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

(→‎Solution 2)
(→‎Solution 2)
 
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Using equations 1 and 2, we get <math>x=(x-6)^2</math>. Solving yields <math>x=9, x=-4</math>. However, <math>-4</math> is extraneous so <math>x=9</math> (also because we know <math>A</math> has to have a positive <math>x</math>-coordinate). Using the second and third equations, we get
 
Using equations 1 and 2, we get <math>x=(x-6)^2</math>. Solving yields <math>x=9, x=-4</math>. However, <math>-4</math> is extraneous so <math>x=9</math> (also because we know <math>A</math> has to have a positive <math>x</math>-coordinate). Using the second and third equations, we get
  
−
<math>\frac{a^y*a^6}{a^y} = \frac{(x-6)^3}{(x-6)^2}</math> <math>\Rightarrow</math> <math>a^6 = x-6</math>.
+
<math>\frac{a^{y+6}}{a^y} = \frac{(x-6)^3}{(x-6)^2}</math> <math>\Rightarrow</math> <math>a^6 = x-6</math>.
  
 
Plugging in <math>9</math> for <math>x</math> yields <math>a^6 = 3</math>, or <math>a= \sqrt[6]{3} \Rightarrow \boxed{\text{A}}</math>.
 
Plugging in <math>9</math> for <math>x</math> yields <math>a^6 = 3</math>, or <math>a= \sqrt[6]{3} \Rightarrow \boxed{\text{A}}</math>.

Latest revision as of 17:09, 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 1

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 the points $B$ and $C$ are

$(x-6, y)$

and

$(x-6, y+6)$

respectively.

Substituting these coordinates into the equations given yields $y = \log_{a}x,$ $y = 2\log_{a}x-6,$ and $y+6 = 3\log_{a}x-6$. Rearranging a bit, we get the following equations:

$1) a^y = x$

$2) a^y = (x-6)^2$

$3) a^{y+6} = (x-6)^3$

Using equations 1 and 2, we get $x=(x-6)^2$. Solving yields $x=9, x=-4$. However, $-4$ is extraneous so $x=9$ (also because we know $A$ has to have a positive $x$-coordinate). Using the second and third equations, we get

$\frac{a^{y+6}}{a^y} = \frac{(x-6)^3}{(x-6)^2}$ $\Rightarrow$ $a^6 = x-6$.

Plugging in $9$ for $x$ yields $a^6 = 3$, or $a= \sqrt[6]{3} \Rightarrow \boxed{\text{A}}$.

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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