AMC12 2014 B

AMC12 2014 B · Q20

AMC12 2014 B · Q20. It mainly tests Linear inequalities, Logarithms (rare).

For how many positive integers $x$ is $\log_{10}(x - 40) + \log_{10}(60 - x) < 2$?

有正整数 $x$ 多少个满足 $\log_{10}(x - 40) + \log_{10}(60 - x) < 2$？

(A) 10 10

(B) 18 18

(D) 20 20

(E) infinitely many 无穷多个

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The domain of $\log_{10}(x-40)+\log_{10}(60-x)$ is $40<x<60$. Within this domain, the inequality $\log_{10}(x-40)+\log_{10}(60-x)<2$ is equivalent to each of the following: $\log_{10}((x-40)(60-x))<2$, $(x-40)(60-x)<10^2=100$, $x^2-100x+2500>0$, and $(x-50)^2>0$. The last inequality is true for all $x\ne50$. Thus the integer solutions to the original inequality are $41,42,\ldots,49,51,52,\ldots,59$, and their number is 18.

答案（B）：$\log_{10}(x-40)+\log_{10}(60-x)$ 的定义域是 $40<x<60$。在此定义域内，不等式 $\log_{10}(x-40)+\log_{10}(60-x)<2$ 等价于以下各式：$\log_{10}((x-40)(60-x))<2$，$(x-40)(60-x)<10^2=100$，$x^2-100x+2500>0$，以及 $(x-50)^2>0$。最后一个不等式对所有 $x\ne50$ 都成立。因此原不等式的整数解为 $41,42,\ldots,49,51,52,\ldots,59$，共有 18 个。

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