Compare: Which is bigger? $3^{\sqrt2}$ or $2^{\sqrt 3}$ Compare: Which is bigger?
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$ $\Large{3^{\sqrt2}}\quad$ or  $\quad \Large{2^{\sqrt 3}}$

My attempts:
Let,  $\quad$ $a=3^{\sqrt 2}$ and $b=2^{\sqrt 3}$
We have

$$a^\sqrt{3}=3^\sqrt{6}\\ b^\sqrt{3}=8&lt;3^{\sqrt{4}}&lt;3^\sqrt{6}=a^\sqrt{3} $$
$$a^\sqrt{3}&gt;b^\sqrt{3} \Longrightarrow a&gt;b$$

Question-1:

Is my solution correct?

Question-2:

I saw the following solution was approved on another site.  But, the last line seems wrong to me. Am I right?
The handwritten one is simply the wrong approach.  If you raise $a$ and $b$ to different powers you simply will not (usually) be able to conclude anything by comparing the results.  How can you?  You raised them to different powers so whatever inequality you get doesn't relate to the original values in any way!

Everything it came up with is true (but not helpful)  up to $a^{\sqrt 3} &gt; b^{\sqrt 2}$ this has been shown[1].  But the conclusion just doesn't make sense.

$a^{\sqrt 3} &gt; b^{\sqrt 2} \not \implies a =b$(!!!!) 

Why would it?  $\sqrt 2 \ne \sqrt 3$. 

This is like claiming $3^3 &gt; 4^2 \implies 3 &gt; 4$.

(Or simply if $a = 2.99$ and $b = 3.01$ then  $2.99^{\sqrt 3} \approx  6.6663278104352795010160893782624$ while $3.01^{\sqrt 2}\approx 4.7511115647726528859943327888079$.

(So $a^{\sqrt 3} &gt; b^{\sqrt 2} $ but $a &lt; b$.)

$a^{bigger number} &gt; b^{smallernumber}$ doesn't tell us anything.[2]

We could try

$a^{\sqrt 3} &gt; b^{\sqrt 2} \implies a &gt; b^{\frac {\sqrt 2}{\sqrt 3}}$

or 

$a^{\sqrt 3} &gt; b^{\sqrt 2} \implies a^{\frac {\sqrt 3}{\sqrt 2}} &gt; b$

but we can't get anywhere from there.

....

But yours is fine.

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[1](It is nothing more than claiming $3^{\sqrt 6} &gt; 2^{\sqrt 6}$. 

[2] However if we got the other result $a^{smaller power} &gt; b^{biggerpower}$ (and if we know $a, b &gt; 1$) we can conclude $a &gt; b$ because 

a) $a^{biggerpower} &gt; a^{smallerpower} &gt; b^{bigger power}$

b) $a^{smallerpower} &gt; b^{biggerpower} &gt; b^{smallerpower}$.

But if the powers and the inequality are in disagreement, we don't know which "overpowers" the other.

This is assuming $a, b &gt; 1$.  If $a,b &lt; 1$ the exact OPPOSITE applies.

And if you don't know if $a,b$ are bigger smaller than $1$ then you can't compare different powers at all because you don't know anything.  That's another reason the handwritten one was the wrong approach.
Yes because $(x,y)\mapsto x^y$ is strictly increasing in both variables on $(1,\infty)\times (0,\infty)$.
No: $a^{\sqrt3}&gt;b^{\sqrt 2}$ does not imply $a&gt;b$. See $a=b=1.5$ for comparison.