Difference between revisions of "2007 AMC 12A Problems/Problem 23"

Revision as of 20:32, 3 July 2013

Problem

Square $ABCD$ has area $36,$ and $\overline{AB}$ is parallel to the x-axis. Vertices $A,$ $B$ , and $C$ are on the graphs of $y = \log_{a}x,$ $y = 2\log_{a}x,$ and $y = 3\log_{a}x,$ respectively. What is $a?$

$\mathrm{(A)}\ \sqrt [6]{3}\qquad \mathrm{(B)}\ \sqrt {3}\qquad \mathrm{(C)}\ \sqrt [3]{6}\qquad \mathrm{(D)}\ \sqrt {6}\qquad \mathrm{(E)}\ 6$

Solution

Let $x$ be the x-coordinate of $B$ and $C$ , and $x_2$ be the x-coordinate of $A$ and $y$ be the y-coordinate of $A$ and $B$ . Then $2\log_ax= y \Longrightarrow a^{y/2} = x$ and $\log_ax_2 = y \Longrightarrow x_2 = a^y = \left(a^{y/2}\right)^2 = x^2$ . Since the distance between $A$ and $B$ is $6$ , we have $x^2 - x - 6 = 0$ , yielding $x = -2, 3$ .

However, we can discard the negative root (all three logarithmic equations are underneath the line $y = 3$ and above $y = 0$ when $x$ is negative, hence we can't squeeze in a square of side 6). Thus $x = 3$ .

Substituting back, $3\log_{a}x - 2\log_{a}x = 6 \Longrightarrow a^6 = x$ , so $a = \sqrt[6]{3}\ \ \mathrm{(A)}$ .

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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Difference between revisions of "2007 AMC 12A Problems/Problem 23"

Revision as of 20:32, 3 July 2013

Problem

Solution

See also