Given a real symmetric negative definite matrix $\mathbf{A}$, consider a system of linear ODEs given by
$$ \dot{\mathbf{x}} = \mathbf{A} \left( \mathbf{x} - \mathbf{x}_\star \right),$$
whose solution is given by
$$ \mathbf{x}(t) = \mathbf{x}_\star + \exp(t \mathbf{A})\left( \mathbf{x}(0) - \mathbf{x}_\star \right).$$
Let $\epsilon \in (0, 1)$. How do we solve the following for $t$?
$$\lVert \mathbf{x}(t) - \mathbf{x}_\star \rVert \leq \epsilon,$$
Should one just consider the largest eigenvalue of $\mathbf{A}$ (corresponding to the slowest decay rate)?
Partial answer:
$$ \|\boldsymbol{x}(t)-\boldsymbol{x}_{\star}\|=\|\exp(\boldsymbol{A}t)(\boldsymbol{x}(0)-\boldsymbol{x}_{\star})\|\leq\|\exp(\boldsymbol{A}t)\|\,\|\boldsymbol{x}(0)-\boldsymbol{x}_{\star}\| $$
I am assuming by $\|\cdot\|$ you mean the Euclidean two-norm. If so, then the norm of the matrix exponential comes down to finding the largest eigenvalue (by magnitude) of $\exp(\boldsymbol{A}t)$. Towards this goal it is important to recognize that if $\boldsymbol{A}$ is diagnolizable with eigenvalues $\lambda_{1},\lambda_{2},\dots,\lambda_{N}$ then the eigenvalues of $\exp(\boldsymbol{A}t)$ are $\exp(\lambda_{1}t),\exp(\lambda_{2}t),\dots,\exp(\lambda_{N}t)$. With this we have
$$ \|\boldsymbol{x}(t)-\boldsymbol{x}_{\star}\| \leq\|\boldsymbol{x}(0)-\boldsymbol{x}_{\star}\|\max_{i=1,2,\dots,N}|\exp(\lambda_{i}t)| $$
Are you able to finish the problem from here?