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

m (Added MAA box)
(Solution)
Line 9: Line 9:
 
Setting the first two equations equal to each other, <math>\log_3 x = \log_x 3</math>.
 
Setting the first two equations equal to each other, <math>\log_3 x = \log_x 3</math>.
  
Solving this, we get <math>\left(3, 1\right)</math> and <math>\left(\frac{1}{3}, 1\right)</math>.
+
Solving this, we get <math>\left(3, 1\right)</math> and <math>\left(-\frac{1}{3}, 1\right)</math>.
  
 
Similarly with the last two equations, we get <math>\left(3, -1\right)</math> and <math>\left(\frac{1}{3}, -1\right)</math>.
 
Similarly with the last two equations, we get <math>\left(3, -1\right)</math> and <math>\left(\frac{1}{3}, -1\right)</math>.
Line 20: Line 20:
  
 
After trying all pairs, we have a total of <math>5</math> solutions <math>\rightarrow \boxed{\textbf{(D)} 5}</math>
 
After trying all pairs, we have a total of <math>5</math> solutions <math>\rightarrow \boxed{\textbf{(D)} 5}</math>
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=A|num-b=15|num-a=17}}
 
{{AMC12 box|year=2016|ab=A|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:30, 6 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 $\left(3, 1\right)$ 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 $\rightarrow \boxed{\textbf{(D)} 5}$

See Also

2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS