Let $\mathbf{A}$ be a real matrix that belongs to $SE(3)$ i.e. has the following form: $$ \mathbf{A} = \begin{pmatrix} \mathbf{R} & \mathbf{t}\\ \mathbf{0}_{1\times 3} & 1 \end{pmatrix} $$ With $\mathbf{R} \in SO(3)$ ($\mathbf{R}$ represents a rotation matrix), and $\mathbf{t} \in \mathbb{R}^3$ ($\mathbf{t}$ represents a translation vector).
Given this I am wondering if there is a way to simplify the following expression: $$\exp\left( k \log\left( \mathbf{A} \right) \right)$$ Where $\{k \in \mathbb{R}\,|\, 0\leq k \leq 1\}$, $\exp(\cdot)$ is the matrix exponential and $\log(\cdot)$ is the matrix logarithm.
Such that $k$ does not appear at the middle of the expression, so that it can be expressed as: $$ f(k) \exp(\log(\mathbf{A})) = f(k) \mathbf{A} \quad\text{ or}\quad \exp(\log(f(k)\mathbf{A})) = f(k) \mathbf{A}$$
I've found closed form solutions for $\exp(\cdot)$ and $\log(\cdot)$ here (at p.47). For example, for the exponential map:
$$\exp(\mathbf{X}) = \begin{pmatrix} \exp(\mathbf{w}^\wedge) & \mathbf{V}\mathbf{v}\\ \mathbf{0}_{1\times3} & 1 \end{pmatrix}$$ Where $\mathbf{w}\in \mathbb{R}^3, \mathbf{v}\in \mathbb{R}^3$ and: $$\begin{align} \exp(\mathbf{w}^\wedge) &= \mathbf{I}_{3\times3} + \frac{\sin{\theta}}{\theta}\mathbf{w}^\wedge + \frac{1-\cos\theta}{\theta^2}(\mathbf{w}^\wedge)^2\\ \mathbf{V} &= \mathbf{I}_{3\times3} + \frac{1-\cos\theta}{\theta^2}\mathbf{w}^\wedge + \frac{\theta - \sin\theta}{\theta^3}(\mathbf{w}^\wedge)^2\\ \theta &= \lVert \mathbf{w} \lVert \end{align}$$
But knowing that there is not a closed-form solution for $\cos(k\theta)$ as noted in this post, this makes me think that this simplification is not feasible.