7. As shown in the figure, a cross-section of the cube $A B C D-E F G H$ passes through vertices $A, C$ and a point $K$ on edge $E F$, dividing the cube into two parts with a volume ratio of $3: 1$. Then the value of $\frac{E K}{K F}$ is $\qquad$.
Let the section be $A K M C$. From $\triangle A B C \sim \triangle K F M \Rightarrow K F=F M$, let $K F=F M=x$. Thus, $V_{K F M-A B C}=\frac{1}{3}\left(\frac{1}{2}+\frac{1}{2} x^{2}+\frac{1}{2} x\right)=\frac{1}{4} \Rightarrow \frac{1}{2} x^{2}+\frac{1}{2} x+\frac{1}{2}=\frac{3}{4}$ $\Rightarrow 2 x^{2}+2 x-1=0 \Rightarrow x=\frac{\sqrt{3}-1}{2}$, hence $1-x=\frac{3-\sqrt{3}}{2}$. Therefore, $\frac{E K}{K F}=\frac{1-x}{x}=\frac{3-\sqrt{3}}{\sqrt{3}-1}=\sqrt{3}$.