Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 2)
Line 27: Line 27:
 
Then <math>\log_b a = \log_a a / \log_a b = 1/ \log_a b</math>
 
Then <math>\log_b a = \log_a a / \log_a b = 1/ \log_a b</math>
  
−
<math>\log_\frac(1,a) b = \log_a 1/a / \log_a b = -1/ \log_a b</math>
+
<math>\log_\frac{1}{a} b = \log_a \frac{1}{a} / \log_a b = -1/ \log_a b</math>
  
−
<math>\log_(1/b) a = -\log_a b</math>
+
<math>\log_\frac{1}{b} a = -\log_a b</math>
 +
 
 +
Therefore, the system of equations can be simplified to:
 +
 
 +
<math>y = t</math>
 +
 
 +
<math>y = -t</math>
 +
 
 +
<math>y = \frac{1}{t}</math>
 +
 
 +
<math>y = -\frac{1}{t}</math>
 +
 
 +
where <math>t = \log_3 x</math>.
 +
 
 +
Graphing this system of easy-to-graph functions will generate a total of
  
 
==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 18:52, 7 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 $\rightarrow \boxed{\textbf{(D)} 5}$

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 $t = \log_3 x$.

Graphing this system of easy-to-graph functions will generate a total of

See Also

2016 AMC 12A (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_16&oldid=75673"