How can I compare numbers raised to a square root For example: $3^\sqrt5$ versus $5^\sqrt3$
I tried to write numbers as this:
$3^{5^{\frac{1}{2}}}$ and then as
$3^{\frac{1}{2}^5}$
But this method gives the wrong answer because $a^{(b^c)} \ne a^{bc}$

user134824's approach, but with the $\sqrt{3}$ instead, is a bit nicer:
$$
\begin{align}
3^\sqrt{5} &\text{ vs. } 5^\sqrt{3}\\
(3^\sqrt{5})^\sqrt{3} &\text{ vs. } (5^\sqrt{3})^\sqrt{3}\\
3^{\sqrt{5}\sqrt{3}} &\text{ vs. } 5^{\sqrt{3}\sqrt{3}}\\
3^\sqrt{15} &\text{ vs. } 5^3\\
\end{align}
$$
Now notice that since $\sqrt{15} < 4$, $3^\sqrt{15} < 3^4$ (via monotonicity of $3^x$), and since $3^4 < 5^3$ ($81 < 125$), we can say $3^\sqrt{15}<5^3$ (via transitivity), and subsequently $3^\sqrt{5} < 5^\sqrt{3}$ (via monotonicity of $x^\dfrac{1}{\sqrt{3}}, x \ge 0$).
