Very basically, a huge number of things in nature could be described in terms of ordinary differential equations up to the 2nd order, and oscillators are a prime example of this. Considering "motion" mentioned there, I would explain it in terms of mechanics, for circuitry one would need to replace coordinates with charges etc. The beginning is probably overly detailed for you but I'll list it for the sake of completeness.
Let's start with what happens to a body with an initial velocity in some viscous media assuming that the counteracting force is proportional to the body velocity at a given time:
$$F = ma = m\ddot x = m\dot v = -s*v, v(t=0) = v_0$$
This equation is solved like so:
$$\dfrac{\dot v}{v} = -\dfrac{s}{m}, \int\dfrac{\dot v}{v}dt=\int-\dfrac{k}{m}dt$$
$$ln(v)=-\dfrac{s}{m}t + C, v=v_0e^{-\dfrac{s}{m}t}.$$
The value of the constant is given by the initial condition. This describes damping.
By contrast, an ideal oscillator (no friction pendulum) is described by
$$\ddot x + \omega^2x = 0,$$
in the case of the mass on a spring $\omega^2 = \dfrac{k}{m}$. The solution then takes the form
$$x = Acos(\omega t + \phi_0),$$
$A$ being the maximum amplitude and $\phi_0$ depending on the initial velocity, if any. Obtaining this solution is bit more convoluted and often is explained in the theory of differential equations by "this solution works and there could be only one solution that does in that case".
What happens if you introduce friction to the harmonic oscillator?
Simple, these two dynamics combine! (Actually not so simple, the frequency changes as well). First, you have oscillations, second, you have damping. Damping, as you can see above, follows the exponential law. Therefore it is natural to describe it in terms of logarithms.
Now it is helpful to observe few more things:
- The time scales of damping and oscillating are regulated by different forces and, therefore, time constants. With too much friction, there won't be even a single oscillation - wiki has a helpful section on that. Essentially, the angular frequency of a damped oscillator is $\omega_0\sqrt{1-\eta^2},$ where $\eta$ is called the damping factor governing the entire system's dynamic. At $\eta > 1$ there are no oscillations, the pendulum just exponentially slows down.
- If the oscillator is underdamped (see above), one could calculate how much energy (or its amplitude equivalent) would be lost per oscillation. Again, the amplitude loss is exponential, but all the physical quantities governing it (such as media viscosity!) reside in the exponent, thus it makes sense to use logarithm there.
Finally, to the actual question.
The ballistic galvanometer starts at a zero position. We are concerned with the amplitude of its throw. Damping, however, is typically described starting with the maximum amplitude - the ballistic galvanometer never quite reaches the corresponding "ideal case" amplitude we want to know. Therefore, what we do is we measure how much is it damped and how quickly the amplitude decays: instead of having nice and clean
$$\theta = \theta_{max}cos(\omega t)*e^{-2\lambda \omega t},$$ where the factor of 2 comes from half period measurements we have
$$\theta = \theta_{max}cos(\omega t + \phi)*e^{-2\lambda \omega t}; \theta(t=0) = 0, \theta(t=t_{1max})=\theta_{1max}$$
We have measured $\theta_{1max}, \theta_{2max}, ..., \theta_{Nmax}$ and want to know $\theta_{max}$. These both values are coming from the same equation (down to phase), but $\theta(t=0)\neq \theta_{max}$, nor is $\theta_{1max}$.
This is the crux of your question. To figure out what $\theta_{max}$ would be in an absence of damping, we observe that the first throw ($\theta_{1max}$) happens in a quarter of oscillation period after the initial position. This means that in that time the amplitude got damped by $e^{-\lambda/2}$. (For simplicity of notation, I drop "max" in "1max", "Nmax" indices in the following).
Over the N periods it got damped by $e^{-2N\lambda}=\dfrac{\theta_{2N + 1}}{\theta_1}$.
Therefore, the "true" maximum throw is given by $$\theta_{max}=\theta_1 * e^{\lambda/2}; \lambda = \dfrac{1}{2N}ln \left(\dfrac{\theta_1}{\theta_{2N+1}}\right); \theta_{max}=\theta_1\sqrt[4N]{\dfrac{\theta_1}{\theta_{2N+1}}}\simeq \theta_1 * \left ( 1 + \dfrac{\lambda}{2}\right )$$
