Question

The minimum value of $\frac{x}{\log \left(x\right)}$ is

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

Answer 1

The correct option is A

$e$

Evaluate the minimum value of the given function

Given : $f\left(x\right)=\frac{x}{\log \left(x\right)}$

Differentiate the function with respect to $x$,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{\log \left(x\right)-x{\displaystyle \frac{1}{x}}}{{\left[\log \left(x\right)\right]}^2}\\ & =& \frac{\log \left(x\right)-1}{{\left[\log \left(x\right)\right]}^2}\end{array}$

For maxima or minima put $f\text{'}\left(x\right)=0$,

$$\begin{gathered} \frac{\log \left(x\right)-1}{{\left[\log \left(x\right)\right]}^2}=0 \\ \Rightarrow \log \left(x\right)=1 \\ \Rightarrow x=e \end{gathered}$$

Differentiate $f\text{'}\left(x\right)$ with respect to $x$,

$$\begin{gathered} f\text{'}\text{'}\left(x\right)=\frac{{\left[\log \left(x\right)\right]}^2.{\displaystyle \frac{1}{x}}-\left(\log \left(x\right)-1\right).{\displaystyle \frac{2\log \left(x\right)}{x}}}{{\left[\log \left(x\right)\right]}^4} \\ \Rightarrow f\text{'}\text{'}\left(e\right)=\frac{1.{\displaystyle \frac{1}{e}}-(1-1).{\displaystyle \frac{2}{e}}}{1} \\ \Rightarrow f\text{'}\text{'}\left(e\right)=\frac{1}{e} \end{gathered}$$

Since value of $f\text{'}\text{'}\left(x\right)$ is positive at $x=e$,.

Therefore, function has minima at $x=e$,

$\begin{array}{rcl}f{\left(x\right)}_{min}& =& \frac{e}{\log \left(e\right)}\\ & =& e\end{array}$

Therefore, the minimum value of the function is $e$.

Hence option $\left(A\right)$ is the correct answer.