The vertices of a quadrilateral lie on the graph of $y=\ln x$, and the $x$-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln\frac{21}{20}$. What is the $x$-coordinate of the leftmost vertex?

1. With leftmost $x$-coordinate $n$: $\dfrac{(n+1)(n+2)}{n(n+3)}=\dfrac{21}{20}$.
2. Cross-multiplying: $20(n+1)(n+2)=21n(n+3)$, i.e. $n^2+3n-40=0$, so $n=5$.
3. The answer is $\boxed5$.