# Difference vs log-difference: do they behave similarly?

Consider two positively-valued time series, $\{X(t),Y(t)>0|t\geq0\}$. Now consider two transformations: $$U(t) = Y(t) - \beta X(t),\\ V(t) = \ln{[Y(t)]} - \ln{[\beta X(t)]},$$ with $\beta>0$ such that $\mathbb{E}[U(t)] = 0$. My interest is in the behaviour of $U(t)$ and $V(t)$, and whether or not they move together. My intuition (and rough simulation) tells me that they do, more or less, provided that $\mathbb{E}[U(t)]=0$ holds.

Is it possible to show that a given percentage change in $U(t)$ from $t$ to $t+1$ is accompanied by an approximately equal percentage change in $V(t)$ from $t$ to $t+1$? That is, $$\frac{U(t+1)}{U(t)}-1 \approx \frac{V(t+1)}{V(t)}-1.$$

$$\exp V(t) = \frac{Y(t)}{\beta X(t)} = 1+\frac{U(t)}{\beta X(t)} = 1 + \frac{\mathbb{E}(X(t))}{\mathbb{E}(Y(t))} \cdot \frac{U(t)}{X(t)},$$
$$V(t) = \ln \Bigg( 1+ \frac{\mathbb{E}(X(t))}{\mathbb{E}(Y(t))} \cdot \frac{U(t)}{X(t)} \Bigg).$$
There is no way to disentangle the functions $U$ and $X$ in this equation without adding an equivalent nuisance function measuring the separate parts. As you can see, the time-series $V$ is effectively measuring ratios of the parts rather than differences of the parts, and so they can only be related if you specify one of the individual parts (or a mathematical equivalent).