# 2016 AMC 12A Problems/Problem 16

## Problem 16

The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$-coordinates lie on two or more of the graphs?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

## Solution

Setting the first two equations equal to each other, $\log_3 x = \log_x 3$.

Solving this, we get $\left(3, 1\right)$ and $\left(\frac{1}{3}, -1\right)$.

Similarly with the last two equations, we get $\left(3, -1\right)$ and $\left(\frac{1}{3}, 1\right)$.

Now, by setting the first and third equations equal to each other, we get $\left(1, 0\right)$.

Pairing the first and fourth or second and third equations won't work because then $\log x \leq 0$.

Pairing the second and fourth equations will yield $x = 1$, but since you can't divide by $\log 1 = 0$, it doesn't work.

After trying all pairs, we have a total of $5$ solutions $\rightarrow \boxed{\textbf{(D)} 5}$

## Solution 2

Note that $\log_b a =\log_c a / \log_c b$.

Then $\log_b a = \log_a a / \log_a b = 1/ \log_a b$

$\log_\frac(1,a) b = \log_a 1/a / \log_a b = -1/ \log_a b$

$\log_(1/b) a = -\log_a b$