# 1950 AHSME Problems/Problem 44

## Problem

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

## Solution

The domain of $\log x$ is the set of all $\underline{positive}$ 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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