7. In the Cartesian coordinate system, it is known that $\odot O_{1}$ and $\odot O_{2}$ intersect at points $P(3,2)$ and $Q$, and the product of the radii of the two circles is $\frac{13}{2}$. If both circles are tangent to the line $l: y=k x$ and the $x$-axis, then the equation of line $l$ is
7. $y=2 \sqrt{2} x$.

By the problem, let the equations of the two circles be
$$
\left(x-a_{i}\right)^{2}+\left(y-r_{i}\right)^{2}=r_{i}^{2}(i=1,2) \text {. }
$$

Then $r_{1} r_{2}=\frac{13}{2}, k>0$.
Since both circles pass through the point $P(3,2)$, we have
$$
\begin{array}{l}
\left(3-a_{i}\right)^{2}+\left(2-r_{i}\right)^{2}=\dot{r}_{i}^{2} \\
\Rightarrow a_{i}^{2}-6 a_{i}+13-4 r_{i}=0 .
\end{array}
$$

Let the slope of $l: y=k x$ be $\alpha(\alpha \in(0, \pi))$. Then $k=\tan \alpha, \tan \frac{\alpha}{2}=\frac{r_{i}}{a_{i}}>0$.
Let $t=\tan \frac{\alpha}{2}=\frac{r_{i}}{a_{i}}$. Then $a_{i}=\frac{r_{i}}{t}$.
Substituting it into equation (1) and rearranging, we get
$$
r_{i}^{2}-6 t r_{i}-4 t^{2} r_{i}+13 t^{2}=0 \text {. }
$$

By Vieta's formulas, we know $r_{1} r_{2}=\frac{13}{2}=13 t^{2}$.
Thus, $t=\frac{\sqrt{2}}{2}, k=\tan \alpha=\frac{2 t}{1-t^{2}}=2 \sqrt{2}$.
Therefore, the equation of the line $l$ is $y=2 \sqrt{2} x$.