X^y=y^x but x will not equal y

x^y=y^x but x will not equal y.

Only 1 solution!

Very easy, if you can't figure this out then you should be classified as borderline intellectually disabled. And don't blame it on well maths is just not important. MATHS is the UNIVERSE!

...

2 and 4

You're borderline intellectually disabled. Sorry to tell you the truth!

shhhh, you're right but don't tell everyone else that, okay??

how would you solve this without guessing

good lord go back to plebbit

Faggot. Making broad generalizations about people like this, you're either a kid or an emotionally-stunted adult. Divest yourself from this way of thinking, because you are just as useless as everyone

You could feasibly write a script to run through it for you.

that's almost like guessing but is there a generalised way to do this

>only one solution
>cartesianfag
>not accounting for imaginary answers
>brainletdetected.png

Yeah, parametric equations make it pretty easy

damn, forgot to write down under real numbers. You got me user you're at an iq around 145, I'd say!

140, I know my worth. Thanks though, math major here. I've been dreaming in non-cartesian planes for months now. Wild time.

I've heard there is even more planes than just i for example they introduce k. Have you explored that realm of numbers yet? I kinda wanna study it but I think my mind will just explode?

or is it J?

Yeah i mean the basics are cartesian, polar, and imaginary but then you get into 3D shit with non-euclidean geometry, and shit gets REAL weird with topology

2 and 4 right?

I know several proofs for real x,y. Here's my favorite.

If a^b = b^a, then a^(1/a) = b^(1/b). Consider the function f(x) = x^(1/x).

Lemma: f(x) has a unique maximum at e.
Proof: Logarithmic differentiation.

Lemma: f(x) has a horizontal asymptote at y=1.
Proof: Take log, use L'Hopital's Rule.

By direct calculation, 2^(1/2)=4^(1/4). Hence there is a solution. However, since f is differentiable on (1,∞) and increasing on (1,e), for n≥5, consider f(n). Then there is a unique c in (1,2) such that f(n) = f(c); however, there are no integers between 1 and 2.

Hence 2^4=4^2 is the only unique natural number solution.

z^a=a^z

a*ln(z)=z*ln(a)

-ln(z)/z=-ln(a)/a

-ln(z)*e^(-ln(z))=-ln(a)/a

ln(1/z)=W[n, -ln(a)/a]

1/z=e^(W[n, -ln(a)/a])

z=-a*W[n, -ln(a)/a]/ln(a)

X*y=x*y

(You)
Oh, I fucked up

>MATHS is the UNIVERSE!
beta

This gets today's fedora award

You can do that, and I can roll cigs. Life is fair I guess

You need to calm down with Rick and Morty user

0 and 1

don't you mean 0 and -0

2^4=4^2 a palendrome

>here is one for ya.
>Using 3/4, 3, 3, and 6 and either division, addition, subtraction, or multiplication, you must get an answer of exactly 24. The order of the numbers does not matter. (so to sum it up, make 24 out of 3/4, 3, 3, 6 using only addition, subtraction, division, or multiplication).

The answers are 2 and 4.

2^4=4^2 (16)