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 $\ln \frac{91}{90}$. What is the $x$-coordinate of the leftmost vertex?
Let the coordinates of these four vertices be $(n, \ln n)$, $(n+1, \ln(n+1))$, $(n+2, \ln(n+2))$, and $(n+3, \ln(n+3))$. Then the are of the quadrilateral will be $$\left|\frac{\left(n\ln(n+1) - (n+1)\ln n\right)+\left((n+1)\ln(n+2) - (n+2)\ln(n+1)\right)+\cdots +\left((n+2)\ln n - n\ln(n+3)\right)}{2}\right|$$
This expression can be rewritten as $$\frac{1}{2}\ln\left|\frac{(n+1)^n(n+2)^{n+1}(n+3)^{n+2}n^{n+3}}{n^{n+1}(n+1)^{n+2}(n+2)^{n+3}(n+3)^{n}}\right|=\ln\frac{(n+1)(n+2)}{n(n+3)}=\ln\frac{91}{90}$$
Note we always have $(n+1)(n+2) > n(n+3)$. Therefore $$\frac{(n+1)(n+2)}{n(n+3)}=\frac{91}{90}\implies n=\boxed{12}$$