2022 AMC 12A Problems/Problem 11

Revision as of 12:36, 12 November 2022 by Jamesl123456 (talk | contribs) (Solution)

Problem

What is the product of all real numbers $x$ such that the distance on the number line between $\log_6x$ and $\log_69$ is twice the distance on the number line between $\log_610$ and $1$?

$\textbf{(A) } 10 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 25 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 81$

Solution

First, notice that there must be two such numbers: one greater than $\log_69$ and one less than it. Furthermore, they both have to be the same distance away, namely $2 \cdot (\log_610 - 1)$. Let these two numbers be $\log_6a$ and $\log_6b$. Because they are equidistant from $\log_69$, we have $(\log_6a + \log_6b)/2 = \log_69$. Using log properties, this simplifies to $\log_6{\sqrt{ab}} = \log_69$. We then have $\sqrt{ab} = 9$, so $ab = \boxed{\textbf{(E)} \, 81}$.

~ jamesl123456