The function is given by \( f(x) = \frac{\log(x)}{x} \). To determine its maximum value, we first compute the derivative of \( f(x) \) with respect to \(x\):
\[f'(x) = \frac{d}{dx}\left(\frac{\log(x)}{x}\right).\]
Applying the quotient rule yields:
\[f'(x) = \frac{x \cdot \frac{1}{x} - \log(x) \cdot 1}{x^2} = \frac{1 - \log(x)}{x^2}.\]
Setting \( f'(x) = 0 \) to identify critical points:
\[\frac{1 - \log(x)}{x^2} = 0 \quad \Rightarrow \quad 1 - \log(x) = 0 \quad \Rightarrow \quad \log(x) = 1 \quad \Rightarrow \quad x = e.\]
To verify that \(x = e\) corresponds to a maximum, we can examine the second derivative or employ the first derivative test. Evaluating \( f(x) \) at \(x = e\):
\[f(e) = \frac{\log(e)}{e} = \frac{1}{e}.\]
Therefore, the maximum value of the function is \(\frac{1}{e}\).