Integer solutions for $x^y=y^x$ Let's say we have the equation: $$x^y=y^x$$ what are the solutions when $x,y \in \Bbb Z$ for ${x}\ne{y}$..   The only one I have found is 2 and 4. There are infinite non-integer solutions but I am interested in the integer ones. Also you can find a new definition for $e$ from this equation $e=x^{\frac{y}{xln(y)}}$ and also $log_x(y)=\frac{y}{x}$
$x^y = y^x$ for integers $x$ and $y$ We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?
Calculating Image boundary / footprint of satellite images using open source tools? How do you change the size of figure drawn with Matplotlib? Today while troubleshooting a service broker problem, I discovered that the database owner was the Windows login of an employee who had left the company. His login had been removed and thus the query notifications were failing.   Supposedly the best practice for dealing with this is to make 'sa' the database owner. We changed it and that cleared out the queue.   My (very elementary) question: what is the database owner and what is its purpose? Additive function $T: \mathbb{R} \rightarrow \mathbb{R}$ that is not linear. A function $T:V \rightarrow W$ is additive if $T(x+y) = T(x) + T(y)$ for every $x, y \in V$. Prove that there exists an additive function $T: \mathbb{R} \rightarrow \mathbb{R}$ that is not linear.  My attempt: Let $T$ be the function $T: \mathbb{R}$ (over the field $\mathbb{Q}$) $\rightarrow \mathbb{R}$ (over the field $\mathbb{R}$). The set $\{1, \sqrt{2}\} \subseteq \mathbb{R}$ is linearly independent for the vector space $\mathbb{R}$ over $\mathbb{Q}$. Then, there must exist a linearly independent set $W \subseteq \mathbb{R}$ (over $\mathbb{Q}$) such that $\{1, \sqrt{2}\} \subseteq W \subseteq \text{span}(W) = \mathbb{R}$ (over $\mathbb{Q}$).  I have been told that the function defined as $T(1) = 1$ and $T(w) = 0$ for all $w \in \text{span}(W) \setminus \{1\}$ is additive but not linear, but I cannot see why this is? I can see why it is not linear, clearly $T(\sqrt{2} \cdot 1) = 0$ but $\sqrt{2} T(1) = \sqrt{2}$. But, why is $T$ additive? For example, $T(1+1) = T(2) =0$ but $T(1) + T(1) = 1+1 = 2$? Is there a mistake somewhere? PHP regular expressions: No ending delimiter '^' found in My options to link large image into AI file My .psd file has a resolution of 6m x 3m @ 150 DPI. This makes it 35433px x 5906px in size. Photoshop CS6 saves the file as .tiff or .psb only. When I first import into Illustrator I get an error that the file is not readable but I have tried many options. Last one is not supported by AI.  So I'm puzzled... is there a way to link large images into Illustrator? I wish to familiarize myself with the capabilities of QGIS up to and beyond QGIS version 3.0.   Can you recommend some good web resources and tutorials that would show me QGIS' capabilities?   I am interested in both using QGIS through its GUI and in developing applications using QGIS. There are a plethora of words for user accounts, like logon, login, signon, and also the action of logging in (or logging on) or signing in.  Are there any usage guidelines here? is "I'll" correct as a short answer? .NET String.Format() to add commas in thousands place for a number pgfplots: lines and markers in different colors With pgfplots, I am creating a simple coordinate plot with error bars like this:    \documentclass[crop,10pt]{standalone} \usepackage[english]{babel} \usepackage{pgfplots}      \begin{document}%         \setlength{\linewidth}{246pt}%              \begin{tikzpicture}             \begin{axis}[             width=\linewidth,             ytick pos = left,             xtick pos = left,             clip=false,             axis on top,             tick align = outside,             yticklabel style={%             /pgf/number format/fixed,             /pgf/number format/precision=1,             /pgf/number format/fixed zerofill         },             ytick scale label code/.code={},             xlabel={X},             ylabel={Y},     ]             \addplot[mark=*]             plot[             error bars/.cd,             y dir=both,             y fixed=0.0001     ]             coordinates{(1,0.0005) (1.1,0.0013) (5,0.0015)};             \end{axis}             \end{tikzpicture}     \end{document}   In the original plot I have a lot more points. I would now like the line and the markers (including the error bars) to have different colors to better distinguish the line from the error bars. Is that somehow possible? The only way I found so far is to plot the data just as a line in one color and then plot it again just as markers in another color. This works, but I guess there must be some better solution to this. Thanks for the support, John I'm an Egyptian citizen, who holds Schengen visa and stay permit in Italy. Can I go to UK for some days with my stay permit and Schengen visa? Is it possible to change your spawn point? Is it possible to change the location you beam to on a planet? I've been to a couple of planets that have no usable resources near the spawn point and everything is a good distance East or West. It would be nice to be able to always beam down to there instead. How are Pokestops chosen? After playing the game for a little while, I've noticed that Pokestops are chosen seemingly at random locations on the map. A few of them are even in inaccessible places; an abandoned church building, inside of an airforce base, etc. How are these locations chosen as Pokestops? Using InDesign CS6, how can I simultaneously edit A4 and A5 versions of a poster? I'm on Adobe Creative Cloud, so I have the latest InDesign, and I'm attempting to design a poster for my choir's next concert. Traditionally, I've done this in Photoshop, in two separate files, however I've recently started using InDesign after converting to it to create wedding albums through Blurb.  I've created an A4 document, and created an alternate-layout for it in A5. I've then lined them up side-by-side, and I've added a black frame to the A4 page, to act as the black background for my page (by the way, feel free to advise on better ways to do this).  My assumption is that I've failed to understand how this works, as I was hoping that I'd magically get a black frame on the A5 version, and that I could use liquid layout to get this to automatically scale (along with all other elements), in order to create the two versions simultaneously.  All help/advice would be greatly appreciated! How to calculate $\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$ I came across this strange limit whilst showing convergence of a series: $$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+3^n}{3^n+4^n}}$$ How can I calculate this limit? Start Mathematica without the menu bar? $A=\begin{pmatrix}0&amp;0&amp;1\\ 0&amp;1&amp;0\\ 1&amp;0&amp;0\end{pmatrix}$  How do I show that there are no solutions for $X^2=A$ in $M_3(R)$?  I tried $X^2=\begin{pmatrix}a_{11}&amp;a_{12}&amp;a_{13}\\ a_{21}&amp;a_{22}&amp;a_{23}\\ a_{31}&amp;a_{32}&amp;a_{33}\end{pmatrix}^2=A$, but I couldn't seem to find anything that shows there are no solutions. How should I approach this? Particular solution of $y''-4y'+5y = 4e^{2x} (\sin x)$ How do I find the particular solution of this second order inhomogenous differential equation? (Using undetermined coefficients).  $y''-4y'+5y = 4e^{2x} (\sin x)$  I can find the generel homogenous solutions but I need help for the particular. Motivated by , can one prove that the order of an element in a finite group divides the order of the group without using Lagrange's theorem? (Or, equivalently, that the order of the group is an exponent for every element in the group?)  The simplest proof I can think of uses the coset proof of Lagrange's theorem in disguise and goes like this: take $a \in G$ and consider the map $f\colon G \to G$ given by $f(x)=ax$. Consider now the orbits of $f$, that is, the sets $\mathcal{O}(x)=\{ x, f(x), f(f(x)), \dots \}$. Now all orbits have the same number of elements and $|\mathcal{O}(e)| = o(a)$. Hence $o(a)$ divides $|G|$.  This proof has perhaps some pedagogical value in introductory courses because it can be generalized in a natural way to non-cyclic subgroups by introducing cosets, leading to the canonical proof of Lagrange's theorem.   Has anyone seen a different approach to this result that avoids using Lagrange's theorem? Or is Lagrange's theorem really the most basic result in finite group theory?