For $x> 0$, the minimum value of $\frac{x}{\log_e x}$ is

$e$

The correct answer is Option (3) → $e$

Given: $x > 0$

Objective: Minimize the expression $\frac{x}{\log_e x}$

Let $f(x) = \frac{x}{\log_e x}$

Using calculus to find the minimum.

Let $f(x) = \frac{x}{\ln x}$

Differentiate using quotient rule:

$f'(x) = \frac{\ln x \cdot 1 - x \cdot \frac{1}{x}}{(\ln x)^2}$

$= \frac{\ln x - 1}{(\ln x)^2}$

Set $f'(x) = 0$:

$\ln x - 1 = 0$

$\ln x = 1$

$x = e$

Check the second derivative to confirm minimum:

$f''(x) = \frac{[(\ln x)^2 \cdot \frac{1}{x}] - [2(\ln x)(\ln x - 1) \cdot \frac{1}{x}]}{(\ln x)^4}$

At $x = e$, $\ln x = 1$

Denominator is positive, numerator becomes positive ⇒ $f''(x) > 0$

So, $x = e$ is a point of minimum.

Minimum value = $\frac{e}{\ln e} = \frac{e}{1} = e$