The function $f(x) = \frac{\log (\pi + x)}{\log (e + x)}$ is (a) increasing on $(0, \infty)$ (b) decreasing on $(0, \infty)$ (c) increasing on $(0, \frac{\pi}{e})$, decreasing on $\left(\frac{\pi}{e}, \infty\right)$ (d) decreasing on $(0, \frac{\pi}{e})$, increasing on $\left(\frac{\pi}{e}, \infty\right)$

Key Concepts Needed

1. Natural Logarithm ($\ln$)
   • Grows slowly and is defined for positive numbers.
2. Derivative and Monotonicity
   • If $f'(x)>0$ for every $x$ in an interval, $f$ is increasing there.
   • If $f'(x)<0$, $f$ is decreasing.
3. Quotient Rule for Derivatives
   • For $f(x)=\dfrac{u(x)}{v(x)}$,
     $f'(x)=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^{2}}$
4. Behaviour of the function $g(s)=s\ln s$
   • For $s>e^{-1}$, we have $g'(s)=\ln s + 1>0$, so $g(s)$ is strictly increasing.

Chain of Thought to Solve

1. Identify parts:
   $u(x)=\ln(\pi+x),\; v(x)=\ln(e+x)$

2. Compute derivatives:
   $u'(x)=\frac{1}{\pi+x},\quad v'(x)=\frac{1}{e+x}$

3. Apply quotient rule:

   f'(x) &= \frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^{2}} \\ &= \frac{\dfrac{1}{\pi+x}\,\ln(e+x)-\dfrac{1}{e+x}\,\ln(\pi+x)}{[\ln(e+x)]^{2}} \end{aligned}$$ Only the numerator matters for the sign.

4. Clear denominators for sign study:

   $\text{Sign of }f'(x)=\text{Sign of }\Big[(e+x)\ln(e+x)-(\pi+x)\ln(\pi+x)\Big]$

   Define

   $h(x)=(e+x)\ln(e+x)-(\pi+x)\ln(\pi+x)$

5. Compare the two terms using monotonicity of $g(s)=s\ln s$:

   • $g(s)$ is increasing for $s>e^{-1}$.

   • For every $x>0$, we have $e+x < \pi+x$ because $\pi>e$.

   • Since $g$ is increasing,

     $g(e+x)<g(\pi+x)\;\Longrightarrow\;h(x)<0.$

6. Conclusion: $h(x)$ is always negative $\Rightarrow f'(x)<0$ for every $x>0$. Therefore $f(x)$ is strictly decreasing on $(0,\infty)$.

Hence the correct option is (b).

What is the question?

We have a rule that turns any number $x$ (bigger than $0$) into another number:

$f(x)=\frac{\text{log of }(\pi + x)}{\text{log of }(e + x)}$

You must tell whether the rule makes bigger outputs when you give it bigger inputs ("increasing") or smaller outputs ("decreasing").

Kid-friendly picture

Imagine two buckets:

• Bucket A always has $\pi$ marbles more than $x$ marbles.
• Bucket B always has $e$ marbles more than $x$ marbles.

We look at the log of each bucket (think of log as a special magnifying glass that grows slowly).

Then we make a fraction:
"zoomed size of Bucket A" ÷ "zoomed size of Bucket B".

The game: as you pour more marbles (increase $x$), does the fraction climb up or slide down?

Because Bucket A always starts with more marbles than Bucket B ($\pi>e$) and the magnifying glass behaves in a certain way, the fraction actually slides down all the time.
So the rule is decreasing.
(Option b)

Step-by-Step Solution

1. Define

   $f(x)=\frac{\ln(\pi+x)}{\ln(e+x)}, \qquad x\in(0,\infty)$

2. Differentiate using the quotient rule

   $f'(x)=\frac{\big(\dfrac{1}{\pi+x}\big)\ln(e+x)-\big(\dfrac{1}{e+x}\big)\ln(\pi+x)}{[\ln(e+x)]^{2}}$

3. Focus on the numerator (denominator is positive):

   $N(x)=\frac{\ln(e+x)}{\pi+x}-\frac{\ln(\pi+x)}{e+x}$

4. Multiply by $(\pi+x)(e+x)$ to avoid fractions:

   $h(x)=(e+x)\ln(e+x)-(\pi+x)\ln(\pi+x)$

5. Use monotonicity of $g(s)=s\ln s$

   • For $s>e^{-1}$, $g'(s)=\ln s + 1>0 \;\Rightarrow\; g(s)$ increases.

   • Because $e+x < \pi+x$ (since $\pi>e$) for every $x>0$, we get

     $g(e+x)<g(\pi+x) \;\Rightarrow\; h(x)<0.$

6. Sign of derivative:

   $f'(x)=\frac{h(x)}{[\ln(e+x)]^{2}}<0 \quad \forall\;x>0.$

7. Conclusion: $f(x)$ is strictly decreasing on $(0,\infty)$.

Correct option: (b).