Question

The minimum value of $\frac{x}{\log \, x}$ is

• A. $e$
• B. $\frac{1}{e}$
• C. $e^2$
• D. $e^3$

Answer 1

Answer: (A)

$e$

Answer 2

Let $f(x) = \frac{x}{\log \, x}$
On differentiating w.r.t. $x$, we get
$f'(x) = \frac{\log \, x - 1}{(\log \, x )^2}$
For maxima and minima, put $f'(x) = 0$
$\log \, x - 1 = 0$
$\Rightarrow \, x = e$
Now ,
$f''(x)=\frac{(\log x)^{2} \cdot \frac{1}{x}-(\log x-1) \cdot \frac{2 \log x}{x}}{(\log x)^{4}}$
$\Rightarrow f" \left(e\right) = \frac{\frac{1}{e} - 0}{1} = \frac{1}{e} > 0$
$\therefore f(x)$ is minimum at $x = e$.
Hence, minimum value of f (x) at x = e is
$f\left(e\right) = \frac{e}{\log e } = e$