Is there a proof for the Determinant of the Log-Euclidean-Tensor-Interpolation yielding the geometric mean of the Determinants of its sampling points? Given a set of $i \in \mathbb{N}$ symmetric-positive-definite-tensors $\boldsymbol{A}_i$ and the corresponding weights $w_i$ with $w_i>0$ and $\sum_i w_i = 1$.
The Log-Eucledian-Mean is defined as
$ \bar{\boldsymbol{A}} = \exp(\sum_i w_i\log(\boldsymbol{A}_i))$.
After some testing on random sets, I assume the Determinant of the interpolated value is equal to the geometric mean of the Determinants of its sampling tensors.
$\det{\bar{\boldsymbol{A}}} = \prod_i \left(\det{\boldsymbol{A}_i} \right)^{w_i}$ 
Does anyone know, whether a nice proof exists for this identity?
TIA

Using some well-known properties connecting exponents, determinants, traces, and logarithms, as well as multiplicativity of determinant and linearity of trace, this boils down to some boring algebra:
$$\det \bar A = \det \exp\sum\limits_i w_i \log(A_i)=\exp\operatorname{tr}\sum\limits_i w_i \log(A_i) = \\
= \exp \sum\limits_i \operatorname{tr} w_i \log(A_i) = \prod \left(\exp\operatorname{tr}\log A_i\right)^{w_i}=\prod \left(\det\exp \log A_i\right)^{w_i} = \\
= \prod \left(\det A_i\right)^{w_i}$$
