1965 AMC 12 第 21 题

Question

可以选取

x\gt\dfrac23

，使得

\log_{10}(x^2+3)-2\log_{10}x

的值：

It is possible to choose

x\gt\dfrac23

in such a way that the value of

\log_{10}(x^2+3)-2\log_{10}x

is:

为负数

negative

为零

zero

为一

one

小于任意给定的正数

smaller than any positive number that might be specified

大于任意给定的正数

greater than any positive number that might be specified

将对数合并为

\log_{10}(1+\frac{3}{x^2})

Combine the logarithms into

\log_{10}(1+\frac{3}{x^2})

令

x

无限增大

Let

x

grow without bound

该表达式等于

\begin{gathered} \log_{10}\left(\frac{x^2+3}{x^2}\right)\\ =\log_{10}\left(1+\frac3{x^2}\right) \end{gathered}\text{。}

它始终为正，但随着

x

增大而趋近于

0

。因此可以使它小于任意给定的正数。

因此，正确答案是 D。

The expression is

\begin{gathered} \log_{10}\left(\frac{x^2+3}{x^2}\right)\\ =\log_{10}\left(1+\frac3{x^2}\right). \end{gathered}

It is always positive, but tends to

0

as

x

grows. Therefore it can be made smaller than any specified positive number.

Thus, the correct answer is D.

其他年份的第 21 题