Order a sequence problem The question simply asks to arrange these from smallest to largest.  Without using a calculator what rationale could I use on a test if I were to see this question?
$$ \sqrt[3]{\pi}, 2^{\sqrt{\pi}}, \sqrt{2}, 3^\pi, \pi^3 $$
The answer is
$$\sqrt{2}<\sqrt[3]{\pi}<2^{\sqrt{\pi}}< \pi^3<3^\pi$$
 A: First of all, if $1 \lt b \lt a$, then $a^b \lt b^a$. This can be a helpful rule when dealing with this kind of problem. This lets us say that
$$\pi^3 \lt 3^\pi$$
and, of course, we can obviously say that
$$\pi^{\frac{1}{3}} \lt \pi^3$$
so we have
$$\pi^{\frac{1}{3}} \lt \pi^3 \lt 3^\pi$$
Also, since $\sqrt \pi \gt \frac{1}{2}$, we have
$$\sqrt 2 \lt 2^{\sqrt \pi}$$
Now we just have to figure out how to merge these inequalities. We can see by approximation that
$$\sqrt 2 \lt 2^{\sqrt \pi} \lt \pi^3 \lt 3^\pi$$
and that
$$\pi^{\frac{1}{3}} \gt \sqrt 2$$
So now we must determine the relationship between $\pi^{\frac{1}{3}}$ and $2^{\sqrt \pi}$. Since we know that $\sqrt \pi$ is greater than one, we can say that $2^{\sqrt \pi}$ is greater than two, and since $\pi^{\frac{1}{3}}$ is less than two, we have $\pi^{\frac{1}{3}}$ between $2^{\sqrt \pi}$ and $\sqrt{2}$:
$$\sqrt 2 \lt \pi^\frac{1}{3} \lt 2^{\sqrt \pi} \lt \pi^3 \lt 3^\pi$$
A: Well we have that $\sqrt[3]{\pi}>\sqrt[3]{3}$ and we have that $\sqrt{2^6}=8<9=\sqrt[3]{3^6}$. 
We also have that $2^{\sqrt{\pi}}>2>\sqrt[3]{\pi}$ this is true since $2^3>\pi$
$$\pi^3>2^2=2^\sqrt{4}>2^\sqrt{\pi}$$
The last one follows since $a^b<b^a$ implies that $b\ln a<a\ln b$ and this is equivalent to$$\frac{\ln a}{a}<\frac{\ln b}{b}$$
This follows because $f(x)=\frac{\ln x}{x}$ is decreasing when $x>e$ so $f(\pi)<f(3)$. This is a fairly standard and useful result you can prove it with derivative.
$$f'(x)=\frac{1-\ln x}{x^2}$$
