The minimum value of displaystylex/(log x)\ (x gt 1) is:

The correct answer is Option (1) → e

Let \(f(x) = \frac{x}{\log x}\). By the quotient rule:

    \( \displaystyle f'(x) = \frac{\log x - x \cdot \frac{1}{x}}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2} \)

    \(f'(x) = 0 \Rightarrow \log x = 1 \Rightarrow x = e\)

For 1 < x < e, log x < 1, so f'(x) < 0; for x > e, f'(x) > 0. By the first derivative test, f has its minimum at x = e:

    \( \displaystyle f(e) = \frac{e}{\log e} = \frac{e}{1} = e \)