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

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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{(A)}</math>.
 
Substituting back, <math>3\log_{a}x - 2\log_{a}x = 6 \Longrightarrow a^6 = x</math>, so <math>a = \sqrt[6]{3}\ \ \mathrm{(A)}</math>.
  
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==Solution 2==
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Notice that all of the graphs <math>y = \log_{a}x,</math> <math>y = 2\log_{a}x,</math> and <math>y = 3\log_{a}x</math> have a domain of <math>(0, \infty]</math>. Also notice that <math>y = \log_{a}x</math> is the furthest to the right, as adding coefficients in front of the <math>\log</math> part only makes the graph steeper. Since <math>A</math> is on the graph of <math>y = \log_{a}x</math> and <math>B</math> is on the graph of <math>y = 2\log_{a}x</math>, <math>\,</math> <math>A</math> must be to the right of <math>B</math>. We are told that <math>\overline{AB}</math> is [[parallel]] to the [[x-axis]]. Let <math>A</math> be the point <math>(x, y)</math>. Then
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2007|num-b=22|num-a=24|ab=A}}
 
{{AMC12 box|year=2007|num-b=22|num-a=24|ab=A}}

Revision as of 16:54, 21 July 2020

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

Solution 2

Notice that all of the graphs $y = \log_{a}x,$ $y = 2\log_{a}x,$ and $y = 3\log_{a}x$ have a domain of $(0, \infty]$. Also notice that $y = \log_{a}x$ is the furthest to the right, as adding coefficients in front of the $\log$ part only makes the graph steeper. Since $A$ is on the graph of $y = \log_{a}x$ and $B$ is on the graph of $y = 2\log_{a}x$, $\,$ $A$ must be to the right of $B$. We are told that $\overline{AB}$ is parallel to the x-axis. Let $A$ be the point $(x, y)$. Then

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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