1961 AHSME Problems/Problem 19

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Problem

Consider the graphs of $y=2\log{x}$ and $y=\log{2x}$. We may say that:

$\textbf{(A)}\ \text{They do not intersect}\qquad \\ \textbf{(B)}\ \text{They intersect at 1 point only}\qquad \\ \textbf{(C)}\ \text{They intersect at 2 points only} \qquad \\ \textbf{(D)}\ \text{They intersect at a finite number of points but greater than 2} \qquad \\ \textbf{(E)}\ \text{They coincide}$

Solution

Substitute $y$ into the other equation. \[2\log{x} = \log{2x}\] Using a logarithm property, \[2\log{x} = \log{2} + \log{x}\] Subtract both sides by $\log{x}$ to get \[\log{x}=\log{2}\] Thus, $x=2$. Since both $2\log{x}$ and $\log{2x}$ are functions, there is only one corresponding y-value, so the answer is $\boxed{\textbf{(B)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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