# 2023 AMC 12A Problems/Problem 6

## Problem

Points $A$ and $B$ lie on the graph of $y=\log_{2}x$. The midpoint of $\overline{AB}$ is $(6, 2)$. What is the positive difference between the $x$-coordinates of $A$ and $B$?

$\textbf{(A)}~2\sqrt{11}\qquad\textbf{(B)}~4\sqrt{3}\qquad\textbf{(C)}~8\qquad\textbf{(D)}~4\sqrt{5}\qquad\textbf{(E)}~9$

## Solution 1

Let $A(6+m,2+n)$ and $B(6-m,2-n)$, since $(6,2)$ is their midpoint. Thus, we must find $2m$. We find two equations due to $A,B$ both lying on the function $y=\log_{2}x$. The two equations are then $\log_{2}(6+m)=2+n$ and $\log_{2}(6-m)=2-n$. Now add these two equations to obtain $\log_{2}(6+m)+\log_{2}(6-m)=4$. By logarithm rules, we get $\log_{2}((6+m)(6-m))=4$. By raising 2 to the power of both sides, we obtain $(6+m)(6-m)=16$. We then get $$36-m^2=16 \rightarrow m^2=20 \rightarrow m=2\sqrt{5}$$. Since we're looking for $2m$, we obtain $(2)(2\sqrt{5})=\boxed{\textbf{(D) }4\sqrt{5}}$

~amcrunner (yay, my first AMC solution)

## Solution 2

We have $\frac{x_A + x_B}{2} = 6$ and $\frac{\log_2 x_A + \log_2 x_B}{2} = 2$. The first equation becomes $x_A + x_B = 12,$ and the second becomes $\log_2(x_A x_B) = 4,$ so $x_A x_B = 16.$ Then \begin{align*} \left| x_A - x_B \right| & = \sqrt{\left( x_A + x_B \right)^2 - 4 x_A x_B} \\ & = \boxed{\textbf{(D) } 4 \sqrt{5}}. \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

## Solution 3

Basically, we can use the midpoint formula

assume that the points are $(x_1,y_1)$ and $(x_2,y_2)$

assume that the points are ($x_1$,$\log_{2}(x_1)$) and ($x_2$,$\log_{2}(x_2)$)

midpoint formula is ($\frac{x_1+x_2}{2}$,$\frac{\log_{2}(x_1)+\log_{2}(x_2)}{2}$)

thus $x_1+x_2=12$ $x_2=12-x_1$ and $\log_{2}(x_1)+\log_{2}(x_2)=4$ $\log_{2}(x_1)+\log_{2}(12-x_1)=\log_{2}(16)$

$\log_{2}((12x_1-x_1^2)/16)=0$

since $2^0=1$ so,

$12x_1-x_1^2=16$

$12x_1-x_1^2-16=0$ for simplicity lets say $x_1 = x$

$12x-x^2=16$. We rearrange to get $x^2-12x+16=0$.

put this into quadratic formula and you should get

$x_1=6+2\sqrt{5}$ Therefore, $x_1=6+2\sqrt{5}-(6-2\sqrt{5})$

which equals $6-6+4\sqrt{5}=\boxed{\textbf{(D) }4\sqrt{5}}$

## Solution 4

Similar to above, but solve for $x = 2^y$ in terms of $y$:

$(2^{y}+2^{2+(2-y)})/2= 6$

$2^y + 2^{4-y} = 12$

$(2^y)^2 + 2^4 = 12(2^y)$

$x^2 -12x + 16 = 0$

Distance between roots of the quadratic is the discriminant: $\sqrt{{12}^2 - 4(1)(16)} = \sqrt{80} = \boxed{\textbf{(D) }4\sqrt{5}}$

~oinava

## Video Solution 1

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

## Video Solution 2 (🚀 Under 3 min 🚀)

~Education, the Study of Everything