Difference between revisions of "1991 AIME Problems/Problem 4"
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==Solution 2== | ==Solution 2== | ||
Notice that the equation is satisfied twice for every sine period (which is <math>\frac{2}{5}</math>), except in the sole case when the two equations equate to <math>0</math>. In that case, the equation is satisfied twice but only at the one instance when <math>y=0</math>. Hence, it is double-counted in our final solution, so we have to subtract it out. We then compute: | Notice that the equation is satisfied twice for every sine period (which is <math>\frac{2}{5}</math>), except in the sole case when the two equations equate to <math>0</math>. In that case, the equation is satisfied twice but only at the one instance when <math>y=0</math>. Hence, it is double-counted in our final solution, so we have to subtract it out. We then compute: | ||
− | <math>32 | + | <math>32 \cdot \frac{5}{2} \cdot 2 - 1 = \boxed {159}</math> |
== See also == | == See also == |
Latest revision as of 16:01, 26 November 2019
Contents
Problem
How many real numbers satisfy the equation ?
Solution
The range of the sine function is . It is periodic (in this problem) with a period of .
Thus, , and . The solutions for occur in the domain of . When the logarithm function returns a positive value; up to it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of ) of the sine curve and another curve that is , so there are values (the subtraction of 6 since all the “intersections” when must be disregarded). When , there is exactly touching point between the two functions: . When or , we can count more solutions. The solution is .
Solution 2
Notice that the equation is satisfied twice for every sine period (which is ), except in the sole case when the two equations equate to . In that case, the equation is satisfied twice but only at the one instance when . Hence, it is double-counted in our final solution, so we have to subtract it out. We then compute:
See also
1991 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.