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). Thus, we can choose a real $x>\frac{2}{3}$ such that the given expression is $\fbox{\textbf{(D) }smaller than any positive number that might be specified}$ .

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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1965 AHSME Problems/Problem 21

Problem 21

Solution

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