Consider the equation of motion for a simple harmonic oscillator,
$y''(x)+y(x) = 0, \hspace{5ex} y(0) = 0, \hspace{5ex} y'(0) = 1$ with solution $y(x) = \sin x.$
Change coordinates, let $x(t) = \log t$ and $f(t) = y( x(t)) = \sin(\log t)$. The zeros of $f(t)$ occur when $\log t_n = n \pi$, that is, when $t_n = (e^\pi)^n$. Notice that $\frac{d}{d x} = \frac{d t}{d x} \frac{d}{dt} = t \frac{d}{dt} \hspace{5ex} \textrm{and so}\hspace{5ex} \frac{d^2}{dx^2} = \left(t \frac{d}{dt}\right)\left(t \frac{d}{dt}\right) = t^2 \frac{d^2}{dt^2} + t \frac{d}{dt}.$ Therefore, the function $f(t)$ satisfies the differential equation $t^2 f''(t) + t f'(t) + f(t) = 0, \hspace{5ex} f(1) = 0,\hspace{5ex} f'(1) = 1.$
As indicated in the comments, if instead we choose $x(t) = \pi \log_T t$, the zeros will be at $t_n = T^n$. The differential equation satisfied by $f(t) = \sin(\pi \log_T t)$ is then $ t^2 f''(t) + t f'(t) + \left(\frac{\pi}{\log T}\right)^2 f(t) = 0,$ We can think of this roughly as a harmonic oscillator with time-dependent restoring force.
Addendum: Find below a plot of $f(t) = \sin(\pi \log_2 t)$.
f(t) = \sin(\pi \log_2 t)">