# 2022 AMC 12A Problems/Problem 11

## 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 1

Let $a = 2 \cdot |\log_6 10 - 1| = |\log_6 9 - \log_6 x| = |\log_6 \frac{9}{x}|$. $\pm a = \log_6 \frac{9}{x} \implies 6^{\pm a} = b^{\pm 1} = \frac{9}{x} \implies x = 9 \cdot b^{\pm 1}$ $9b^1 \cdot 9b^{-1} = \boxed{81}$.

~ oinava

## Solution 2

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 3 (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)

~Education, the Study of Everything

## Video Solution 1 (Understand the question first)

~Math-X

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