# Which is larger, $70^{71}$ or $71^{70}$? [duplicate]

This question already has an answer here:

Yet another question of which is larger: $70^{71}$ or $71^{70}$. I solved it by observing that $f(x)=\frac{\ln(x)}{x}$ is decreasing for all $x>e$ since $f'(x)=\frac{1-\ln(x)}{x^2}<0$ for all $x>e$. Then we have \begin{align*} \frac{\ln(70)}{70}>\frac{\ln(71)}{71} &\iff 71\ln(70)>70\ln(71)\\ &\iff \ln(70^{71})>\ln(71^{70})\\ &\iff e^{\ln(70^{71})}>e^{\ln(71^{70})}\\ &\iff 70^{71}>71^{70}. \end{align*} I briefly search for similar problems along this idea, but didn't really find much in other ideas about how to solve this. So I open this to the group: how else might this be proved?

EDIT: So this problem is a specific case of the more general problem I apparently duplicated. However, we can extend the duplicate problem by allowing real number inputs as opposed to just integer inputs. Thus $x^{(x+1)}>(x+1)^x$ for all reals $x>e$. This is proved following the exact same proof I gave above with the appropriate substitutions: \begin{align*} \frac{\ln(x)}{x}>\frac{\ln(x+1)}{x+1} &\iff (x+1)\ln(x)>x\ln(x+1)\\ &\iff \ln(x^{(x+1)})>\ln((x+1)^{x})\\ &\iff e^{\ln(x^{(x+1)})}>e^{\ln((x+1)^{x})}\\ &\iff x^{(x+1)}>(x+1)^{x}. \end{align*} Thanks for the input everyone!

## marked as duplicate by Watson, Claude Leibovici, Najib Idrissi, Daniel W. Farlow, Lee MosherJun 13 '16 at 13:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• I think $f(x)$ is decreasing for '$x>e$'. – zxcvber Jun 13 '16 at 11:06
• Of course. I will edit it accordingly. – Laars Helenius Jun 13 '16 at 11:08
• – Henry Jun 13 '16 at 11:26

## 3 Answers

$$\left(1+\dfrac1{70}\right)^{70}<e<70$$

• Bravo.    – dbanet Jun 13 '16 at 11:50

Comparing $70^{71}$ and $71^{70}$ is the same as comparing $70^{1/70}$ or $71^{1/71}$.

The function $f(x)=x^{1/x}$ has exactly one local maximum, at $x=e$, and so is decreasing for $x>e$.

Therefore, because $70>e$, we have $70^{1/70}>71^{1/71}$ and so $70^{71}>71^{70}$.

The same technique proves that $(n+1)^n<n^{n+1}$ for $n\geq 3$.

Another way, $70^{71}=70^{70}\times70$ and $71^{70}=70^{70}\times(\frac{71}{70})^{70}$. Now it's require a little calculation that imply $(\frac{71}{70})^{70}=2.699<70.$ Hence $70^{71}>71^{70}.$