Difference between revisions of "1991 AIME Problems/Problem 4"
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Thus, <math>-1 \le \frac{1}{5} \log_2 x \le 1</math>, and <math>-5 \le \log_2 x \le 5</math>. The solutions for <math>x</math> occur in the domain of <math>\frac{1}{32} \le x \le 32</math>. When <math>x > 1</math> the [[logarithm]] function returns a [[positive]] value; up to <math>x = 32</math> it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of <math>x</math>) of the sine curve and another curve that is <math>< 1</math>, so there are <math>\frac{32}{2} \cdot 10 - 6 = 160 - 6 = 154</math> values (the subtraction of 6 since all the “intersections” when <math>x < 1</math> must be disregarded). When <math>y = 0</math>, there is exactly <math>1</math> touching point between the two functions: <math>\left(\frac{1}{5},0\right)</math>. When <math>y < 0</math> or <math>x < 1</math>, we can count <math>4</math> more solutions. The solution is <math>154 + 1 + 4 = \boxed{159}</math>. | Thus, <math>-1 \le \frac{1}{5} \log_2 x \le 1</math>, and <math>-5 \le \log_2 x \le 5</math>. The solutions for <math>x</math> occur in the domain of <math>\frac{1}{32} \le x \le 32</math>. When <math>x > 1</math> the [[logarithm]] function returns a [[positive]] value; up to <math>x = 32</math> it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of <math>x</math>) of the sine curve and another curve that is <math>< 1</math>, so there are <math>\frac{32}{2} \cdot 10 - 6 = 160 - 6 = 154</math> values (the subtraction of 6 since all the “intersections” when <math>x < 1</math> must be disregarded). When <math>y = 0</math>, there is exactly <math>1</math> touching point between the two functions: <math>\left(\frac{1}{5},0\right)</math>. When <math>y < 0</math> or <math>x < 1</math>, we can count <math>4</math> more solutions. The solution is <math>154 + 1 + 4 = \boxed{159}</math>. | ||
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| + | ==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: | ||
| + | <math>32 * \frac{5}{2} * 2 - 1 = \boxed {159}</math> | ||
== See also == | == See also == | ||
Revision as of 19:33, 15 July 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. 
