


v that's because I thought you can

  • Sep 25, 2005

tevi

tevi

Misa's evil twin

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Err, you've still got an x on both sides though... ~_~

And I don't know... I can't get rid of the e^x...

<edit> Well, err, I assumed it's not right to still have the x... Which is why I said "I can't get rid of the e^x"

Err, how did you get 2x? When you multiply e's, you *add* the indices...

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- same person; no difference at all, just a different sex

  • Sep 25, 2005
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^ try natural logs...I think ^_^'

  • Sep 25, 2005

leosama84

leosama84

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Quote by j0n0This isn't that hard. The answer is xy = jjasdkfjljvviruskjhvisrus.jkdlfuckjfdsljdf

LOOOOOOOOOOOOOL

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hey jasaiyajin, my calculator told me the answer to xy is 9.974058285E-11-1.335133474E-10i, now will you please reveal the steps now??! lol

okay, here's my third/forth whatever try (after tevi's)
xy= 12e^[(4X+3Xi)+(-5X+4Xi)
xy= 12e^(-X+7Xi)
xy= 12e^ (-25+175i)
xy= 9.974058285E-11-1.335133474E-10i

tevi: "Err, how did you get 2x? When you multiply e's, you *add* the indices..."
lol, yeah, I wasn't thinking, I forgot how to do these. I had to set myself an example in order to think that's the way to do it, lol oh lord....

  • Sep 25, 2005

jasaiyajin

jasaiyajin

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tavi got it... lol

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  • Sep 25, 2005
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except...x is on both sides...making it incorrect...or is it just fine for here?

  • Sep 26, 2005

j0n0

j0n0

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no taffystar, you still can define a variable that is dependant on itself. These kind of equations or typically called differeantial equations or trancendental equations (two are not the same). The operations that were done to get xy aren't illigal in math, it was just basic algebra manipulations. The only thing happens was the x = e^((4+3i)x),- a trancendental equation- made the xy also a trancendental equation.

An eye for an eye brings justice, but it is compassion that changes a man.

Another point of view doesn't necessarily make yours more or less right.

  • Sep 30, 2005
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meh, my teacher always told us to get all the x's on one side, even with differential functions and definitely with transcendental functions. Maybe not incorrect, persay, maybe...incomplete

  • Oct 06, 2005

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