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 $a^{(b^c)} \ne a^{bc}$
 A: Doing a problem like this takes a combination of knowledge about rules of exponentiation and skill at estimation.
First, raise both $3^{\sqrt 5}$ and $5^{\sqrt 3}$ to the power $\sqrt 5$, which gives the numbers $3^5=243$ and $5^{\sqrt{15}}$ using the rule $(a^b)^c = a^{bc}$. This has the effect of removing one of the square roots from an exponent: now we just need to estimate $5^{\sqrt 15}$. And $\sqrt{15}>3.5$, so
$$
5^{\sqrt{15}}>5^{3.5}=5^3\sqrt 5 = 125\sqrt 5.
$$
Since $\sqrt 5>2$, we have that $5^{\sqrt{15}}>125\cdot2=250 > 243 = 3^5$.
So
$$
\fbox{$5^{\sqrt 3} > 3^{\sqrt 5}$}.
$$
A: $$5^\sqrt{3}>5^{5/3}=\sqrt[3]{3125}>\sqrt[3]{2187}=3^{7/3}>3^{\sqrt{5}}$$
Another option is to take roots, and compare $3^{1/\sqrt{3}}$ with $5^{1/\sqrt{5}}$.  This function increases for $1<x<7.39$, then decreases.
A: 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 $x^\dfrac{1}{\sqrt{3}}, x \ge 0$).
A: Consider 
$f(x)=x^{\sqrt{5}}-5^{\sqrt{x}}$
$f(\sqrt{5})=0 $
$f'(x)=(\sqrt{5}-1)x-\frac {log(5)}{2 \sqrt{x}}e^{\sqrt{x}log5}$
so $ f(x) < 0$ for $x$ less than $\sqrt{5}$ so we have $$ 5^{\sqrt{3}} > 3^{\sqrt{5}}$$
A: $$3^{\sqrt 5} \lt 5^{\sqrt3}$$
$$\sqrt 5 \cdot \ln(3) \lt \sqrt3 \cdot \ln(5)$$
$$\sqrt{5 \over 3} \cdot \log_5 (3) \lt \sqrt{5 \over 3}  \cdot {3 \over 4} \lt 1$$
because $\log_a(b) \lt 1$ for $a \gt b$, $\ln(3) \lt {3 \over 2}$, and $\ln(5) \lt 2$
A: Hint: Consider the function $$f(x)={\ln x\over \sqrt x}$$ and see where it is increasing or decreasing.
