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Revision as of 17:38, 11 March 2007

Problem

How many real numbers $x^{}_{}$ satisfy the equation $\frac{1}{5}\log_2 x = \sin (5\pi x)$?

Solution

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The range of the sine function is $-1 \le y \le 1$. It is periodic (in this problem) with a period of $\frac{2}{5}$.

Thus, $-1 \le \frac{1}{5} \log_2 x \le 1$, and $-5 \le \log_2 x \le 5$. The solutions for $x$ occur in the domain of $\frac{1}{32} \le x \le 32$. When $x > 1$ the logarithm function returns a positive value; up to $x = 32$ it will pass through the sine curve. There are exactly 10 intersections of five periods (every two integral values of $x$) of the sine curve and another curve that is $< 1$, so there are $\frac{32}{2} \cdot 10 - 6 = 160 - 6 = 154$ values (the subtraction of 6 since all the “intersections” when $x < 1$ must be disregarded). When $y = 0$, there is exactly $1$ touching point between the two functions: $(\frac{1}{5},0)$. When $y < 0$ or $x < 1$, we can count $4$ more solutions. The solution is $154 + 1 + 4 = 159$.

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
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