i graphed \$x^y=y^x\$ and also it is a union the the heat y=x, v some various other curve. Therefore my an initial question is, how do ns derive that various other curve? My next question is, why don"t I get the very same graph as soon as I plot \$x^1/x=y^1/y\$? \$egingroup\$ If everyone cares, it seems that the two components intersect in ~ the suggest \$(e,e)\$. \$endgroup\$
David ns was talking around the two components on the graph of \$x^y=y^x\$. Are you talking about the two curves or the two components of the first curve? and I used Wolfram. \$endgroup\$
Let \$y=inter-base.netcalRx\$, \$inter-base.netcalR\$ a variable. Then,\$\$x^inter-base.netcalRx=(inter-base.netcalRx)^x\implies inter-base.netcalRxln x=xln (inter-base.netcalRx)=xln inter-base.netcalR+xln x\implies (inter-base.netcalR-1)ln x=ln inter-base.netcalR\$\$We can select \$inter-base.netcalR=1\$, or, if \$inter-base.netcalR>1\$,\$\$x=expleft(dfracln inter-base.netcalRr-1 ight)=inter-base.netcalR^1/(inter-base.netcalR-1)\$\$I.e.,\$\$y=inter-base.netcalR^inter-base.netcalR/(inter-base.netcalR-1)\$\$The non-straight graph is the graph because that \$y\$ because that which \$inter-base.netcalR eq 1\$.

You are watching: (x y) or (y x)

Since \$inter-base.netcalRx eq dfrac1x\$, girlfriend don"t obtain the very same graph.

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edited might 8 "14 at 1:57
answered may 8 "14 at 1:45
user122283user122283
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\$egingroup\$ Isn't that just a line? \$endgroup\$
–user142299
might 8 "14 at 1:48
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NicholasStull \$xinter-base.netcalR=inter-base.netcalR^inter-base.netcalR/(inter-base.netcalR-1)\$. \$endgroup\$
–user122283
might 8 "14 in ~ 1:52
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NotNotLogical The graph i m sorry the OP is talking about is an implicit plot, and the graph is offered by the over equation. \$endgroup\$
–user122283
might 8 "14 at 1:53
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SanathDevalapurkar so that's a single point? Or a line? I'm not clear top top what that is. \$endgroup\$
–user142299
may 8 "14 at 1:54
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Here is the way I an initial saw it,many year ago:

If \$x^y = y^x\$with \$x > 0\$ and \$y > 0\$,let \$r = y/x\$,so \$y = rx\$.We will derive a parameterizationfor \$x\$ and \$y\$in terms of \$r\$.

Then\$x^rx = (rx)^x\$or,taking the \$x^th\$ root,\$x^r = rx\$or\$x^r-1 = r\$or,if \$r e 1\$,\$x = r^1/(r-1)\$and\$y = rx = r^1+1/(r-1)= r^r/(r-1)\$.

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Letting \$r\$ walk throughthe reals \$> 1\$gives allude \$(x, y)\$ with\$x^y = y^x\$.

Another method to look at thisis to think about the curve\$v = u^1/u\$.For each \$v\$ such that\$1  