1965 AHSME Problems/Problem 21

Problem 21

It is possible to choose $x > \frac {2}{3}$ in such a way that the value of $\log_{10}(x^2 + 3) - 2 \log_{10}x$ is

$\textbf{(A)}\ \text{negative} \qquad \textbf{(B) }\ \text{zero} \qquad \textbf{(C) }\ \text{one} \\ \textbf{(D) }\ \text{smaller than any positive number that might be specified} \\ \textbf{(E) }\ \text{greater than any positive number that might be specified}$

Solution

$\fbox{D}$

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1965 AHSME Problems/Problem 21

Problem 21

Solution

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