Question

Show that the function $f\left(x\right)=\frac{\log (\pi +x)}{\log (e+x)}$ is a decreasing function in the interval $]0,\infty [$.

Answer 1

$x\in$ ]$0,\infty$[
$\Rightarrow x>0$

Given $f\left(x\right)=\frac{\log (\pi +x)}{\log (e+x)}$

$\frac{dy}{dx}=\frac{\log (e+x)\frac{1}{\pi +x}-\log (\pi +x).\frac{1}{e+x})}{{\left[\log (e+x)\right]}^2}$

$=\frac{(e+x)\log (e+x)-(\pi +x)\log (\pi +x)}{(\pi +x)(e+x){\left[\log (e+x)\right]}^2}$

Since, $e,\pi >0$ and $x>0$

$e+x>0$ , $\pi +x>0$

So, denominator is positive

Since, $e<\pi$ and $x>0$

$e+x<\pi +x$

So, numerator is negative.
So, $\frac{dy}{dx}<0$
Hence $f\left(x\right)$ is decreasing when $x>0$