<s>[INST] How to prove the inequalities between $20^{70^2},30^{60^2},40^{50^2}$ Let $$M=\{ 20^{70^2}, 30^{60^2},40^{50^2}\}$$.
What number is the greatest and which is the smallest?

I thought about beginning by assuming certain inequalities and trying to prove them, for example: $$20^{70^2} < 30^{60^2}$$
However, I do not know exactly what do from here. I took logarithms and expanded the expressions, this took me nowhere. I could get no obvious identity out of them, is there a clever way to do this?
 [/INST] Compare
$$
20^{4900} \mbox{ and } 30^{3600} \mbox{ and } 40^{2500};
$$
i.e.
$$
(20^{49})^{100} \mbox{ and } (30^{36})^{100} \mbox { and } (40^{25})^{100};
$$
therefore there is enough to compare
$$
20^{49} \mbox{ and } 30^{36} \mbox { and } 40^{25};
$$
i.e.
$$
2^{49}\cdot 10^{49} \mbox{ and } 3^{36}\cdot 10^{36} \mbox { and } 4^{25}\cdot 10^{25};
$$
and (dividing by $10^{36}$)
$$
\color{red}{2^{49}\cdot 10^{13}} \mbox{ and } 
\color{green}{3^{36}} \mbox { and } 
\color{blue}{4^{25} / 10^{11}}.
$$
Now, 

$\color{green}{3^{36}} = 9^{18} = 9^5 \cdot 9^{13} < 16^5 \cdot 10^{13} = 2^{20} \cdot 10^{13} \color{red}{< 2^{49}\cdot 10^{13}}$;

and

$\color{green}{3^{36}} = 3^{28}\cdot 3^8 = 9^{14}\cdot 3^8 \color{blue}{> 4^{14} \cdot (4/10)^{11}}$;
hence
$$
20^{70^2}> 30^{60^2} > 40^{50^2}.
$$
</s>