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

Revision as of 18:59, 14 September 2007

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 $2\log_ax \displaystyle = y \Longrightarrow a^{y/2} = x$ 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:

$x^2 - x - 6 = 0$
$x = -2, 3$

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 $x = 3$ .

$\displaystyle 3\log_{a}x - 2\log_{a}x = 6$
$\log_{a}\frac{x^3}{x^2} = 6$
$a^6 = x$

So $a = \sqrt[6]{3}\ \ \mathrm{(D)}$ .

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

@@ Line 16: / Line 16: @@
 == See also ==
-{{AMC12 box|year=2007|num-b=22|num-a=24}}
+{{AMC12 box|year=2007|num-b=22|num-a=24|ab=A}}
 [[Category:Introductory Algebra Problems]]
 [[Category:Introductory Geometry Problems]]

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

Revision as of 18:59, 14 September 2007

Problem

Solution

See also