Problem:
How many real numbers x satisfy the equation 51log2x=sin(5πx)?
Solution:
Since ∣sinθ∣≤1 for all real θ, we need only consider those values of x for which
∣∣∣∣∣51log2x∣∣∣∣∣≤1
This inequality is satisfied by all values of x between 321 and 32, inclusive. We first consider 321≤x<1. For such x we have −1≤51log2x<0, while sin5πx≤0 only for 51≤x≤52 and 53≤x≤54. It follows that the graphs of y=51log2x and y=sin5πx meet at 4 points for 321≤x<1. (See accompanying figure.) When 1<x≤32, we have 0<51log2x≤1 while sin5πx≥0 only for 52k≤x≤52k+1(k=3,4,…,79). The graphs of the two functions intersect at 2 points on each of these 77 intervals, giving 154 points of intersection for 1<x≤32. Since both functions take on the value 0 when x=1, we have a total of 4+154+1=159 solutions to the equation.
The problems on this page are the property of the MAA's American Mathematics Competitions