# Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger?

Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger?

I tried an induction approach. First I showed that if $b=3$ then any $a \geq4$ satisfied $a^b<b^a$.

Then using that as my base case I tried to show that given any pair of positive integers $a,b$ satisfying $a>b>2$ and $a^b<b^a$, then $(a+1)^{b+1}<(b+1)^{a+1}$ - but that is where I got stuck.

Any help would be appreciated.

• what if you took root $1/ab$ from both sides and studied $y=x^{1/x}$. Jun 4, 2013 at 3:19
• If you let $a=b+k$, then you want to solve $(b+k)^b < b^{(b+k)}$, which is equivalent to $(1+\frac{k}{b})^b < b^k$. If you know the definition that $e^k$ is the limit of the increasing sequence $(1+\frac{k}{n})^n$, then this implies $b>e$, or equivalently that $b\ge3$.
– user641
Jun 4, 2013 at 7:56
• $a\gt b\gt 2$, does not state everything. If specifically the order with respect to $e$ is said, like $e\gt a\gt b\gt 2$ or $a\gt b\gt e\gt 2$, then it would be more clear.
– user249332
Dec 15, 2015 at 16:55
• Jan 2, 2020 at 9:54

Suppose for a moment: $$a^b = b^a.$$ Taking natural log: $$b\ln a = a\ln b,$$ which is $$\frac{\ln a}{a} = \frac{\ln b}{b} .$$ Now consider the function: $$f(x) = \frac{\ln x}{x},$$ where $$f'(x) = \frac{1 - \ln x}{x^2} < 0 \quad \text{ if }\;x>e.$$ I believe you could take it from here.

• Thanks so much for the help, I had realized that calculus would make pretty short work of this, and I see how your proofs go. However, I'm not supposed to use it. Apparently there is some sort of induction/bounding argument that does it, just for integers. By the way, thanks for letting me finish off your proof, it's much more informative than otherwise. Jun 4, 2013 at 3:42
• @JohnMarty I see, but I am not an expert in number theory. I will try to think of a number theoretic answer and edit my post then. Jun 4, 2013 at 3:43

The result follows easily using calculus. Here's an elementary approach, which uses the fact that $a, b$ are positive integers.

Consider $n \geq 3$. Then $$(n+1)^n=\sum_{i=0}^{n}{\binom{n}{i}n^{n-i}}=1+n^2+\sum_{i=0}^{n-2}{\binom{n}{i}n^{n-i}}<n^n+\sum_{i=0}^{n-2}{n^n}=n^{n+1}$$ since $\binom{n}{i} \leq n^i$, and $1+n^2<n^n$ for $n \geq 3$.

Therefore $n^{\frac{1}{n}}>(n+1)^{\frac{1}{n+1}}$ for $n \geq 3$. This immediately implies that $a^{\frac{1}{a}}<b^{\frac{1}{b}}$, so $a^b<b^a$.

Hint:

Which is larger, $b\ln a$ or $a\ln b$?

Which is larger, $\dfrac{\ln a}{a}$ or $\dfrac{\ln b}{b}$?

How does $\dfrac{\ln x}{x}$ behave as $x$ increases? Looks like a job for the derivative.

• As I mentioned to Shuhao Cao, I'm not supposed to use calculus for this question. Any hint on how to proceed with a number theory proof would be great. Jun 4, 2013 at 4:00

Another elementary approach:
First notice that it suffices to prove that $n^{n+1}>(n+1)^n$ for $n\ge 3$. We divide both sides by $n^n$ to turn it into $n>(1+1/n)^n$. As a last step, show by induction that $(1+1/n)^n\le n$ for $n\ge3$.

The case $n=3$ is clear. Now, if $(1+1/k)^k\le k$, then $(1+1/(k+1))^{k+1}\le (1+1/(k))^{k+1}\le k(1+1/(k))\le k+1$.

Q.E.D.
Inform me if anything needs improvements. Thanks in advance.

Most of the answers submitted are fabulous, but let's endeavor to make this as simple as humanly possible, leveraging intuition primarily.

~Begin by asking yourself: Which is larger, $2^3$ or $3^2$? $~8<9$.

~Continue by asking: Which is larger, $3^4$ or $4^3$? $~81>64$. --> This is a reversal of the inequality because the 'weight' of the exponent operation begins to dominate as numbers becomes larger.

~Conclude by asking: Which is larger, $4^5$ or $5^4$? $~1024>625$. --> The disparity is more extreme in absolute terms and likely grows as n increases to ∞.

---The trend suggests strongly that $a>b>2$. Of course, more rigor is needed to definitively confirm that the trend does not reverse, but intuition leaves little doubt.