2003 AMC 12B 第 17 题

先试着解答 2003 AMC 12B 第 17 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 2003 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。

17.

若 log⁡(xy3)=1\log(xy^3) = 1 且 log⁡(x2y)=1\log(x^2y) = 1,log⁡(xy)\log(xy) 是多少?
If log⁡(xy3)=1\log(xy^3) = 1 and log⁡(x2y)=1,\log(x^2y) = 1, what is log⁡(xy)?\log(xy)?
−12-\dfrac{1}{2}
00
12\dfrac{1}{2}
35\dfrac{3}{5}
11
答案:D
知识点:对数方程组
小提示:
令 X=log⁡xX = \log x,Y=log⁡yY = \log y;方程会变成关于 XX 和 YY 的线性方程
Let X=log⁡xX = \log x and Y=log⁡y;Y = \log y; the equations become linear in XX and YY
大提示:
解 X+3Y=1X + 3Y = 1 与 2X+Y=12X + Y = 1,再求 X+YX + Y
Solve X+3Y=1X + 3Y = 1 and 2X+Y=1,2X + Y = 1, then compute X+YX + Y
解答:
令 X=log⁡xX = \log x,Y=log⁡yY = \log y,则 X+3Y=1X + 3Y = 1 且 2X+Y=1。2X + Y = 1\text{。}
解得 X=25X = \dfrac{2}{5},Y=15Y = \dfrac{1}{5},所以 log⁡(xy)=X+Y=35。 \log(xy) = X + Y = \frac{3}{5}\text{。}
因此,正确答案是 D。
Let X=log⁡xX = \log x and Y=log⁡y.Y = \log y. Then X+3Y=1X + 3Y = 1 and 2X+Y=1.2X + Y = 1.
Solving gives X=25X = \dfrac{2}{5} and Y=15,Y = \dfrac{1}{5}, so log⁡(xy)=X+Y=35. \log(xy) = X + Y = \frac{3}{5}.
Thus, the correct answer is D.
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