The maximum value of $\frac{\log x^3}{3 x}$ occurs at x = ________.

e

The correct answer is Option (1) → e

$f(x)=\frac{\log x^3}{3x}$

for critical points, $f'(x)=0$

$⇒f'(x)=\frac{\frac{1}{x^3}×3x^2×3x-3\log x^3}{(3x)^2}$

$=\frac{9-9\log x}{(3x)^2}=\frac{1-\log x}{x}=0$

$⇒\log x=1$

$⇒x=e$

Now for $x=e$ to be maximum, $f''(x)<0$,

$f''(x)=\frac{-\frac{1}{x}×x-(1-\log x)}{x^2}=\frac{-2+\log x}{x^2}$

$f''(x=e)=\frac{-2\log e}{e^2}=\frac{-2+1}{e^2}=\frac{-1}{e^2}<0$

$∴x=e$ → Maximum occurs