Put the following numbers in ascending order:
$$\log_{2}{3}$$
$$\sqrt{e}$$
$$(\sqrt{2})^2$$
$$2^{\sqrt{2}}$$
My success so far (in ascending order): $$(\sqrt{2})^2 = \log_{2}{1}$$ $$2^\sqrt{2} = \log_{2}{\sqrt{2}}$$ $$\log_{2}{3}$$
UPD: I messed up with exp to log transitions. Besides, it seems that the chosen approach (to make all functions to log) was wrong.
Please don't just give an answer but explain. And don't recommend calculator or super-advanced formulas solution.
The first two lines of "your success" are wrong. $\log_2 1=0$, but $(\sqrt 2)^2=2$, so $\log_2 (\sqrt2^2)=1$. $\log_2(2^{\sqrt 2})=\sqrt 2$, not the other way around. Since $\log_2$ is monotonic you can sort numbers by their base $2$ logs, but I don't think that is useful for the numbers you are given.
I would start with $(\sqrt 2)^2=2, 2^{\sqrt 2} \gt 2$, then note that $\log_2 3 \lt \log_2 4 = 2$, so $\log_2 3$ is first so far. Also $\sqrt e \lt \sqrt 4=2$ so we are down to comparing $\log_2 3$ and $\sqrt e$. My first thought is that $e \gt 2.56,$ so $\sqrt e \gt 1.6$. It is true that $\log_2 3 \lt 1.6$ but I don't have an easy way to show that.