1965 AHSME 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

\begin{array}{l} \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} \end{array}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.