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

Options

A. increasing in `[0, pi]`
B. decreasing in `(0, oo)`
C. increasing of `[0, (pi)/(e ))` and decreasing on `[(pi)/ (e ),oo)`
D. decreasing in `[0, (pi)/(e )]` and increasing on `[(pi)/(e ),oo]`

Answer

To analyze the function \( f(x) = \frac{\log(\pi + x)}{\log(e + x)} \), we will follow these steps:

### Step 1: Differentiate the function
We need to find the derivative \( f'(x) \) using the quotient rule. The quotient rule states that if you have a function \( \frac{u}{v} \), then its derivative is given by:

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

Here, let \( u = \log(\pi + x) \) and \( v = \log(e + x) \).

### Step 2: Find \( u' \) and \( v' \)
Now we differentiate \( u \) and \( v \):

\[ u' = \frac{1}{\pi + x} \quad \text{and} \quad v' = \frac{1}{e + x} \]

### Step 3: Apply the quotient rule
Substituting \( u \), \( v \), \( u' \), and \( v' \) into the quotient rule formula:

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

### Step 4: Simplify the expression
Now, we simplify the numerator:

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

### Step 5: Analyze the sign of \( f'(x) \)
To determine whether \( f(x) \) is increasing or decreasing, we need to analyze the sign of \( f'(x) \).

For \( x > 0 \):
- We know that \( \pi \) (approximately 3.14) is greater than \( e \) (approximately 2.7).
- Therefore, \( \pi + x > e + x \) for \( x > 0 \).

This implies:
- \( \frac{1}{\pi + x} < \frac{1}{e + x} \)

Thus, we can conclude:
- \( \log(e + x) > \log(\pi + x) \) because the logarithm is an increasing function.

### Step 6: Conclusion about the function
Since \( f'(x) < 0 \) for \( x > 0 \), we conclude that \( f(x) \) is a decreasing function for \( x \in (0, \infty) \).

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

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