2016 AMC 12A Problems/Problem 16

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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?

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


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

$\log_\frac{1}{a} b = \log_a \frac{1}{a} / \log_a b = -1/ \log_a b$

$\log_\frac{1}{b} a = -\log_a b$

Therefore, the system of equations can be simplified to:

$y = t$

$y = -t$

$y = \frac{1}{t}$

$y = -\frac{1}{t}$

where $t = \log_3 x$. Note that all values of $t$ correspond to exactly one positive $x$ value, so all $(t,y)$ intersections will correspond to exactly one $(x,y)$ intersection in the positive-x area.

Graphing this system of functions will generate a total of $5$ solutions $\rightarrow \boxed{\textbf{(D)} 5}$

See Also

2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 12 Problems and Solutions

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