Exchanging Exponentials

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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 $𝑓 (𝑎, 𝑏) > 0$ .

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

It is a natural thought to use exponentials like $𝑎 𝑎$ $𝑎 𝑎$ or $𝑏 𝑏$ $𝑏 𝑏$ 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 $𝑒 𝑏 l n (𝑎) < 𝑒 𝑎 l n (𝑏)$ is true, which is in fact simply $𝑏 l n (𝑎) < 𝑎 l n (𝑏)$ . It is a simple matter to see that this is equivalent to $𝑎 𝑏 < l n (𝑎) l n (𝑏)$ . However, you can’t see how can this solve the problem. A traditional way is to rewrite the inequality like the following:

l n (𝑎) 𝑎 < l n (𝑏) 𝑏 .

l n (𝑎) 𝑎 < l n (𝑏) 𝑏 .

The next step is to let $𝑓 (𝑥) ≔ l n (𝑥) 𝑥$ , and study the property of this function on the interval $(1, + i n f)$ . 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 $𝑎 𝑏 < l n (𝑎) l n (𝑏)$ holds. Let’s draw a picture of $𝑥 \mapsto 𝑥$ and $𝑥 \mapsto l n 𝑥$ on a same figure. Hence we are actually comparing the the slope of the line passing through $(𝑎, l n (𝑎))$ $(𝑎, l n (𝑎))$ and $(𝑏, l n (𝑏))$ $(𝑏, l n (𝑏))$ . Since these are lines, we may let the one through the points $(𝑎, l n (𝑎))$ $(𝑎, l n (𝑎))$ and $(𝑏, l n (𝑏))$ $(𝑏, l n (𝑏))$ be $𝑦 = 𝑚 * 𝑥 + 𝑛$ . 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 $𝑎 < 𝑒 l n (𝑘) 𝑘 - 1$ , where $𝑘 = 𝑏 𝑎$ And the inequality never holds if $𝑎 \geq 𝑒$ .

Proof. First proof that $𝑎 𝑏 < 𝑏 𝑎 \Rightarrow 𝑎 < 𝑒 l n (𝑘) 𝑘 - 1$ . Since $𝑎 𝑏 < 𝑏 𝑎$ , we obtain $𝑒 𝑏 l n (𝑎) < 𝑒 𝑎 l n (𝑏)$ . By the results above, we have

𝑎 𝑏 > l n (𝑎) l n (𝑏) .

𝑎 𝑏 > l n (𝑎) l n (𝑏) .

Set points $𝐴 (𝑎, l n (𝑎))$ $𝐴 (𝑎, l n (𝑎))$ and $𝐵 (𝑏, l n (𝑏))$ $𝐵 (𝑏, l n (𝑏))$ in a plane. Let the line through $𝐴$ $𝐴$ and $𝐵$ $𝐵$ 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 $𝑎 < 𝑒 l n (𝑘) 𝑘 - 1$ . Substitute the position of $𝐴$ $𝐴$ , $𝐵$ $𝐵$ into $A B : 𝑦 = 𝑚 𝑥 + 𝑛$ , we obtain

𝑚 = l n (𝑎) - l n (𝑏 𝑎) 𝑏 𝑎 - 1 .

𝑚 = l n (𝑎) - l n (𝑏 𝑎) 𝑏 𝑎 - 1 .

𝑚 = l n (𝑎) - l n (𝑏 𝑎) 𝑏 𝑎 - 1 .

Let $𝑘 = 𝑏 𝑎$ , then $𝑚 = l n (𝑎) - l n (𝑘) 𝑘 - 1$ . Thus, $𝑎 < 𝑒 l n (𝑘) 𝑘 - 1$ .

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

□

View Typst source

#import "@preview/noteworthy:0.4.0": noteworthy
#import "/themes/site/notes.typ": *
#show: web-equations
#import "../shared.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.
]