2025 AMC 12B 第 24 题

A solution needs $-1 \le \log_{20} x \le 1,$ so $x$ lies in $\left[\tfrac{1}{20}, 20\right]$

On each monotonic half-wave of $\sin(20\pi x)$ the slowly rising log crosses once; count the branches carefully at both ends

Since $|\sin(20\pi x)|\le1,$ every solution lies in $\left[\tfrac1{20},20\right].$ For $x\lt1$ the logarithm is negative, so only the $10$ negative sine lobes in $\left[\tfrac1{20},1\right]$ can contribute. Each has one crossing on each side of its minimum; on the last lobe, the second crossing is the endpoint $x=1.$ Hence this part contributes $20$ solutions.

For $x\gt1,$ only positive lobes contribute. There are $190$ of them: one in each interval from $1+\tfrac{k}{10}$ to $1+\tfrac{k}{10}+\tfrac1{20}$ for $0\le k\le189.$ Every one after the first has one crossing on each side of its maximum. The first has only one new crossing because its left endpoint is the already-counted solution $x=1.$ Thus $x\gt1$ contributes $1+2\cdot189=379$ more solutions.

For completeness, the claimed uniqueness on each half-lobe follows by differentiating the difference of the two sides. On a falling positive half-lobe it is strictly decreasing. On a rising positive half-lobe its derivative increases and then decreases once, so the difference crosses from negative to positive only once. Applying the same argument to $-\sin(20\pi x)+\log_{20}x$ handles a negative lobe. The total is $20+379=399.$

Thus, the correct answer is D.