# 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

By the rules of logarithms, $\log_{10}(x^2+3)-2\log_{10} x=\log_{10}(\frac{x^2+3}{x^2})=\log_{10}(1+\frac{3}{x^2})$. As $x$ goes to infinity, $1+\frac{3}{x^2}$ gets arbitrarily close to $1$ (without ever reaching it), so $\log_{10}(1+\frac{3}{x^2})$ gets arbitrarily close to $\log_{10}(1)=0$ (without ever reaching it). Furthermore, because $1+\frac3{x^2} > 1$, $\log(1+\frac3{x^2})$ is never negative. Thus, we can choose a real $x>\frac{2}{3}$ such that the given expression is $\fbox{\textbf{(D) }smaller than any positive number that might be specified}$.