Let $G$ be a Lie group with group operation ★: $G \times G \rightarrow G$ and Lie algebra $Lie(G)$. Each $\omega \in Lie(G)$ generates a left-invariant vector field $V$$\omega$ on $G$, and that the exponential map describes the integral curves of the vector field. Specifically, the integral curve $\gamma$: $\mathbb R \rightarrow G$ of the left-invariant vector field $V$$\omega$ that starts at the point $x \in G$ at time $t = 0$ is given by:
$$\gamma(t) \triangleq x ★\text{exp}(t\omega)$$
This equation provides a prescription for "moving around" on the Lie group $G$ along
the "direction" determined by $\omega$. Say a point $x \in G$ lies in the image of $G's$ exponential map, we write $\log(x)$ to denote one of $x$'s preimages, so that:
$$x = \text{exp}(\log(x))$$
If $G's$ exponential map is surjective, then there is always at least one choice of $\log(x) \in Lie(G)$ that will satisfy $x = $exp$(\log(x))$. Now suppose that $x, y \in G$ and that $G$'s exponential map is surjective. Using $\gamma(t) \triangleq x ★$exp$(t\omega)$, I want to derive a formula for a curve $\gamma:[0,1] \rightarrow G$ such that $\gamma (0) = x$ and $\gamma (1) = y$