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Let $x,y$ be positive real numbers with $1<x<y$ and $x^y = y^x$. Prove ${\bf carefully}$ that $x<e<y$.

My attempt to the solution.

Notice that if $x^y = y^x$ we can write it as $\dfrac{\ln x }{x} = \dfrac{ \ln y }{y} $ because $x,y$ are both more than 1.We know that the function $f(t) = \dfrac{ \ln t }{t}$ is maximal at $t = e$. Thus, $\dfrac{ \ln y}{y} \leq \dfrac{ \ln e }{e} = \dfrac{1}{e} $ and so

$$y \geq e \ln y $$

Now, if $e \geq y$, then $\ln y \leq 1$. I can't seem to see a contradiction from here. Any hint??

Next, Notice that $\dfrac{\ln x }{x} < \dfrac{1}{e}$ then clearly $\boxed{x<e}$. But, I am still stuck on showing $y>e$.

Community
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James
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3 Answers3

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Hint Show that $f(t)=\frac{\ln(t)}{t}$ is strictly increasing on $(0,e)$ and strictly decreasing on $(e, \infty)$.

Now, if you assume by contradiction that $y<e$ then $0<x<y<e$ and $f(x)=f(y)$ which is NOT possible. Same way $x>e$ is not possible.

N. S.
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We know that the function $f(t)=\ln t/t$ is maximal at $t=e$. Since $f$ is strictly monotone increase (decrease) at $(0,e)$ ($(e,+\infty)$), $x<e<y$.

Jingeon An-Lacroix
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Let $f(t)=\frac {\ln t} t$. Firstly since $f$ is elementary and does not contain isolated points in its domain, it is continuous and differentiable over its domain. $\frac {df}{dt}=\frac {\frac 1 t t - \ln t}{t^2}=\frac {1-\ln t}{t^2}$. Notice that $\frac {df}{dt}$ has one and only one root over the domain of $f$, and is $t=e$. Therefore, since $f$ is continuous over its domain, $f$ is strictly monotone on interval $(0,e)$ and $(e,+\infty)$. Since $f(x)=f(y)$, they can't be both in $(0,e)$ or $(e,+\infty)$. Since $x \lt y$, $x \lt e \lt y$.

user12986714
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