jpat1023 Posted February 12, 2004 Share Posted February 12, 2004 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. Link to comment Share on other sites More sharing options...
jpat1023 Posted February 12, 2004 Share Posted February 12, 2004 you beat me to it fafolone... Link to comment Share on other sites More sharing options...
fafalone Posted February 12, 2004 Author Share Posted February 12, 2004 I worked it down to this.. yx+1 - xy - y = 0... maybe a little closer than previous solutions. Link to comment Share on other sites More sharing options...
psi20 Posted February 14, 2004 Share Posted February 14, 2004 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? Link to comment Share on other sites More sharing options...
psi20 Posted February 14, 2004 Share Posted February 14, 2004 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. Link to comment Share on other sites More sharing options...
fafalone Posted February 14, 2004 Author Share Posted February 14, 2004 Like I previously said, its no simple algebraic manipulation. Link to comment Share on other sites More sharing options...
JaKiri Posted February 14, 2004 Share Posted February 14, 2004 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. Link to comment Share on other sites More sharing options...
Sayonara Posted February 14, 2004 Share Posted February 14, 2004 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? Link to comment Share on other sites More sharing options...
psi20 Posted February 14, 2004 Share Posted February 14, 2004 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. Link to comment Share on other sites More sharing options...
JaKiri Posted February 14, 2004 Share Posted February 14, 2004 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. Link to comment Share on other sites More sharing options...
psi20 Posted February 14, 2004 Share Posted February 14, 2004 What link? Link to comment Share on other sites More sharing options...
neo_maya Posted February 14, 2004 Share Posted February 14, 2004 psi20 said in post # :What link? This one I think : http://en.wikipedia.org/wiki/Cubic_equation Link to comment Share on other sites More sharing options...
Neurocomp2003 Posted February 14, 2004 Share Posted February 14, 2004 i think the only positive solutions are (x,y) = (t,t) and (x,y) = (2,4)=(4,2) Link to comment Share on other sites More sharing options...
JaKiri Posted February 14, 2004 Share Posted February 14, 2004 Why not READ THE THREAD? Link to comment Share on other sites More sharing options...
aommaster Posted February 15, 2004 Share Posted February 15, 2004 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 Link to comment Share on other sites More sharing options...
fafalone Posted February 15, 2004 Author Share Posted February 15, 2004 hmm... what ever happened to mimetex.. Link to comment Share on other sites More sharing options...
Sayonara Posted February 15, 2004 Share Posted February 15, 2004 It's still on blike.com... Link to comment Share on other sites More sharing options...
wolfson Posted February 17, 2004 Share Posted February 17, 2004 Try: y= -xproductLog(-Log(x)/x)/Log(x) and x= -yproductLog(-Log(y)/y)/Log(y) It works for me. Link to comment Share on other sites More sharing options...
JaKiri Posted February 17, 2004 Share Posted February 17, 2004 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 Link to comment Share on other sites More sharing options...
bloodhound Posted April 12, 2004 Share Posted April 12, 2004 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 1 Link to comment Share on other sites More sharing options...
bloodhound Posted April 16, 2004 Share Posted April 16, 2004 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. Link to comment Share on other sites More sharing options...
wolfson Posted April 16, 2004 Share Posted April 16, 2004 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. Link to comment Share on other sites More sharing options...
J'Dona Posted April 28, 2004 Share Posted April 28, 2004 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. :/ Link to comment Share on other sites More sharing options...
fafalone Posted April 28, 2004 Author Share Posted April 28, 2004 You can attach images to your post and they will be uploaded to our server. Link to comment Share on other sites More sharing options...
J'Dona Posted April 28, 2004 Share Posted April 28, 2004 Wow, thanks. Hopefully this worked then: Notice how the line on the right splits into pixels... not exactly sure what's happening here. It almost looks like the second curve has asymptotes at about e/2 Link to comment Share on other sites More sharing options...
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