View source for 1961 AHSME Problems/Problem 19

== Problem ==

Consider the graphs of <math>y=2\log{x}</math> and <math>y=\log{2x}</math>. We may say that:

<math>\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}   </math>  

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

==See Also==
{{AHSME 40p box|year=1961|num-b=18|num-a=20}}

[[Category:Introductory Algebra Problems]] 

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