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fafalone

y^x = x^y

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No.

 

To solve for y in terms of x means that if you know x you can find y. In your answer you still have to know x and y to find y, so the problem still remains.

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I worked it down to this..

 

yx+1 - xy - y = 0... maybe a little closer than previous solutions.

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when x=0 and y=0, it doesn't work. all points for x=y except when x and y = 0 work. does that mean that it's a line y=x and y can't be 0?

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Hmm, this reminds me of the time I tried to make a formula to solve ax^3 + bx^2 + cx^1 + dx^0 = 0 in terms of x. It's impossible. You have to use laplace transformations or something like that.

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psi20 said in post # :

Hmm, this reminds me of the time I tried to make a formula to solve ax^3 + bx^2 + cx^1 + dx^0 = 0 in terms of x. It's impossible. You have to use laplace transformations or something like that.

 

Actually, no it isn't.

 

You can calculate the cubic roots, but it IS impossible to do it for quartic or above.

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aommaster said in post # :

sorry guys! I was tired when i wrote that! Of couse 1 to the power of two is a hundred!

You mean 1x102 I take it?

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Can you give me the formula to calculate the cubic root? When I tried to make the formula, I had x = cuberoot( something with x etc etc) so it wasn't really a formula.

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psi20 said in post # :

Can you give me the formula to calculate the cubic root? When I tried to make the formula, I had x = cuberoot( something with x etc etc) so it wasn't really a formula.

 

Looking in the link I posted would be a good start.

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Sayonara³ said

You mean 1x102 I take it?

 

I was joking :)

 

lol, when i tried to do it, i got

 

x to the xth root = y to the yth root

 

I dunno how to write it out tho :(

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Try:

 

y= -xproductLog(-Log(x)/x)/Log(x)

 

and

 

x= -yproductLog(-Log(y)/y)/Log(y)

 

It works for me.

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aommaster said in post # :

 

I was joking :)

 

lol, when i tried to do it, i got

 

x to the xth root = y to the yth root

 

I dunno how to write it out tho :(

 

x1/x = y1/y

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anyway. i did this question for integer solution of a^b=b^a given 0<a<b

 

here it is

 

take a function f where f (x)=ln(x)/x

therefore a^b=b^a if and only if f(a)=f(b). now differentiating f(x) u will find that there is a maximum at x=e. therefore the function f is increasing for x<e and decreasing for x>e. therefore its not possible for both a,b <e or a,b >e cos then f(a) will never be equal to f(b) for a<b

so the only solutions possible is a<e and b>e . now we have 0<a<e. and a is an integer. therefore a=1 or a=2. putting in a=1 we get b=1 which is not a solution as we need a<b. the only other soultion is a=2. puting in a=2 and solving f(b)=f(2) by similar method or however u want we get b =4 . therefore the only solution to a^b=b^a for 0<a<b is a=2,b=4

 

i dont think its possible to find a general solution for x^y=y^x. something do with trascendental numbers and fucntions

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It's like asking why |x| always equals a positive.

 

i dont wanna be picky, but just to correct that statement

 

|x| is non-negative. it doesnt have to be always equal to positive because it can also take the value of 0.

 

many pple confuse positive with non-negative when claiming x^2 is always positive and such.

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it doesnt have to be always equal to positive because it can also take the value of 0

 

Perfecto, as you say people do confuse only positive without the addition of 0.

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I know this topic hasn't bee touched for a couple weeks, but there's something interesting that I found. I'm sure you guys have drawn a graph and noticed this, but just in case...

 

I drew the graph in a trail version I downloaded of a program called Autograph (we use it at my college). If you put the graph of y^x = x^y into it, you get a graph with the line y = x (not including the origin, I would assume) but it has a second curve on it. The curve crosses the points (2, 4) and (4, 2), and interestingly, it crossed the line y = x at (e, e). You probably all know this so I'm not contributing anything (I read this thread before and didn't see any mention but I may have missed it), but that suggests something about the equation of the second curve, if there is a separate equation for it. I'm only even making a bother over this because this was an equation I saw ages ago (from this thread, actually!) and I'm just curious. It's the first thread I really read on these forums. :)

 

I'd post a link to a picture of it, but I can't find any good web picture hosts. :/

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You can attach images to your post and they will be uploaded to our server.

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Wow, thanks. :) Hopefully this worked then:

Notice how the line on the right splits into pixels... not exactly sure what's happening here. :confused:

It almost looks like the second curve has asymptotes at about e/2

yxxyfull.JPG

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