Someone explain this to me.

thats just how exponential functions work famalam, e shows up everywhere

>>8551198
nice non-explanation

>>8551176
take the log base x of both sides, then apply the change of base formula to rewrite your equation in terms of natural logarithms and see what you get.

>>8551176
I wish I could find an equation to that weird curve there but unfortunately I can't.

Anyways, there is a simple explanation.

Instead of working with two functions lets deal with constants. Set x=2 and get
2^y = y^2

There is the trivial solution y=2 but notice that y=4 also works. Do the same analysis now with e.

e^x = x^e

The only solution there is x=e so there is no magical second solution which, if you notice, is what defines that second curve.

>>8551255
>The only solution there is x=e so there is no magical second solution
But whyyyyyy

>>8551255
>I wish I could find an equation to that weird curve there but unfortunately I can't.

Do you mean Lambert function?

>>8551255
But why isn't there. It's been proven that (2,4) and (4,2) are the only whole number solutions if x != y. But why is it that every single positive number has two other positive numbers that when raised to eachother's power will equal the first number. Why is e the exception?

>>8551176
x^y = y^x is the same as e ^ y ln(x) = e ^ x ln(y)

>>8551176
There are 2 functions on the left and they are represented on a graph on the right. Need it simpler? It's an image.

I think it's because

y^x = x^y; y = e

ln[e^x = x^e]

x = ln(x^e)

if x = e

e = ln(e^e)

e = e

>>8551261
Because the function fundamentally denotes a rate of change for the system.

When you multiply 2 by itself, it grows by a factor of 2. Do that 4 times, you get 16. Do the opposite for 4; it grows by a factor of 4 with each new iteration, except that you only get to the 2nd power before reaching 16.

The Euler constant e is the one number whose rate of change as it grows is the same as the number itself. That's its defining qualitative characteristic.

>>8551176
You'll notice that the intersection is also at x=e. The basic reason why is because the slope of e^x at any x value is always e^x which is a unique quality. You can figure it out from there.

>>8551838
wow that's nuts

>>8552328
have you not taken calculus 1?

if not, do it :^)

>>8551176
>e^e=e^e
Wow no way.
What exactly do you need explained?

>>8551255
(x^y-y^x)/(x-y)=0

Another cool thing is that the maximum of the function x^(1/x) is when x is at e

>>8552328
https://youtu.be/AuA2EAgAegE?t=7m10s

>>8551176
You're question is sort of vague, but I've been having some fun with this, so I'll give it a go.
I assume you're looking for (x,y) solutions to the equation y^x = x^y, and are curious as to why there are two solutions for a given x (or y) unless x (or y) is e, and that you're also curious as to how this e arises.

We have
x^y = y^x
y/x = ln(y)/ln(x)
Some of the solutions are obvious - namely, when y = x. Indeed, in your graph, we see that line drawn. So let's consider when y/x != 1; let's also impose that y and x are both strictly positive.
Then y/x = C > 0, and we obtain the equations
y/x = C
ln(y)/ln(x) = log_x(y) =C
or, with some manipulation,
y = Cx
y = x^C
Plugging in, we have
x^C = Cx
x^(C-1) = C
x = C^(1/(C-1))
We consider when C != 1; indeed, we've already considered when C = 1.
All of our remaining solutions are of the form
(C^(1/(C-1)), C^(C/(C-1))).
How neat! It is when C approaches 1 that our pair approaches (e,e); can you show why?

I still can't figure out what the other red line is though, even though I've just found all of the points on it. I think it can be nicely modeled as x^f(x) for some function f(x) where f(2) = 2, f(e) = 1, f(4) = 1/2, lim_(z->inf) f(z) = 0, and lim_(z->1) f(z) = inf.

>>8551176
It's a graph
You're welcome

>>8553985
neat

File: s.jpg (336KB, 1360x616px) Image search: [Google]

336KB, 1360x616px

I broked it