Difference between revisions of "2016 AMC 12A Problems/Problem 16"
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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}, | + | Similarly with the last two equations, we get <math>\left(3, -1\right)</math> and <math>\left(\frac{1}{3}, 1\right)</math>. |
Now, by setting the first and third equations equal to each other, we get <math>\left(1, 0\right)</math>. | Now, by setting the first and third equations equal to each other, we get <math>\left(1, 0\right)</math>. | ||
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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> | ||
+ | ==Solution 2== | ||
+ | |||
+ | Note that <math>\log_b a =\log_c a / \log_c 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 \frac{1}{a} / \log_a b = -1/ \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>. Note that all values of <math>t</math> correspond to exactly one positive <math>x</math> value, so all <math>(t,y)</math> intersections will correspond to exactly one <math>(x,y)</math> intersection in the positive-x area. | ||
+ | |||
+ | Graphing this system of easy-to-graph functions will generate 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}} |
Latest revision as of 18:58, 7 February 2016
Contents
Problem 16
The graphs of and are plotted on the same set of axes. How many points in the plane with positive -coordinates lie on two or more of the graphs?
Solution
Setting the first two equations equal to each other, .
Solving this, we get and .
Similarly with the last two equations, we get and .
Now, by setting the first and third equations equal to each other, we get .
Pairing the first and fourth or second and third equations won't work because then .
Pairing the second and fourth equations will yield , but since you can't divide by , it doesn't work.
After trying all pairs, we have a total of solutions
Solution 2
Note that .
Then
Therefore, the system of equations can be simplified to:
where . Note that all values of correspond to exactly one positive value, so all intersections will correspond to exactly one intersection in the positive-x area.
Graphing this system of easy-to-graph functions will generate a total of solutions
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.