The function `f(x)=(log(pi+x))/(log(e+x))`s is

Options

A. increasing in `(0,oo)`
B. decreasing on `(0,oo)`
C. increasing on `(0,pi/e)", decreasing on"(pi/e,oo)`
D. None of these

Answer

To determine whether the function \( f(x) = \frac{\log(\pi + x)}{\log(e + x)} \) is increasing or decreasing, we need to find its derivative and analyze the sign of the derivative.

### Step 1: Find the derivative \( f'(x) \)

Using the quotient rule for differentiation, which states that if \( f(x) = \frac{g(x)}{h(x)} \), then

\[ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} \]

we can identify \( g(x) = \log(\pi + x) \) and \( h(x) = \log(e + x) \).

1. **Differentiate \( g(x) \)**:
\[ g'(x) = \frac{1}{\pi + x} \]

2. **Differentiate \( h(x) \)**:
\[ h'(x) = \frac{1}{e + x} \]

Now substituting these into the quotient rule:

\[ f'(x) = \frac{\left(\frac{1}{\pi + x}\right) \log(e + x) - \log(\pi + x) \left(\frac{1}{e + x}\right)}{(\log(e + x))^2} \]

This simplifies to:

\[ f'(x) = \frac{\log(e + x)}{(\pi + x)} - \frac{\log(\pi + x)}{(e + x)} \]

### Step 2: Analyze the sign of \( f'(x) \)

To determine where \( f'(x) \) is positive or negative, we need to compare the two fractions:

\[ \frac{\log(e + x)}{\pi + x} \quad \text{and} \quad \frac{\log(\pi + x)}{e + x} \]

Since \( e \approx 2.718 \) and \( \pi \approx 3.14 \), we know that \( \pi > e \). Thus, for \( x > 0 \):

1. **For \( x = 0 \)**:
\[ f'(0) = \frac{\log(e)}{\pi} - \frac{\log(\pi)}{e} \]
Since \( \log(e) = 1 \) and \( \log(\pi) > 1 \), we can see that \( f'(0) < 0 \).

2. **For \( x > 0 \)**:
As \( x \) increases, both \( \log(e + x) \) and \( \log(\pi + x) \) increase. However, since \( \pi + x > e + x \) for all \( x > 0 \), we can conclude that \( f'(x) < 0 \) for \( x > 0 \).

### Conclusion

Since \( f'(x) < 0 \) for \( x > 0 \), the function \( f(x) \) is decreasing on the interval \( (0, \infty) \).

Thus, the final answer is that the function \( f(x) \) is decreasing in the interval \( (0, \infty) \).

---