The problem asks us to find the volume of the solid generated by rotating the region bounded by the curve $y = \log x$, a line $l$ passing through the origin and tangent to the curve $C$, and the $x$-axis around the $x$-axis.

1. Problem Description

The problem asks us to find the volume of the solid generated by rotating the region bounded by the curve $y = \log x$, a line $l$ passing through the origin and tangent to the curve $C$, and the $x$-axis around the $x$-axis.

2. Solution Steps

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

The derivative of $y = \log x$ is $y' = \frac{1}{x}$.

Let $(x_1, \log x_1)$ be the point of tangency.

The equation of the tangent line $l$ is 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 $\frac{1}{x_1} = \frac{1}{e}$.

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

Or $x = ey$.

We are given the curve $y = \log x$. We can rewrite this as $x = e^y$.

The intersection of $y=\frac{1}{e}x$ and $x$-axis gives $x=0$.

The intersection of $y=\log x$ and $x$-axis gives $x=1$.

The intersection point between the curve $C$ and the line $l$ is $(e,1)$.

The volume $V$ can be calculated using the disk method, rotating the region around the $x$-axis.

$V = \pi \int_{0}^{1} ((e^y)^2 - (ey)^2) dy$

$V = \pi \int_{0}^{1} (e^{2y} - e^2y^2) dy$

$V = \pi [\frac{1}{2}e^{2y} - \frac{e^2}{3}y^3]_{0}^{1}$

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

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

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

3. Final Answer

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