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

Options

A. increasing on `(0,oo)`
B. decreasing on`(0,oo)`
C. increasing on `(0,(pi)/e)`, decreasing on `((pi)/e,oo)`
D. decreasing onn `(0,(pi)/e)`, increasing on `((pi)/e,oo)`

Explanation

To determine whether the function \( f(x) = \frac{\log(\pi + x)}{\log(e + x)} \) is increasing or decreasing, we will follow these steps:

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

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

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

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

### Step 2: Differentiate \( h(x) \) and \( g(x) \)

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

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

### Step 3: Substitute into the quotient rule formula

Now we substitute \( h'(x) \), \( h(x) \), \( g'(x) \), and \( g(x) \) into the quotient rule:

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

### Step 4: Simplify the derivative

This simplifies to:

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

Combining the fractions gives:

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

### Step 5: Analyze the sign of \( f'(x) \)

To determine if \( f(x) \) is increasing or decreasing, we need to analyze the sign of \( f'(x) \):

1. **For \( x > 0 \)**, we know \( \pi \) (approximately 3.14) is greater than \( e \) (approximately 2.71). Thus, \( \pi + x > e + x \).
2. Therefore, \( \log(\pi + x) > \log(e + x) \).

This implies that the numerator \( \log(e + x)(e + x) - \log(\pi + x)(\pi + x) \) will be negative, as the larger logarithmic term is multiplied by a smaller quantity.

### Conclusion

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

### Final Answer

The function \( f(x) = \frac{\log(\pi + x)}{\log(e + x)} \) is **monotonically decreasing** for \( x > 0 \).

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