Let $S$ be a symmetric space of non-compact type, specifically let $S= SL(n)/SO(n)$ for some $n\geq 2$. I know that such spaces are geodesically complete, simply connected, and have non-positive curvature. Therefore, by the Cartan-Hadamard Theorem, the Riemannian exponential map $Exp_p$ at any point $p \in S$ is a global diffeomorphism to the appropriately dimensioned Euclidean space.
Edit: Is it correct that $Exp_e(v)=\pi\circ \exp$ where $\exp$ is the matrix exponential and $\pi:SL(n)\rightarrow SL(n)/SO(n)=S$?
For a linear group like $SL(n)$, the Lie theory exponential agrees with the matrix exponential $\mathrm{exp}$. However, the Riemannian exponential map doesn't always agree with the Lie theoretic exponential map. For example, if you choose a nonzero nilpotent matrix $N$, then $c(t) = \pi \circ \exp(tN)$ is not a geodesic in the associated symmetric space. Fortunately it is easy to answer your question in terms of something called the "Cartan decomposition."
When you pick a point in a symmetric space $G/K$, you can define a Cartan decomposition of the Lie algebra $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$. Here $\mathfrak{k}$ is the Lie algebra of $K$, and this decomposition is not a Lie algebra decomposition, rather it is a vector space decomposition with the bracket satisfying $$ [\mathfrak{k},\mathfrak{k}] \subset \mathfrak{k}, \, [\mathfrak{k},\mathfrak{p}] \subset \mathfrak{p}, \, [\mathfrak{p},\mathfrak{p}] \subset \mathfrak{k} .$$
In your example, the natural basepoint to choose is the identity coset $[SO(n)]$. Then $\mathfrak{k} = \mathfrak{so}(n)$ is the Lie algebra of skew-symmetric matrices, and $\mathfrak{p}$ is the vector space of symmetric traceless matrices.
Theorem. For any $X \in \mathfrak{p}$, $c(t) = \pi \circ \exp(tX)$ is a geodesic. Morever, every geodesic through the basepoint arises this way.
So in your example, the geodesics through $[SO(n)]$ are exactly $\pi \circ \exp (tX)$ for $X$ a symmetric traceless matrix.