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

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== Solution ==
 
== Solution ==
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Let <math>x</math> be the x-coordinate of <math>B</math> and <math>C</math>, and <math>x_2</math> be the x-coordinate of <math>A</math> and <math>y</math> be the y-coordinate of <math>A</math> and <math>B</math>. Then <math>2\log_ax \displaystyle = y \Longrightarrow a^{y/2} = x</math> and <math>\log_ax_2 = y \Longrightarrow x_2 = a^y = \left(a^{y/2}\right)^2 = x^2</math>. Since the distance between <math>A</math> and <math>B</math> is 6:
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Let <math>x</math> be the x-coordinate of <math>B</math> and <math>C</math>, and <math>x_2</math> be the x-coordinate of <math>A</math> and <math>y</math> be the y-coordinate of <math>A</math> and <math>B</math>. Then <math>2\log_ax= y \Longrightarrow a^{y/2} = x</math> and <math>\log_ax_2 = y \Longrightarrow x_2 = a^y = \left(a^{y/2}\right)^2 = x^2</math>. Since the distance between <math>A</math> and <math>B</math> is <math>6</math>, we have <math>x^2 - x - 6 = 0</math>, yielding <math>x = -2, 3</math>.
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<div style="text-align:center;"><math>x^2 - x - 6 = 0 </math><br /><math>x = -2, 3</math></div>
 
  
 
However, we can discard the negative root (all three [[logarithm]]ic equations are underneath the line <math>y = 3</math> and above <math>y = 0</math> when <math>x</math> is negative, hence we can't squeeze in a square of side 6). Thus <math>x = 3</math>.
 
However, we can discard the negative root (all three [[logarithm]]ic equations are underneath the line <math>y = 3</math> and above <math>y = 0</math> when <math>x</math> is negative, hence we can't squeeze in a square of side 6). Thus <math>x = 3</math>.
  
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<div style="text-align:center;"><math>\displaystyle 3\log_{a}x - 2\log_{a}x = 6</math><br /><math>\log_{a}\frac{x^3}{x^2} = 6 </math><br /><math>a^6 = x</math></div>
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Substituting back, <math>3\log_{a}x - 2\log_{a}x = 6 \Longrightarrow a^6 = x</math>, so <math>a = \sqrt[6]{3}\ \ \mathrm{(D)}</math>.
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So <math>a = \sqrt[6]{3}\ \ \mathrm{(D)}</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 17:25, 2 January 2008

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

See also

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
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