Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 12A Problems/Problem 11"

(Created page with "==Problem== What is the product of all real numbers <math>x</math> such that the distance on the number line between <math>\log_6x</math> and <math>\log_69</math> is twice th...")
 
Line 4: Line 4:
  
 
<math>\textbf{(A) } 10 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 25 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 81</math>
 
<math>\textbf{(A) } 10 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 25 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 81</math>
 +
 +
==Solution==
 +
First, notice that there must be two such numbers: one greater than <math>\log_69</math> and one less than it. Furthermore, they both have to be the same distance away, namely <math>2 \cdot (\log_610 - 1)</math>. Let these two numbers be <math>\log_6a</math> and <math>\log_6b</math>. Because they are equidistant from <math>\log_69</math>, we have <math>(\log_6a + \log_6b)/2 = \log_69</math>. Using log properties, this simplifies to <math>\log_6{\sqrt{ab}} = \log_69</math>. We then have <math>\sqrt{ab} = 9</math>, so <math>ab = \boxed{\textbf{(E)} \, 81}</math>.

Revision as of 12:35, 12 November 2022

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 $ab = \boxed{\textbf{(E)} \, 81}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_11&oldid=181040"