2009 AMC 12B 第 15 题

先试着解答 2009 AMC 12B 第 15 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2009 AMC 12B 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

15.

假设 0<r<30 \lt r \lt 3。下面是五个关于 xx 的方程。哪个方程的解 xx 最大?
Assume 0<r<3.0 \lt r \lt 3. Below are five equations for x.x. Which equation has the largest solution x?x?
3(1+r)x=73(1 + r)^x = 7
3(1+r10)x=73(1 + \frac{r}{10})^x = 7
3(1+2r)x=73(1 + 2r)^x = 7
3(1+r)x=73(1 + \sqrt{r})^x = 7
3(1+1r)x=73(1 + \frac{1}{r})^x = 7
答案:B
知识点:对数不等式
小提示:
解出每个方程得 x=log⁡(73)log⁡(1+f(r))x = \dfrac{\log(\frac{7}{3})}{\log(1 + f(r))};底数越小,xx 越大。
Solving each gives x=log⁡(73)log⁡(1+f(r))x = \dfrac{\log(\frac{7}{3})}{\log(1 + f(r))}; a smaller base makes xx larger
大提示:
比较 r, r10, 2r, r, 1rr,\ \frac{r}{10},\ 2r,\ \sqrt{r},\ \frac{1}{r} 在 0<r<30 \lt r \lt 3 时的大小,找出最小者。
Compare the values r, r10, 2r, r, 1rr,\ \frac{r}{10},\ 2r,\ \sqrt{r},\ \frac{1}{r} for 0<r<30 \lt r \lt 3 and find the smallest
解答:
每个方程都给出 x=log⁡(73)log⁡(1+f(r))x = \dfrac{\log(\frac{7}{3})}{\log(1 + f(r))},所以正数 f(r)f(r) 越小,解越大。
当 0<r<30 \lt r \lt 3 时,有 r10<r<2r\dfrac r{10}\lt r\lt2r,并且因为 r<100r\lt100,还有 r10<r\dfrac r{10}\lt\sqrt r。又因为 r2<9<10r^2\lt9\lt10,所以 r10<1r\dfrac r{10}\lt\dfrac1r。因此五个量中 r10\frac{r}{10} 最小,方程 (B) 的解最大。
所以正确答案是 B。
Each equation gives x=log⁡(73)log⁡(1+f(r)),x = \dfrac{\log(\frac{7}{3})}{\log(1 + f(r))}, which is largest when the positive quantity f(r)f(r) is smallest.
For 0<r<3,0 \lt r \lt 3, we have r10<r<2r\dfrac r{10}\lt r\lt2r and r10<r\dfrac r{10}\lt\sqrt r because r<100.r\lt100. Also r2<9<10,r^2\lt9\lt10, so r10<1r.\dfrac r{10}\lt\dfrac1r. Thus r10\frac{r}{10} is the smallest of the five quantities, and equation (B) has the largest solution.
Thus, the correct answer is B.
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