Equation $c^x = x^n$, $n \in \mathbb{I}^+$
Column 1:
(A) $n = 1$
(B) $n = 2$
(C) odd $n \geq 3$
(D) even $n \geq 4$
Column 2 (Number of real roots):
(p) 3
(q) 2
(r) 1
(s) 0
Step-by-Step Solution
Key Concept: Transform $e^x = x^n$ into $f(x) = x^n e^{-x}$ and analyze critical points using $f'(x) = x^{n-1}(n-x)e^{-x}$ to determine when $f(n) = n^n e^{-n}$ exceeds, equals, or falls below 1, which directly counts the number of intersection points.
Given $f(x) = \frac{x^n}{e^x}$ with $f(0) = 0$. Computing $f'(x) = \frac{x^{n-1}(n-x)}{e^x}$, we find $f$ is increasing for $0 a$ where $a = n$. The equation $f(x) = 1$ has two solutions when $f(a) > 1$, one solution when $f(a) = 1$, and no solutions when $f(a) < 1$. Since $\lim_{x \to \infty} f(x) = 0$, the maximum of $f$ occurs at $x = n$.
Correct Answer: [A-s] [B-r] [C-q] [D-p]