Difference between revisions of "1950 AHSME Problems/Problem 44"

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\textbf{(D)}\ \text{Cuts neither axis} \qquad\\
 
\textbf{(D)}\ \text{Cuts neither axis} \qquad\\
 
\textbf{(E)}\ \text{Cuts all circles whose center is at the origin}</math>
 
\textbf{(E)}\ \text{Cuts all circles whose center is at the origin}</math>
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== Solution ==
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The domain of <math>\log x</math> is the set of all positive reals, so the graph of <math>y=\log x</math> clearly doesn't cut the <math>y</math>-axis. It therefore doesn't cut every line perpendicular to the <math>x</math>-axis. It does however cut the <math>x</math>-axis at <math>(1,0)</math>. In addition, if one examines the graph of <math>y=\log x</math>, one can clearly see that there are many circles centered at the origin that do not intersect the graph of <math>y=\log x</math>. Therefore the answer is <math>\boxed{\textbf{(C)}\ \text{Cuts the }x\text{-axis}}</math>.
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{{image}}
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== See Also ==
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{{AHSME 50p box|year=1950|num-b=43|num-a=45}}
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[[Category:Introductory Algebra Problems]]

Revision as of 11:23, 23 April 2012

Problem

The graph of $y\equal{}\log x$ (Error compiling LaTeX. Unknown error_msg)

$\textbf{(A)}\text{Cuts the }y\text{-axis} \qquad\\ \textbf{(B)}\ \text{Cuts all lines perpendicular to the }x\text{-axis} \qquad\\ \textbf{(C)}\ \text{Cuts the }x\text{-axis} \qquad\\ \textbf{(D)}\ \text{Cuts neither axis} \qquad\\ \textbf{(E)}\ \text{Cuts all circles whose center is at the origin}$

Solution

The domain of $\log x$ is the set of all positive reals, so the graph of $y=\log x$ clearly doesn't cut the $y$-axis. It therefore doesn't cut every line perpendicular to the $x$-axis. It does however cut the $x$-axis at $(1,0)$. In addition, if one examines the graph of $y=\log x$, one can clearly see that there are many circles centered at the origin that do not intersect the graph of $y=\log x$. Therefore the answer is $\boxed{\textbf{(C)}\ \text{Cuts the }x\text{-axis}}$.


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See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 43
Followed by
Problem 45
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All AHSME Problems and Solutions