Question

If $f\left(x\right)={\log }_{x}1/9-{\log }_3{x}^2(x>1)$, then max $f\left(x\right)$ is equal to

Answer 1

Let, $y={\log }_{x}1/9-{\log }_3{x}^2(x>1)$
$={\log }_x\left(3^{-2}\right)-2{\log }_{3}x=-2{\log }_{x}3-2{\log }_{3}x$
$=-2({\log }_{x}3+{\log }_{3}x)$
Now since $x>1\Rightarrow {\log }_{3}x,{\log }_{x}3>0$
Thus using, $A.M\ge G.M$
$\Rightarrow \frac{{\log }_{3}x+{\log }_{x}3}{2}\ge \sqrt{{\log }_{x}3{\log }_{3}x}=1,[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
Hence $(-y{)}_{max}=2\times 2=2$