A matrix of positive determinant that is not an exponential

Setup. $\det B=2>0$, so $B\in GL_2^+(\mathbb{R})$. Suppose $e^A=B$ for some $A\in M_2(\mathbb{R})$.

$A$ commutes with $B$. $A$ commutes with each power $A^k$, hence with the sum $\sum A^k/k!=e^A=B$.

$A$ is diagonal. Write $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$. Then
$$AB=\begin{pmatrix}-a&-2b\\-c&-2d\end{pmatrix},\qquad BA=\begin{pmatrix}-a&-b\\-2c&-2d\end{pmatrix}.$$
$AB=BA$ forces $-2b=-b$ and $-c=-2c$, i.e. $b=c=0$. So $A=\mathrm{diag}(a,d)$ is diagonal. (This is where the distinct eigenvalues of $B$ are used.)

Contradiction. For diagonal $A$, $e^A=\mathrm{diag}(e^a,e^d)$, whose entries are positive. But $B$ has negative diagonal entries.

Conclusion. $B\in GL_2^+(\mathbb{R})$ is not in the image of $\exp$.

Remark. Over $\mathbb{C}$, $B=\exp\mathrm{diag}(i\pi,\ \ln2+i\pi)$, so every invertible complex matrix is an exponential. Over $\mathbb{R}$, a matrix is an exponential iff each Jordan block with a negative eigenvalue occurs an even number of times, which fails for $B$ (two *different* negative eigenvalues). By contrast $-I=e^{\pi J}$ is an exponential.