The problem asks us to find the volume of the solid generated by revolving the region enclosed by the curve $y = \log x$, the line $l$ which passes through the origin and is tangent to the curve $C$ defined by $y = \log x$, and the $x$-axis around the $x$-axis.

1. Problem Description

The problem asks us to find the volume of the solid generated by revolving the region enclosed by the curve $y = \log x$, the line $l$ which passes through the origin and is tangent to the curve $C$ defined by $y = \log x$, and the $x$-axis around the $x$-axis.

2. Solution Steps

First, we need to find the equation of the tangent line $l$.

Let the point of tangency be $(x_1, \log x_1)$. The derivative of $y = \log x$ is $y' = \frac{1}{x}$. Thus, the slope of the tangent line at $x = x_1$ is $\frac{1}{x_1}$.

The equation of the tangent line $l$ is then given by

$y - \log x_1 = \frac{1}{x_1} (x - x_1)$.

Since the line passes through the origin $(0,0)$, we have

$0 - \log x_1 = \frac{1}{x_1} (0 - x_1)$

$-\log x_1 = -1$

$\log x_1 = 1$

$x_1 = e$

The point of tangency is $(e, 1)$. The slope of the tangent line is $\frac{1}{e}$.

Therefore, the equation of the tangent line is $y = \frac{1}{e}x$.

Also, $x = ey$.

The region is bounded by $y = \log x$ or $x = e^y$, $y = \frac{1}{e}x$ or $x = ey$, and the $x$-axis (i.e., $y=0$).

The intersection of $y = \log x$ and $y = \frac{x}{e}$ is $(e, 1)$.

The $x$-axis is the line $y=0$. The intersection of $y = \log x$ with the $x$-axis is $(1, 0)$. The intersection of the tangent line $y = \frac{x}{e}$ with the $x$-axis is $(0, 0)$. However, the region is from $x = 1$ to $x=e$.

The volume of the solid generated by revolving the region around the $x$-axis is given by the washer method.

The volume $V$ is given by

$V = \pi \int_0^1 ((ey)^2 - (e^y)^2) dy$

$V = \pi \int_0^1 (e^2 y^2 - e^{2y}) dy$

$V = \pi \left[ \frac{e^2 y^3}{3} - \frac{e^{2y}}{2} \right]_0^1$

$V = \pi \left[ (\frac{e^2}{3} - \frac{e^2}{2}) - (0 - \frac{1}{2}) \right]$

$V = \pi \left[ \frac{2e^2 - 3e^2}{6} + \frac{1}{2} \right]$

$V = \pi \left[ -\frac{e^2}{6} + \frac{1}{2} \right]$

$V = \pi \left[ \frac{3 - e^2}{6} \right]$

$V = \frac{\pi}{6} (3 - e^2)$

3. Final Answer

The volume is $\frac{\pi}{6}(3-e^2)$.