Difference between revisions of "1953 AHSME Problems/Problem 21"

(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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==Solution==
 
==Solution==
 
We know that <math>x^2-3x+6=10^1</math>, after expanding the logarithm. Taking the 10 to the other side, <math>x^2-3x-4=0</math>. Factoring, we get <math>(x-4)(x+1)=0</math>, so <math>x=\boxed{4,-1\Rightarrow \text{(D)}}</math>.
 
We know that <math>x^2-3x+6=10^1</math>, after expanding the logarithm. Taking the 10 to the other side, <math>x^2-3x-4=0</math>. Factoring, we get <math>(x-4)(x+1)=0</math>, so <math>x=\boxed{4,-1\Rightarrow \text{(D)}}</math>.
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==See Also==
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{{AHSME 50p box|year=1953|num-b=18|num-a=20}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 21:27, 15 January 2018

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 $x=\boxed{4,-1\Rightarrow \text{(D)}}$.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AHSME Problems and Solutions

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