Difference between revisions of "2016 AMC 12A Problems/Problem 16"

(Created page with "==Problem 16== The graphs of <math>y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,</math> and <math>y=\log_x \dfrac{1}{3}</math> are plotted on the same set of axes. How many p...")
 
(Solution)
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==Solution==
 
==Solution==
  
Setting the first two equations equal to each other, <math>\log_3 x = log_x 3</math>.
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Setting the first two equations equal to each other, <math>\log_3 x = \log_x 3</math>.
  
 
Solving this, we get <math>(3, 1)</math> and <math>(\frac{1}{3}, 1)</math>.
 
Solving this, we get <math>(3, 1)</math> and <math>(\frac{1}{3}, 1)</math>.

Revision as of 18:28, 4 February 2016

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 $(3, 1)$ and $(\frac{1}{3}, 1)$.

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

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

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 $\rightarrow \boxed{\textbf{(D)} 5}$

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