Does $\exp(\ln(I+A))=I+A$ when $\|A\|<1$? For matrices, I know certain equalities like $e^{A+B}=e^Ae^B$ aren't always true. I'm curious, do $\exp$ and $\ln$ serve as inverses?
I saw earlier that if $\|A\|<1$, then $\ln(I+A)$ converges. My question is, does
$$
\exp(\ln(I+A))=I+A
$$
when $\|A\|<1$? Certainly $\exp(\ln(I+A))$ converge since $\ln(I+A)$ converges and $\exp(\cdot)$ converges for all matrices, but do we necessarily get the original matrix back?

Another more analytical approach than quid's fine answer is the following:
In the unit ball of $\text{Mat}_n(\mathbb{C})$ the diagonalizable matrices (that are in the ball) are dense. It is trivial to verify the equation 
$$\exp(\log(1+A))=1+A$$
holds for $A$ in the unit ball, and diagonalizable. Then, you can use the fact that both $X\mapsto \exp(\log(1+X))$ and $X\mapsto 1+X$ are continuos functions from the unit ball to $\text{Mat}_n(\mathbb{C})$. Since the latter space is metrizable (Hausdorff) and the diagonalizable matrices are dense, it follows that $X\mapsto \exp(\log(1+X))$ and $X\mapsto 1+X$ must agree on the unit ball.
