2023 AMC 12A Problem 6

Below is the professionally curated solution for Problem 6 of the 2023 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12A solutions, or check the answer key.

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Concepts:logarithm midpoint algebraic manipulation

Difficulty rating: 1350

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 $B?$

$2\sqrt{11}$

$4\sqrt{3}$

$8$

$4\sqrt{5}$

$9$

Solution:

Let the $x$ -coordinates be $x_1$ and $x_2.$ The midpoint gives $x_1+x_2=12,$ and the average of the $y$ -values gives $\log_2 x_1+\log_2 x_2=4,$ so $x_1x_2=2^4=16.$

Then $|x_1-x_2|=\sqrt{(x_1+x_2)^2-4x_1x_2}=\sqrt{144-64}=\sqrt{80}=4\sqrt5.$

Thus, the correct answer is D.

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