Answer: 159

Small Hint:

The bound $\lvert\sin(5\pi x)\rvert\leq1$ restricts $x$ to a finite interval

Big Hint:

Separate the positive and negative half-waves of the sine function and count two crossings on each eligible full half-wave

Solution:

Let $h(x)=\sin(5\pi x)-\frac15\log_2x.$ Any root lies in $[2^{-5},2^5]=[\frac{1}{32},32].$

For $x<1,$ a root can occur only on a negative half-wave of the sine. There are two such half-waves, $(\frac{1}{5},\frac{2}{5})$ and $(\frac{3}{5},\frac{4}{5}).$ On each, $h$ is positive at both endpoints and negative at the midpoint, so there are two roots. Uniqueness on the descending half follows from monotonicity; on the ascending half, $$\begin{aligned}h''&=-25\pi^2\sin(5\pi x)\\&\quad+\frac1{5x^2\ln2}>0,\end{aligned}$$ so $h'$ increases from a negative value at the midpoint and vanishes once. The function $h$ therefore first decreases and then increases from a negative midpoint value to a positive endpoint value, giving exactly one root on this half. Thus there are $4$ roots below $1.$

Also $x=1$ is a root. For $1<x<32,$ roots can occur only on positive half-waves $(\frac{2m}{5},\frac{2m+1}{5})$ with $m=3,$ $4,$ $\ldots,$ $79.$ There are $77$ of these. The value of $h$ is negative at each endpoint and positive at the midpoint. The right half is strictly decreasing. On the left half, $$\begin{aligned}h'''&=-125\pi^3\cos(5\pi x)\\&\quad-\frac2{5x^3\ln2}<0,\end{aligned}$$ so $h'$ is strictly concave. It is positive at the left endpoint and negative at the midpoint, so it changes sign exactly once. Consequently, $h$ rises to one maximum and then falls to a still-positive midpoint, giving exactly one crossing on the left half. Hence every such half-wave contributes exactly two roots. Therefore the total number is $$4+1+2(77)=159.$$