AMC12 2007 12A, Problem 23

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 $a?$

Solution: 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 $a= \sqrt[6]{3} \Rightarrow \fbox{\sqrt [6]{3}}$.

Answer: \sqrt [6]{3}