It is possible to choose x>23 in such a way that the value of log_10(x^2+3)-2log_10x is:

1965 AMC 12 Problem 21

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21.

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:

negative

zero

one

smaller than any positive number that might be specified

greater than any positive number that might be specified

Combine the logarithms into

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

Let

x

grow without bound

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

x

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

Thus, the correct answer is D.