Solution.
We first note that we can eliminate all negative values of
$k$ since then the right-hand side of the given equation would be
negative, while the left-hand side would be positive.
So $k>0$. Let $a$ be a solution to the equation; that
is, $e^{2a}=k\sqrt{a}$. Because the curves will just touch at
the single solution, they will have a common tangent at $x=a$.
Thus the curves $y=e^{2x}$ and $y=k\sqrt{x}$ have the same
slope at $x=a$. Therefore
2e^{2a}=\frac{k}{2\sqrt{a}}
Solving the two equations $e^{2a}=k\sqrt{a}$ and
$2e^{2a}=k/(2\sqrt{a})$, we get
2e^{2a}=2k\sqrt{a} = \frac{k}{2\sqrt{a}}
\Longrightarrow 2\sqrt{a} = \frac1{2\sqrt{a}} \Longrightarrow a =
\textstyle\frac14
and $k=e^{2(1/4)}/\sqrt{1/4} = \sqrt{e}/(1/2) = 2\sqrt{e}$.