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>(3, 1)</math> and <math>(\frac{1}{3}, 1)</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>(3, -1)</math> and <math>(\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>(1, 0)</math>. | + | Now, by setting the first and third equations equal to each other, we get <math>\left(1, 0\right)</math>. |
Pairing the first and fourth or second and third equations won't work because then <math>\log x \leq 0</math>. | Pairing the first and fourth or second and third equations won't work because then <math>\log x \leq 0</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> | ||
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| + | ==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: | ||
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| + | <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 functions will generate a total of <math>5</math> solutions <math>\rightarrow \boxed{\textbf{(D)} 5}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2016|ab=A|num-b=15|num-a=17}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 12:05, 19 July 2021
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 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. 
