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?

For a more elementary approach: you could play around with the factors.
$$20^{70^2} = 2^{4900} \cdot 10^{4900} = 32^{980}  \cdot 10^{1300}\cdot 10^{3600}, $$
$$30^{60^2} = 3^{3600} \cdot 10^{3600} = 27^{980} \cdot 3^{660} \cdot 10^{3600}.$$
Comparing each factor, it is now obvious that $20^{70^2} > 30^{60^2}$.
Similarly,
$$30^{60^2} = 3^{3600} \cdot 10^{3600} = 3^{268} \cdot 3^{3332} \cdot 10^{1100}\cdot 10^{2500} = 3^{268} \cdot 9^{1666} \cdot 10^{1100}\cdot 10^{2500},$$
$$40^{50^2} = 4^{2500} \cdot 10^{2500} = 2^{5000} \cdot 10^{2500} = 2 \cdot 8^{1666} \cdot 10^{2500},$$
hence $30^{60^2} > 40^{50^2}$.
The trick here is always to find powers of $2$ and $3$ with small difference: $2^5 > 3^3$ and $2^3 < 3^2$.
