1953 AHSME Problems/Problem 21

Problem

If $\log_{10} (x^2-3x+6)=1$ , the value of $x$ is:

$\textbf{(A)}\ 10\text{ or }2 \qquad \textbf{(B)}\ 4\text{ or }-2 \qquad \textbf{(C)}\ 3\text{ or }-1 \qquad \textbf{(D)}\ 4\text{ or }-1\\ \textbf{(E)}\ \text{none of these}$

Solution

We know that $x^2-3x+6=10^1$ , after expanding the logarithm. Taking the 10 to the other side, $x^2-3x-4=0$ . Factoring, we get $(x-4)(x+1)=0$ , so $x=\boxed{4,-1\Rightarrow \text{(D)}}$ .

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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1953 AHSME Problems/Problem 21

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