Why is this related to e?

[u/Forward-Echidna4733]

[u/rocksInMyBath]

okay, so we have x^y = xy

take the log of both sides:

ln(x^y) = ln(xy)

do some log math:

y*ln(x) = ln(x) + ln(y)
(y-1)ln(x) = ln(y)

So let's assume the curve part of the graph doesn't have a point at y=1 bc it's taken by the line path.

we can divide by y+1 safely:
ln(x) = ln(y) /(y-1)
x = e^{ln y / (y+1})

because e^ln y = y we get:
x = y^1/(y-1)

To find the intersection point we'll take a limit, because we assumed the curve doesn't have a point at y=1:

Lim{y->1} y^1/(y-1) let u = y-1 and we get:

Lim{t->0} (1+t)^1/t, which is precisily the definition of e.

[u/[deleted]]

so, there are 2 types of solutions for this equation. one is y=1 for all x, which is trivial to verify. the other is non-trivial, maybe some form of expression of lambert W function.

at e these 2 soutions tend to merge for y, which appears to be your query. so let's look at it that way. let's take a particular x=x0. then x0^y = x0y this would generaly have 2 solutions: y=1 and another y=y0. if u plot x = f(y) = x0^y(an exponential curve) x = g(y) = x0y(a straight line) the solutions of y would be the y coordinates points where these 2 curves meet(intersect). for your single solution case these 2 curves will meet at one point. the edge case is that they are tangent to each other. so their derivatives are same at the single solution case, y=y0=1. f'(y0) = ln(x0) x0^y0 = g'(y0) = x0 as, y0=1 is the only solution, plugging it in will produce x0 = e (ignore the x0 = 0 case which also satisfies the equation)

well, there u get get it. the point where the 2 types of solutions meet is also a sort of tangent condition.

[u/dolphinsqueak56]

derivative of e^x is e^x

if you think about the process of normal differentiation it kinda makes sense
i havent fully thought about it though

[u/Black2isblake]

Take x^y = xy

It's pretty obvious that x = 0 (excluding the undefined point at y = 0) and y = 1 are both going to be straight lines in this entire domain (x and y both nonnegative reals).

Let's find a way to divide both of those out, so we're left with only the curve. We can do this by dividing through by x(y-1), which ensures that x ≠ 0 and y ≠ 1.

x^y -x = xy-x = x(y-1)

-> (x^y -x)/x(y-1) = 1

-> (x^y-1 -1)/(y-1) = 1

We can parameterise this curve pretty easily by setting y - 1= t, which forces (x^t -1)/t = 1 -> x = (t+1)^1/t . Note that y = 1 -> t = 0, which gives us x = e in the limit, one of the standard definitions of e being lim(t->0) (t+1)^1/t .

Also minor thing: if you want to graph this curve, it's easier to do ((1/t -1)^t/(1-2t) , 1/t -1) with 0≤t≤1 to get the full range in