Step 1: Set up the sign test.
The function is increasing wherever its slope is positive. The slope of $f$ is $f^{\prime}(x)$, so we study the sign of $f^{\prime}(x)$ on $(0, \infty)$.
Step 2: Find the slope function.
Write $f(x) = u \cdot v$ with $u = x$ and $v = \log x$. Then $u^{\prime} = 1$ and $v^{\prime} = \frac{1}{x}$.
\[ f^{\prime}(x) = u^{\prime}v + u v^{\prime} = \log x + 1 \]
Step 3: Find the turning point.
Set $f^{\prime}(x) = 0$. Then $\log x = -1$, so $x = \frac{1}{e}$. This point splits the domain into two parts.
Step 4: Test a point in each part.
Take $x = \frac{1}{e^2}$ in $\left(0, \frac{1}{e}\right)$. Here $f^{\prime}(x) = -2 + 1 = -1 < 0$, so $f$ is decreasing.
Take $x = 1$ in $\left(\frac{1}{e}, \infty\right)$. Here $f^{\prime}(1) = 0 + 1 = 1 > 0$, so $f$ is increasing.
Final Answer:
The slope is positive only to the right of $\frac{1}{e}$.
\[ \boxed{\left(\frac{1}{e}, \infty\right)} \]