Prove that the function

case I

When $a>1$

let $x_{1}, x_{2} \in(0, \infty)$

We have, $x_{1}

$\Rightarrow \log _{\mathrm{e}} \mathrm{x}_{1}<\log _{\mathrm{e}} \mathrm{x}_{2}$

$\Rightarrow \mathrm{f}\left(\mathrm{x}_{1}\right)<\mathrm{f}\left(\mathrm{x}_{2}\right)$

So, $f(x)$ is increasing in $(0, \infty)$

case II

When $0

$f(x)=\log _{a} x=\frac{\log x}{\log a}$

when $a<1 \Rightarrow \log a<0$

let $x_{1}

$\Rightarrow \log x_{1}<\log x_{2}$

$\Rightarrow \frac{\log x_{1}}{\log a}>\frac{\log x_{2}}{\log a}[\because \log a<0]$

$\Rightarrow \mathrm{f}\left(\mathrm{x}_{1}\right)>\mathrm{f}\left(\mathrm{x}_{2}\right)$