The function f(x) = frac{log(pi + x)}{log(e + | vedclass

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(D) Given $f(x) = \frac{\log(\pi + x)}{\log(e + x)}$.Applying the quotient rule,$f'(x) = \frac{\frac{1}{\pi + x} \log(e + x) - \log(\pi + x) \cdot \fr

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decreasing on $(0, \infty)$

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(D) Given $f(x) = \frac{\log(\pi + x)}{\log(e + x)}$.Applying the quotient rule,$f'(x) = \frac{\frac{1}{\pi + x} \log(e + x) - \log(\pi + x) \cdot \frac{1}{e + x}}{\{\log(e + x)\}^2}$.$f'(x) = \frac{(e + x) \log(e + x) - (\pi + x) \log(\pi + x)}{(\pi + x)(e + x) \{\log(e + x)\}^2}$.Let $g(x) = (e + x) \log(e + x) - (\pi + x) \log(\pi + x)$.Then $g'(x) = \log(e + x) + 1 - (\log(\pi + x) + 1) = \log(e + x) - \log(\pi + x)$.Since $\pi > e$,for $x > 0$,$\pi + x > e + x$,so $\log(\pi + x) > \log(e + x)$.Thus,$g'(x) Since $g(0) = e \log e - \pi \log \pi = e - \pi \log \pi  0$.Therefore,$f'(x) < 0$ for all $x \in (0, \infty)$,implying $f(x)$ is decreasing on $(0, \infty)$.

The function $f(x) = \frac{\log(\pi + x)}{\log(e + x)}$ is

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