Exchanging Exponentials

Inequalities/All notes14 Sep 2026

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1. The Start

The problem we are discussing here can be formalized with the following proposition.

Proposition 1.1

Given $1<𝑎<𝑏$, ${𝑎}^{𝑏}<{𝑏}^{𝑎}$ if and only if $𝑓\left(𝑎,𝑏\right)>0$.

Here, the map $𝑓\left(𝑎,𝑏\right)$ is our purpose.

It is a natural thought to use exponentials like ${𝑎}^{𝑎}$ or ${𝑏}^{𝑏}$ to solve the problem. However, this is not effective. Here is an example.

Solution

${𝑎}^{𝑎}<{𝑎}^{𝑏}<{𝑏}^{𝑏}$, while ${𝑏}^{𝑎}<{𝑏}^{𝑏}$.

Hence we need some other tools, instead of pure inequalities. Another natural thought to solve the problem, is to use the natural logarithm. Thus the problem is asking when ${𝑒}^{𝑏ln\left(𝑎\right)}<{𝑒}^{𝑎ln\left(𝑏\right)}$ is true, which is in fact simply $𝑏ln\left(𝑎\right)<𝑎ln\left(𝑏\right)$. It is a simple matter to see that this is equivalent to $\frac{𝑎}{𝑏}<\frac{ln\left(𝑎\right)}{ln\left(𝑏\right)}$. However, you can’t see how can this solve the problem. A traditional way is to rewrite the inequality like the following:

$$\frac{ln\left(𝑎\right)}{𝑎}<\frac{ln\left(𝑏\right)}{𝑏}.$$

The next step is to let $𝑓\left(𝑥\right)≔\frac{ln\left(𝑥\right)}{𝑥}$, and study the property of this function on the interval $\left(1,+inf\right)$. This, as you can see, is pretty boring. After drawing many pictures, which filled half of my crafts, I discovered the following, which is much more interesting.

2. The Climax

We have already shown that the original problem is equivalent to find when the equality $\frac{𝑎}{𝑏}<\frac{ln\left(𝑎\right)}{ln\left(𝑏\right)}$ holds. Let’s draw a picture of $𝑥\mapsto 𝑥$ and $𝑥\mapsto ln𝑥$ on a same figure. Hence we are actually comparing the the slope of the line passing through $\left(𝑎,ln\left(𝑎\right)\right)$ and $\left(𝑏,ln\left(𝑏\right)\right)$. Since these are lines, we may let the one through the points $\left(𝑎,ln\left(𝑎\right)\right)$ and $\left(𝑏,ln\left(𝑏\right)\right)$ be $𝑦=𝑚\ast 𝑥+𝑛$. By Karamata’s inequality, we know that the original inequality holds if and only if $𝑛<0$.

3. The End

The following theorem is the final result of this note.

Theorem 3.1

Given $1<𝑎<𝑏$, ${𝑎}^{𝑏}<{𝑏}^{𝑎}$ if and only if $𝑎<{𝑒}^{\frac{ln\left(𝑘\right)}{𝑘-1}}$, where $𝑘=\frac{𝑏}{𝑎}$ And the inequality never holds if $𝑎\ge 𝑒$.

Proof

First proof that ${𝑎}^{𝑏}<{𝑏}^{𝑎}\Rightarrow 𝑎<{𝑒}^{\frac{ln\left(𝑘\right)}{𝑘-1}}$. Since ${𝑎}^{𝑏}<{𝑏}^{𝑎}$, we obtain ${𝑒}^{𝑏ln\left(𝑎\right)}<{𝑒}^{𝑎ln\left(𝑏\right)}$. By the results above, we have

$$\frac{𝑎}{𝑏}>\frac{ln\left(𝑎\right)}{ln\left(𝑏\right)}.$$

Set points $𝐴\left(𝑎,ln\left(𝑎\right)\right)$ and $𝐵\left(𝑏,ln\left(𝑏\right)\right)$ in a plane. Let the line through $𝐴$ and $𝐵$ be $𝑦=𝑚𝑥+𝑛$. It is clear that $𝐴\left(𝑎,𝑚𝑎+𝑛\right)$, $𝐵\left(𝑏,𝑚𝑏+𝑛\right)$. Thus, by Karamata’s inequality, the original inequality holds if and only if $𝑛<0$. So we only need to prove that $𝑛<0$ if and only if $𝑎<{𝑒}^{\frac{ln\left(𝑘\right)}{𝑘-1}}$. Substitute the position of $𝐴$, $𝐵$ into $\text{AB}:𝑦=𝑚𝑥+𝑛$, we obtain

$$𝑚=ln\left(𝑎\right)-\frac{ln\left(\frac{𝑏}{𝑎}\right)}{\frac{𝑏}{𝑎}-1}.$$

Let $𝑘=\frac{𝑏}{𝑎}$, then $𝑚=ln\left(𝑎\right)-\frac{ln\left(𝑘\right)}{𝑘-1}$. Thus, $𝑎<{𝑒}^{\frac{ln\left(𝑘\right)}{𝑘-1}}$.

Finally, we prove that the inequality never holds if $𝑎\ge 𝑒$. It is obvious that $ln\left(𝑘\right)\le 𝑘-1$. Clearly, since $𝑏>𝑎$, $𝑘>1$. Thus we obtain $\frac{ln\left(𝑘\right)}{𝑘-1}\le 1$. Hence, unless $𝑎<𝑒$, otherwise the inequality is false, which is equivalent to our goal.⁠∎

View Typst source

#import "../shared.typ": *
#import "@preview/noteworthy:0.4.0": *
#import "/themes/site/notes.typ": *

#show: tylenotes.with(
  title: "Exchanging Exponentials",
  date: "2026-09-14",
  tags: ("inequalities"),
  summary: "When exponentials get larger by exchanging the two parameters",
)

= The Start

The problem we are discussing here can be formalized with the following proposition.
#proposition[
  Given $1 < a < b$, $a^b < b^a$ if and only if $f(a,b)>0$.
]
Here, the map $f(a,b)$ is our purpose.

It is a natural thought to use exponentials like $a^a$ or $b^b$ to solve the problem.
However, this is not effective. Here is an example.
#solution[
  $a ^ a < a ^ b < b ^ b$, while $b ^ a < b ^ b$.
]
Hence we need some other tools, instead of pure inequalities.
Another natural thought to solve the problem, is to use the natural logarithm.
Thus the problem is asking when $e^(b ln(a)) < e ^ (a ln(b))$ is true,
which is in fact simply $b ln(a) < a ln(b)$.
It is a simple matter to see that this is equivalent to $a / b < ln(a) / ln(b)$.
However, you can't see how can this solve the problem.
A traditional way is to rewrite the inequality like the following:
$ ln(a) / a < ln(b) / b. $
The next step is to let $f(x) := ln(x) / x$,
and study the property of this function on the interval $(1, +inf)$.
This, as you can see, is pretty boring. After drawing many pictures,
which filled half of my crafts, I discovered the following, which is much more interesting.

= The Climax

We have already shown that the original problem is equivalent to find
when the equality $a / b < ln(a) / ln(b)$ holds.
Let's draw a picture of $x |-> x$ and $x |-> ln x$ on a same figure.
Hence we are actually comparing the the slope of the line passing through $(a, ln(a))$ and $(b, ln(b))$.
Since these are lines, we may let the one through the points $(a, ln(a))$ and $(b, ln(b))$
be $y = m * x + n$. By Karamata's inequality, we know that the original
inequality holds if and only if $n < 0$.

= The End

The following theorem is the final result of this note.

#theorem[
  Given $1 < a < b$, $a^b < b^a$ if and only if $a < e ^ ((ln(k))/(k-1))$, where $k=b/a$
  And the inequality never holds if $a >= e$.
]

#proof[
  First proof that $a^b < b^a => a < e^(ln(k) / (k-1))$.
  Since $a^b < b^a$, we obtain $e^(b ln(a)) < e^(a ln(b))$.
  By the results above, we have $ a/b > ln(a) / ln(b). $
  Set points $A(a, ln(a))$ and $B(b, ln(b))$ in a plane.
  Let the line through $A$ and $B$ be $y = m x + n$.
  It is clear that $A(a, m a + n)$, $B(b, m b + n)$.
  Thus, by Karamata's inequality, the original inequality holds if and only if $n<0$.
  So we only need to prove that $n < 0$ if and only if $a < e^(ln(k)/(k-1))$.
  Substitute the position of $A$, $B$ into $"AB": y = m x + n$, we obtain
  $ m = ln(a) - ln(b/a)/(b/a - 1). $
  Let $k = b/a$, then $m = ln(a) - ln(k)/(k-1)$. Thus, $a < e^(ln(k)/(k-1))$.

  Finally, we prove that the inequality never holds if $a >= e$.
  It is obvious that $ln(k) <= k-1$. Clearly, since $b > a$, $k > 1$.
  Thus we obtain $ln(k)/(k-1) <= 1$.
  Hence, unless $a < e$, otherwise the inequality is false,
  which is equivalent to our goal.
]