# Is nᵐ>mⁿ if m>n?

I remember playing with my calculator when I was young. I really liked big numbers so I'd punch big numbers like $$20^{30}$$ to see how big it really is.

On such a quest, I did observe that $$20^{30}$$ is greater than the value of $$30^{20}$$. In fact, in many cases, I found that $$n^m>m^n$$ if $$m>n$$.

Is this a general fact? If so, can it be proved?

• – Henry Aug 21 '17 at 7:34
• It is true if $m \gt n \gt e$, but note $1^m \lt m^1$ (for $m \gt 1$) and $2^3 \lt 3^2$ and $2^4=4^2$ – Henry Aug 21 '17 at 7:35
• $2^3=8<3^2=9$, so Henry's condition is needed. – Evargalo Aug 21 '17 at 7:37
• While checking the big numbers you liked, you missed the small ones which make the difference. We have $3 > 2$. But $3^2 = 9$ is not less than $2^3 = 8$. So, your statement is not true for any real value. However, as @Henry says, it will be true for $m > n > e$. – Aniruddha Deshmukh Aug 21 '17 at 7:37
• My bad I guess. In my attempts to verifying for bigger numbers I completely overlooked the smaller ones. – Pritt Balagopal Aug 21 '17 at 7:38

For a positive integer $m$, consider the function $f(x)=m^x/x^m$. And $g(x)=\ln f(x)=x\ln m-m\ln x$.

Then $$g'(x)=\ln m-\frac mx$$ which is positive for $x>m/\ln m$. Then $g$ is increasing in $(m/\ln m,\infty)$. For $m>e$ we have $m>m/\ln m$ and $g(x)>0$ for $x>m$. Then, as it has been said in comments,

$$n>m>e\implies m^n>n^m$$

• I guess I didn't properly answer the question and just turned the proposition around. If anyone deserves a tick (or at least +1) then @ajotatxe deserves it. – George N. Missailidis Aug 21 '17 at 8:45

This is not an answer I just need to show the graph. I forgot to label axis. Horizontal is $n$ and the other is $m$

Hope you like it • Thanks, but I already knew the graph. :) Anyway it'll be helpful for other users visiting this question from google. (+1) – Pritt Balagopal Aug 21 '17 at 11:06
• This would benefit from a description of the boundary curves, one being $m=n$ and the other being the other solutions to $\dfrac{\log m}{m} = \dfrac{\log n}{n}$ with $m,n >1$, crossing at $(e,e)$ – Henry Aug 21 '17 at 11:25
• @PrittBalagopal: Are you sure you understood what this graph is saying? It shows the regions in which $n^m > m^n$. It answers your question negatively, as there are areas where $m>n$ and yet $m^n>n^m$ (the white area to the left). But if $n>e$ (that's the vertex point where the two shaded areas meet) you have $m>n \Rightarrow n^m>m^n$. – Meni Rosenfeld Aug 21 '17 at 15:08
• @MeniRosenfeld Ohh, so that's what it means. Thanks for patching it up – Pritt Balagopal Aug 21 '17 at 15:58

You are proposing that $n^m > m^n \iff n > m$. However, there are many examples where this is not entirely true.

If $n = 2 \land m = 3 \implies n^m < m^n : n < m$

If $n = 2 \land m = 4 \implies n^m = m^n : n < m$

And obviously if $n = 1 \land m > 1 \implies n^m < m^n : n < m$

But perhaps what you are trying to say is that:

If $n > m \implies n^m > m^n$ because it seems like $n < m$ in these contradicting examples above. I mean, why do these seem to be the only contradicting examples? With the examples above, we know that $n \neq m \neq 0 \lor 1 \because n^m < m^n$. So moving from $1$ to $2$ where $n = 2$ and $m > 2$, we find a little shift in the equality signs.

For the first example, $n^m < m^n$

For the second example, $n^m = m^n$

And it seems that if $m > 2^2 = 4$ then your theory is true where $n^m > m^n$. And it seems like the reason your theory looks true on the condition that $m > 2^2$ is because we must find the first $n^m \lor m^n : n \land m > 1$ (because $1 > 0$ which is obviously $2^2$).

In summary, the theory is not that "if $n^m > m^n$ then $n > m$" but is instead that:

$$\text{if} \qquad n > m \implies n^m > m^n : n \land m \in \mathbb{W}$$

(since $\mathbb{W}$ is the set of all numbers $\ge 0$ aka "Whole Numbers")

• Comment below the things you may not understand here so I can edit the post and tell you whatever it is that you don't understand. – George N. Missailidis Aug 21 '17 at 8:14
• Thanks for your explanation. You clarified a part of my question, so I didn't accept this one. (+1 tho) :) – Pritt Balagopal Aug 21 '17 at 11:08
• OK thanks for letting me know. I figured anyway. But I think it's not that we must find the first $n^m : n \land m > 1$ but we should find the first $n^m > e$, perhaps, according to what everyone else is saying. I appreciate a (+1) tho so thank you :) – George N. Missailidis Aug 21 '17 at 21:45

In simple terms, for integers you can start with smallest no. i.e. (1,2), (2,3), (2,4).

• $1^2 < 2^1$,
• $2^3 < 3^2$ and
• $2^4 = 4^2$.

In all the above cases $n^m > m^n$ was false for all m>n. By observing the pattern for all n>=2 and m>4 we have $n^m > m^n$ true. Consider

• (2,5) $\implies$ 32 > 25 or
• (3,4) $\implies$ 81 > 64 or
• (4,100) $\implies$ (1.6 * 10^60) > 100000000 and so on...

So basically even a small number but with large exponent/power is greater than a big number with small exponent as observed above, except for some cases. Bigger exponent matters more than a big base number.

As for the proof part you can take log of $n^m$ and $m^n$.
As the function $\frac{\log x}{x}$ is a decreasing function for x > e($\approx$ 2.718)

$\implies$ $\frac{\log n}{n} > \frac{\log m}{m}$ (for m>n)

So, (as mentioned in above answers also)
m > n > e $\implies$ $n^m > m^n$.

• math.meta.stackexchange.com/q/5020/306553 please use mathjax. – Siong Thye Goh Aug 21 '17 at 20:14
• forgot to add '$'. Added. – Dark_Byt3 Aug 21 '17 at 20:22 • If you think mathjax is complicated, advice: don't read the whole thing, and only read what is necessary for you. Trust me, I may be a mathy person but I found the mathjax pretty confusing. – George N. Missailidis Aug 21 '17 at 21:50 One more attempt: Consider the function:$f(x) := \dfrac{x}{\log x}$,$x \gt e$(say), is strictly increasing, since$f'(x) = \dfrac{\log x - 1}{(\log x)^2} \gt 0$for$x \gt e$.$f(x_1) \lt f(x_2)$for$ x_1 \lt x_2$. With$x_1 = n$,$ x_2= m $,$ n \lt m $,$m,n$positive integers$\dfrac{n}{\log n} \lt \dfrac{m}{\log(m)}$;$n \log (m) \lt (m) \log n$;$\log (m)^n \lt \log (n)^m$;$\exp(\log (m)^n) \lt \exp (\log (n)^m)$;$m^n \lt n^m$for$m\gt n.\$