$$ { ${x^y=y^x}$ } $$
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$$x^y = y^x$$

Logarithmic differentiation to find dy/dx | Physics Forums

Hello, I know how I need to do the problem, I have figured out an answer as best as I could but I have a feeling the answer is not correct. In fact I'm pretty sure it's not. Anyway, I'm supposed to use the logarithmic differentiation to find dy/dx for the equation: x^y = y^x I know I need...
$$x^y = y^x$$

$x^y = y^x$ for integers $x$ and $y$

Say x^y = y^x , and x > y > 0 . Taking logs, y \log x = x \log y ; rearranging, (\log x)/x = (\log y)/y . Let f(x) = (\log x)/x ; then this is f(x) = f(y) . Now, f^\prime(x) = (1-\log x)/x^2 , so f is increasing for x<e and decreasing for x>e . So if x^y = y^x has a solution, then x > e > y . So y must be 1 or 2 . But y = 1 doesn't work. y=2 gives x=4 . (I've always thought of this as the ``sta...

Reference desk/Archives/Mathematics/2012 December 5

= December 5 = Define f(n) := \sum_{k=1}^n \ln (n/k). Show that f(n) = O(n). --AnalysisAlgebra (talk) 01:03, 5 December 2012 (UTC) :See Stirling's approximation. Sławomir Biały (talk) 01:09, 5 December 2012 (UTC) Any suggestions? HOOTmag (talk) 10:01, 5 December 2012 (UTC) :Loads like {2,2,3}={4,4,1} Dmcq (talk) 10:50, 5 December 2012 (UTC) :: Are you sure the two sets are equal...? --CiaPan (talk) 10:58, 5 December 2012 (UTC) :1^{n^0} ...

Is $(0,0)$ a solution to $x^y-y^x=0$?

@StefanSmith The question stemmed from a discussion of whether the graph of x^y=y^x contains the origin. Knowing the p...
$$x^y = y^x$$

$x^y = y^x$ for integers $x$ and $y$

This is a classic (and well known problem). The general solution of x^y = y^x is given by \begin{align*}x &= (1+1/u)^u \\ y &= (1+1/u)^{u+1}\end{align*} It can be shown that if x and y are rational, then u must be an integer. For more details, see this and this....

For 17 integers: $a_{i} (i=1,2,\dotsm,17)$, $a_{1}^{a_{2}}=a_{2}^{a_{3}}=a_{3}^{a_{4}}=\dotsm=a_{16}^{\quad a_{17}}=a_{17}^{a_{1}}$.

Hint If x^y=y^x then \frac{\ln(x)}{x} = \frac{\ln(y)}{y} Now, f(x)=\frac{\ln(x)}{x} is increasing on (0,e) and decreasing on (e , \infty) . Deduce from here that x^y=y^x implies that exactly one of the following happens x=y x<e<y y <e<x . Now, assume by contradiction that a_1 \neq a_2 . Show first that a_{j} \neq a_{j+1} for all 1 \leq j \leq 16 . Next, deduce that if a_j <e then a_{j+1}>e ...

$x^{y}=y^{x}$ Intersection Question

I'm looking at this problem, x^{y}=y^{x} and the question is to find the point of intersection of the two curves that form a solution set. I understand how to find the point of intersection using a different method, but I don't understand the following solution. Why is it when "f(x)" is at a minimum or maximum, we have intersection of the two curves? http://math.berkeley.edu/sites/default/files/pages/solutions_1.pdf...

Commutativity of powers in $\Bbb Q$.

The equation x^y=y^x for x< y has just one solution in \Bbb Z ( x=2,y=4 ), and infinitely many in \Bbb R , but what happens with \Bbb Q ? And with \overline{\Bbb Q} ? I have visited Gelfond-Schneider theorem but it doesn't seem help here. My try: Let (a,b)=(p,q)=1 where (x,y) denotes the gcd of x and y . If \left(\frac ab\right)^{p/q}=\left(\frac pq\right)^{a/b} then \left(\frac ab\right)^{bp}=\left(\frac pq\rig...
$$x^y = y^x$$

Find the all possible real solutions of $x^y=y^x$

Applying the logarithm to both sides gives y \log x = x \log y, and rearranging gives \frac{\log x}{x} = \frac{\log y}{y} , so we get a nontrivial solution (i.e., one for which x \neq y ) for any value assumed twice by f(x) := \frac{\log x}{x} . More explicitly, if x and y are distinct values such that f(x) = f(y) , then by construction we have a nontrivial solution x^y = y^x . The function f is obviously continuous, an...
$$x^y = y^x$$

How do I solve for $Y$ with $Y = X^{X^Y}$ If the answer set is more than just $(x,y)=\{(0,0), (1,1)\},$ is it graphable?

I remember this question from 40 years ago, and I never found a solution on my own. If Y = X ^ {X ^ Y} : When I attempt to reduce this, it always involves both an X and Y on both LHS and RHS of the equation. I'm assuming that either 1. This is done via complex numbers or 2. This cannot be easily solved (it reminds me of a discussion of x^y=y^x )....
$$x^y = y^x$$

Is it possible to parametrise $x^{\frac{1}{x-1}}=y^{\frac{1}{y-1}}$?

A parameterization for y^\frac{1}{y}=x^\frac{1}{x} is well-known (and not original by me). Write it as x^y = y^x and let y = ry . Then x^{rx}=(rx)^x . Taking x -th roots, x^r = rx . Dividing by x , x^{r-1} - r or x = r^{\frac1{r-1}} and y = rx = rr^{\frac1{r-1}} = r^{1+\frac1{r-1}} = r^{\frac{r}{r-1}} . You might this in your other equation and see where it gets you....

Differentiating $x^y=y^x$

I assume y is a function of x given *implicitly* by x^y=y^x , and you want to find \frac{dy}{dx} ?...

Solution of $x^y=y^x$ and $x^2=y^3$

Solve the given set of equations: x^y=y^x and x^2=y^3 where x,y \in \mathbb{R} Would any other solution exist other that x=y=1 because I think x^2=y^3 will only be true for x=y=1 or x=y=0 ...
$$x^y = y^x$$

Why didn't they simplify $x^y=y^x$ to $x=y$?

Solving x^y = y^x analytically in terms of the Lambert W function This "solution" for x^y=y^x should simplify to y=x , but for some reason no pointed that out in the OP. According to the stack exchange, the answer is y= \frac{-xW(-\frac{ln(x)}{x})}{ln(x)} . However, the term \frac{-ln(x)}{x} itself can be rewritten as \frac{-ln(x)}{x}=-ln(x)e^{-ln(x)} Therefore, the productlog of that expression should simplify as follows...
$$x^y = y^x$$

Find slope at tangent line.

@CarlosV Since you just posted about differentiating x^y = y^x I was under the assumption that this was your context, ...

Self intersection of the implicit curve $x^y-y^x=0$

Watching the graph of the curve defined by x^y=y^x , which contains the line y=x , I noticed that the line intersects the curve itself only at one point that looks to be (e;\;e) . How can I formally prove this?...
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