Sorry if this question sounds silly, and it probably is. I can prove easily that matrix multiplication is noncommutative; however, look at this 'fake' proof:
$$AB=e^{\ln(A)}e^{\ln(B)}=e^{\ln(A)+\ln(B)}$$
$\ln(A)$ and $\ln(B)$ are also matrices, and matrix addition is comutative. So:
$$e^{\ln(A)+\ln(B)}=e^{\ln(B)+\ln(A)}=e^{\ln(B)}e^{\ln(A)}=BA$$
What's the problem?
The definition of a matrix exponential is: $$e^A = \sum_{k\in\mathbb{N}}\frac{1}{k!}A^k$$ hence on one hand: $$e^{(A+B)} = \sum_{k\in\mathbb{N}}\frac{1}{k!}(A+B)^k$$ and on the other hand: $$e^{A}e^{B} = \left(\sum_{k\in\mathbb{N}}\frac{1}{k!}A^k \right) \left(\sum_{k\in\mathbb{N}}\frac{1}{k!}B^k \right) \\ = \sum_{k\in\mathbb{N}} \sum_{i=0}^k \frac{1}{i!(k-i)!}A^iB^{k-i}$$ If $A$ and $B$ commute you can verify by the binomial theorem that: $$\frac{1}{k!}(A+B)^k = \sum_{i=0}^k \frac{1}{i!(k-i)!}A^iB^{k-i}$$ hence $$AB=BA \Rightarrow e^{(A+B)} = e^A e^B$$ But the identity you leverage in your fake proof is not true in general ! Martin Argerami provided a counterexample in his response.
Here is a simple example you can try by hand.
Let $$ A=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\qquad\qquad B=\begin{bmatrix} 0&1\\0&0\end{bmatrix} . $$ Then $$ e^A=\begin{bmatrix} e&0\\0&1\end{bmatrix},\qquad \qquad e^B=\begin{bmatrix} 1&1\\0&1\end{bmatrix}, $$
$\ $
$$ e^Ae^B=\begin{bmatrix} e&e\\0&1\end{bmatrix},\qquad\qquad e^Be^A=\begin{bmatrix} e&1\\0&1\end{bmatrix} \qquad\qquad e^{A+B}=\begin{bmatrix} e&e-1\\0&1\end{bmatrix}. $$
First, let's come up with a way to define the exponential and logarithm of a matrix.
Suppose that a matrix $M$ is diagonalizable, i.e., there exists a non-singular matrix $S$ and a diagonal matrix $D$ such that $M = SDS^{-1}$. Then for any integer $n$, it can be shown that $M^n = SD^nS^{-1}$.
This allows us to use the Taylor series to define the matrix exponential:
$$e^M = \sum_{k=0}^\infty \frac{1}{k!} M^k$$ $$= \sum_{k=0}^\infty \frac{1}{k!} SD^kS^{-1}$$ $$= S(\sum_{k=0}^\infty \frac{1}{k!} D^k)S^{-1}$$ $$= Se^DS^{-1}$$
where $e^D$ denotes a diagonal matrix where the individual diagonal elements are the exponentials of the corresponding elements in $D$.
We can similarly define $\ln{M} = S(\ln D)S^{-1}$. But note that $\ln D$ will be complex if $D$ contains any negative values. And $\ln D$ will be undefined if $D$ has any zeroes on its diagonal (i.e., if $M$ is singular).
If two matrices $A$ and $B$ are simultaneously diagonalizable, with $A = SDS^{-1}$ and $B = SES^{-1}$, then yes, it is true that $\ln AB = S(\ln D + \ln E)S^{-1} = S(\ln E + \ln D)S^{-1} = \ln BA$, and so the two matrices commute. But if they're not simultaneously diagonalizable, then they won't necessarily commute.
Commutative means : $\forall A , B (AB=BA)$
Non-Commutative means : It is not true that $\forall A , B (AB=BA)$ ; In other words $\exists A , B (AB \ne BA)$
Even in Non-Commutative Case, there may be some Examples of $\exists A , B (AB = BA)$
Eg : set $B=I$, and you will get Equality.
Issue(s) in your "fake" Proof is :
(1) There may be some $A$ & $B$ where it is Equal, but not for all $A$ & $B$, in general.
(2) Assuming that $e^A$ & $log(A)$ of Matrix A are Well-Defined for all A.
