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

Revision as of 14:41, 14 December 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 $2(\log_610 - 1)$ . 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

@@ Line 6: / Line 6: @@
 ==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(\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>\frac{\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>.
+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(\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>\frac{\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>.
 ~ jamesl123456
@@ Line 52: / Line 52: @@
 Finding the product of the distinct values,
-<math>x_1x_2 = \boxed{\textbf{(E)} \, 81}</math>
+<math>x_1x_2 = \boxed{\textbf{(E) } 81}</math>
 ~Spektrum

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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