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"

m (→‎Solution)
m (→‎Solution)
Line 6: Line 6:
  
 
==Solution==
 
==Solution==
−
Let <math>a = 2 \cdot |\log_6 10 - 1| = |\log_6 9 - \log_6 x| </math>.
+
Let <math>a = 2 \cdot |\log_6 10 - 1| = |\log_6 9 - \log_6 x|  =  |\log_6 \frac{9}{x}| </math>.
  
−
<math> a = \pm (\log_6 9 - \log_6 x) \implies  \pm a = \log_6 (9/x) \implies {6^a}^{\pm 1} = 9/x \implies x = 9 \cdot b^{\pm 1} </math>  
+
<math> \pm a = \log_6 \frac{9}{x} x \implies 6^{\pm a} = b^{\pm 1} = 9/x \implies x = 9 \cdot b^{\pm 1} </math>  
  
−
<math> (9 / b^1) \cdot (9 / b^{-1}) = 9 \cdot 9 = \boxed{81}</math>.
+
<math> (9b^1) \cdot (9b^{-1}) = 9 \cdot 9 = \boxed{81}</math>.
  
 
~ oinava
 
~ oinava

Revision as of 18:08, 13 March 2023

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

Let $a = 2 \cdot |\log_6 10 - 1| = |\log_6 9 - \log_6 x| = |\log_6 \frac{9}{x}|$.

$\pm a = \log_6 \frac{9}{x} x \implies 6^{\pm a} = b^{\pm 1} = 9/x \implies x = 9 \cdot b^{\pm 1}$

$(9b^1) \cdot (9b^{-1}) = 9 \cdot 9 = \boxed{81}$.

~ oinava

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 $2(\log_610 - 1)$. Let these two numbers be $\log_6a$ and $\log_6b$. Because they are equidistant from $\log_69$, we have $\frac{\log_6a + \log_6b}{2} = \log_69$. 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

Solution 2 (Logarithmic Rules and Casework)

In effect we must find all $x$ such that $\left|\log_6 9 - \log_6 x\right| = 2d$ where $d = \log_6 10 - 1$.

Notice that by log rules \[d = \log_6 10 - 1 = \log_6 \frac{10}{6}\] Using log rules again, \[2d = 2\log_6 \frac{10}{6} = \log_6 \frac{25}{9}\]

Now we proceed by casework for the distinct values of $x$.

Case 1

\[\log_6 9 - \log_6 x_1 = 2d\] Subbing in for $2d$ and using log rules, \[\log_6 \frac{9}{x_1} = \log_6 \frac{25}{9}\] From this we may conclude that \[\frac{9}{x_1} = \frac{25}{9} \implies x_1 = \frac{81}{25}\]

Case 2

\[\log_6 9 - \log_6 x_2 = -2d\] Subbing in for $-2d$ and using log rules, \[\log_6 \frac{9}{x_2} = \log_6 \frac{9}{25}\] From this we conclude that \[\frac{9}{x_2} = \frac{9}{25} \implies x_2 = 25\]

Finding the product of the distinct values, $x_1x_2 = \boxed{\textbf{(E) } 81}$

~Spektrum

Video Solution 1 (Quick and Simple)

https://youtu.be/2sqyO4SlFfc

~Education, the Study of Everything

See Also

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_11&oldid=190921"