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?
 A: A good start is to notice all three inequalities have the form
$$
f(x) = x^{(90-x)^2}
$$
So what you want is the order of $f(20), f(30), f(40)$.
Rather than comparing the numbers directly, it's a good idea to get rid of nested exponentials by taking logarithms.
We can set
$$
g(x) = \ln \ln f(x) = \ln \left[ (90-x)^2 \ln x\right]
= 2 \ln (90-x) + \ln \ln x
$$
Now we have an intuitive answer: $\ln \ln x$ is a VERY slow-growing function and won't change too much from $20$ to $40$, while on the other hand $\ln(90-x)$ will get a bit smaller.
So we expect $g(20) > g(30) > g(40)$, and therefore $f(20) > f(30) > f(40)$.
To actually prove this, we can use calculus to show $g$ is decreasing on the interval $[20,40]$. We have
\begin{align*}
g'(x)
&= \frac{-2}{90-x} + \frac{1}{x \ln x}\\
&= \frac{1}{x(90-x)\ln x} \left( 90-x - 2x\ln x\right)
\end{align*}
and for $x\ge 20$, $\ln x > 2$ so
\begin{align*}
90 - x - 2x \ln x
  &< 90 - (20) - 2(20)(2) \\
  &= -10 \\
  &< 0.
\end{align*}
A: It is enough to switch to logarithms and consider that:
$$ \frac{9}{13}<\log 2<\frac{7}{10},\qquad \frac{78}{71}<\log 3<\frac{11}{10},\qquad \frac{23}{10}<\log(10)<\frac{76}{33}.$$
With crude estimations $70^2\log(20)\approx 14000$, $60^2\log 30\approx 12000$ and $50^2\log(40)\approx 9000$.
A more elegant approach is to study the function $f(x)=x^2\log(90-x)$ over the interval $(0,90)$ and prove that its relative maximum is located at $x_0>70$, so $f(x)$ is increasing on $(0,70]$.
A: 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 $2^3 < 3^2$.
A: 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
$$
20^{70^2}> 30^{60^2} > 40^{50^2}.
$$
A: $40=30^a \;$ where $$a=[\ln(40]/\ln 30=1+[\ln (1+1/3)]/\ln 30<1+[1/3]/\ln 30<1+1/9.$$ Therefore  $\quad 40^{50^2}=30^{a\cdot  50^2}<30^{(1+1/9)50^2}<30^{60^2}.$
$30=20^b \;$ where $$b=[\ln 30]/\ln 20=1+[\ln (1+1/2)]/\ln 20<1+[1/2]/\ln 20<1+1/4.$$ Therefore $\quad 30^{60^2}=20^{b\cdot 60^2}<20^{(1+1/4)60^2}<20^{70^2}.$
A: Intuitively, to get a sense of the size of $a^b$, $b$ matters much more. This isn't a general rule though. But considering decreasing the exponent $70^2$ to $60^2$ versus increasing $20$ to $30$, the former is a greater reduction, especially given that it is an exponent, one would intuitively expect $20^{70^2}>30^{60^2}$. But again, this reasoning will not always hold. 
This answer is probably not very appropriate in this context, but I will post it anyways since someone might find it interesting or useful. Even some precalculus students may find it helpful to developing a deeper intuitive understanding.
More generally, consider $(A+k)^{B^2}$ vs $A^{{(B+k)}^2}$. Taking logarithms we get that $(A+k)^{B^2}<A^{{(B+k)}^2}$ when 
$$
\frac{\ln(A+k)}{\ln A}<\frac{(B+k)^2}{B^2}.
$$
Given that quadratic growth is faster than logarithmic growth, we should expect the relative growth (percentage increase) from $B^2$ to $(B+k)^2$ to be greater than that from $\ln A$ to $\ln (A+k)$ unless $B$ is very large relative to $A$ and $k$ is relatively small. 
This occurs because $B^2$ is already going to be a very large number, so increasing $B$ by a sufficiently small $k$ doesn't result in much of a percentage increase. However, if $A$ is sufficiently small, increasing $A$ by $k$ can result in a larger percentage increase in the logarithmic function.
For $A=20$ and $k=10$, we need around $B>152$ in order to reverse the above inequality.
A similar argument will hold for powers other than $2$, say $(A+k)^{B^n}$ vs $A^{{(B+k)}^n}$ since power growth is always faster than logarithmic growth. For a fixed $A$ and $k$, a larger power requires a larger $B$  to reverse the expected/intuitive inequality given above.
A: $20^{70^2} = 20^{70^2}$
$30^{60^2} = 1.5^{60^2}20^{60^2}$
$40^{50^2} = 2^{50^2}20^{50^2}$
=====
$20^{70^2-60^2}=20^{10*130}=20^{1300} $
$1.5^{60^2}=2.25^{30*60}=5.0625^{90}$
$2^{50^2}20^{50^2-60^2}=2^{2500}2^{-10*110}10^{-1100} =2^{1400}10^{-1100}=2^{300}5^{-1100}>5^{-800}$
======
So $40^{50^2}<30^{60^2}<20^{70^2} $ and by fairly significant margins.
A: Here's a strange way to view it but bear with me:
Obviously if $0< a < b; x < y$ then $a^x < b^y$ but we should develop an intuition for when $0 < a < b; x >y$ how to compare $a^x$ to $b^y$.  
Let $b = k*a; k > 1$ so $ a < b$.  Let $x = my; m > 1$ so $x >y$.
Then $a^x <|=|> b^y \iff$
$\frac{a^x}{b^y} = \frac{a^{my}}{a^yk^y} <|=|> 1 \iff$
$a^{(m-1)y} <|=|> k^y \iff a^{m-1} <|=|> k$
So $20^{70^2} <|=|> 30^{60^2} \iff 20^{\frac 76^2 - 1} <|=|> 1.5$ 
As $20^{13/36}>20^{1/3} > 8^{1/3} = 2 > 1.5$ we have $20^{70^2} > 30^{60^2}$
Likewise $30^{60^2} <|=|> 40^{50^2} \iff 30^{\frac 65^2 - 1} <|=|> 1\frac 13$.
As $30^{11/25} > 30^{2/5} = 900^{1/5} > 32^{1/5} = 2 > 1 \frac 13$ so $30^{60^2} > 40^{50^2}$
In general, we should develop the intuition that increases in exponents are ...er,  exponentially .. more significant than increases in bases. 
