Putting y in function of x on $x^y = y^x$ using logarithmics

Question

I was trying to depict y in function of x:

• $x^y = y^x$
• $log x^y = log y^x$

• $y*(log x) =x*(log y)$

• $(log x)/x = (log y)/y$

I got stuck on this part, any way to continue past it? What is the simplest form of this equation?

Answer 1

Your equation has two solutions. First of all the trivial result: $y = x$. We can find the other, more interesting, solution by postulating the relation: $y = px$, where $p$ is a parameter. Substitution of this expression in your original equation yields:

$$x^{(px)} = (px)^x = p^x x^x$$

Dividing both sides of the equation by $x^x$ leads to: $$x^{(p-1)x} = p^x$$

From which it follows that $$x^{(p-1)} = p$$

Consequently $$x = p^{1/(p-1)}$$

So in terms of the parameter $p$ the solution is $$(x,y) = (p^{1/(p-1)},p^{p/(p-1)})$$

The domain and range of this curve are the real numbers from $1$ to infinity. The shape is hyperbolic. The curve intersects the line $y = x$ in the point $(e, e)$, corresponding to $p$ approaching the limiting value $1$. There are two integer solutions: $(2,4)$ and $(4,2)$.