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## Solutions of x^y=y^x

http://planetmath.org/SolutionsOfXyYxThe equation xy=yx (1) has trivial solutions on the line y=x. For other solutions one has the Theorem. 1∘. The only positive solutions of the equation (1) with 1<x<y are in a parametric form x=(1+u)1u,y=(1+u)1u+1 (2) where u>0. 2∘. The only rational solutions of (1) are x=(1+1n)n,y=(1+1n)n+1 (3) where n=1, 2, 3,… 3∘. Consequently, the only integer solution of (1) is 24= 16= 42. Proof. 1∘...

## Topic entry on rational numbers

http://planetmath.org/TopicEntryOnRationalNumbersrational number • fraction Properties • commensurable numbers • rational algebraic integers Forms • partial fractions • any rational number is a sum of unit fractions • Egyptian fractions • mixed fraction • decimal fraction • decimal expansion • order valuation • continued fraction Where are rational numbers? • rational points on one dimensional sphere •...

## Catalan’s conjecture

http://planetmath.org/CatalansConjectureThe successive positive integers 8 and 9 are integer powers of positive integers (23 and 32), with exponents greater than 1. Catalan’s conjecture (1844) said that there are no other such successive positive integers, i.e. that the only integer solution of the Diophantine equation xm-yn=1 with x>1, y>1, m>1, n>1 is x=n=3,y=m=2. It took more than 150 years before the conjecture was proven. Mihailescu gave i...

## Equation xʸ=yˣ

https://en.wikipedia.org/wiki/Equation_xʸ=yˣIn general, exponentiation fails to be commutative. However, the equation x^y=y^x holds in special cases, such as x=2, y=4. The equation x^y=y^x is mentioned in a letter of Bernoulli to Goldbach (29 June 1728). The letter contains a statement that when x\ne y, the only solutions in natural numbers are (2,4) and (4,2), although there are infinitely many solutions in rational numbers. The reply by Goldbach (31 January 1729) contains gener...

## Solve $x^y \, = \, y^x$ - Answer 2

https://math.stackexchange.com/questions/114476
Possible Duplicate:
$x^y = y^x$ for integers $x$ and $y$
I obtained a question asking for how to solve $\large x^y = y^x$. The given restraints was that $x$ and $y$ were both positive integers.
By a bit of error an trial we quickly see that $x=2$ and $y=4$ is one solution.
My question is: How do one show that $(2\,,\,4)$ is the only non trivial, positive solution to the equation?&...

## Find the distribution and and density functions of $Y=\ln(x)$ - Answer 1

https://math.stackexchange.com/questions/205595Suppose $X$ is an uniform random variable on the interval (0,1). Find the distribution and and density functions of $Y=X^n$ for $n\ge2$.
Since $n\ge2$ I say that $-\infty < y < \infty$. Where I am having trouble is solving $P(X^n\le y)$. My first reaction is to take the $nth$ root of both sides, but that really does not seem right. My other thought is to take the $\ln$ of each side. If the latter is correct would it be $...

## Finding $y'$ by implicit differentiation if $x^y=y^x$ - Question

https://math.stackexchange.com/questions/315591I have it down to $$\ln(y)+(x/y)y' = y'\ln(x)+y/x$$ The problem is factoring out $y'$, which leads to either $$y'=\frac{\ln(y)-(y/x)}{\ln(x)-(x/y)}$$ or to $$y'=\frac{(y/x)-\ln(y)}{(x/y)-\ln(x)}$$ Am I missing something?

## Finding $y'$ by implicit differentiation if $x^y=y^x$ - Answer 1

https://math.stackexchange.com/questions/315594I have it down to $$\ln(y)+(x/y)y' = y'\ln(x)+y/x$$ The problem is factoring out $y'$, which leads to either $$y'=\frac{\ln(y)-(y/x)}{\ln(x)-(x/y)}$$ or to $$y'=\frac{(y/x)-\ln(y)}{(x/y)-\ln(x)}$$ Am I missing something?

## Finding $y'$ by implicit differentiation if $x^y=y^x$ - Answer 2

https://math.stackexchange.com/questions/315595I have it down to $$\ln(y)+(x/y)y' = y'\ln(x)+y/x$$ The problem is factoring out $y'$, which leads to either $$y'=\frac{\ln(y)-(y/x)}{\ln(x)-(x/y)}$$ or to $$y'=\frac{(y/x)-\ln(y)}{(x/y)-\ln(x)}$$ Am I missing something?

## Is this an Equivalence Relation and why? - Answer 1

https://math.stackexchange.com/questions/519132if $I$ is a set of positive integers and relation $\def\R{\mathrel R}\R$ is defined over the set $I$ by $x\R y$ iff
$x^y = y^x$.
Is this an Equivalence Relation and why?