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1953 AHSME Problems/Problem 21

Revision as of 22:26, 15 January 2018 by Rekt4 (talk | contribs) (Created page with "==Problem== If <math>\log_{10} (x^2-3x+6)=1</math>, the value of <math>x</math> is: <math>\textbf{(A)}\ 10\text{ or }2 \qquad \textbf{(B)}\ 4\text{ or }-2 \qquad \textbf{(C)}...")
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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 $x=\boxed{4,-1\Rightarrow \text{(D)}}$.

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