Which is larger, $70^{71}$ or $71^{70}$?

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

See How is $a_n=(1+1/n)^n$ monotonically increasing and bounded by $3$?


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 $70^{71}>71^{70}.$

Tags:

Inequality

Number Comparison

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