Define $E(X)_t=\exp(X_t-\frac{1}{2} \langle X \rangle _t)$ where $(X_t)$ is an adapted continuous semimartingale. Then it is trivial that this is a continuous semimartingale and is the unique solution to the SDE:
$$dZ_t = Z_t dX_t \text{ with } Z_0 = 1$$
To show this I know $2$ solutions:
$1)$ Apply Ito's formula to $\frac{1}{Z_t}$
$2)$ Consider a solution of the form $\exp(X_t+V_t)$ and apply Ito's formula again to get the result.
However, I came across another solution and I'm not really sure why it is true:
Applying Ito to the semimartingale $X_t −\frac{1}{2} \langle X \rangle_t$ and the exponential $x \rightarrow \exp(x)$ shows
$$dE(X)_t=E(X)_td(X_t −\frac{1}{2} \langle X \rangle_t)+\frac{1}{2}E(X)_td \langle X_t −\frac{1}{2} \langle X \rangle_t \rangle=E(X)_tdX_t$$
I'm not really sure about the last inequality. It says that (with a bit of an imagination instead of the dots)
$$d(\dots)+d \langle \dots \rangle =d(\dots)$$
To be honest, I have no idea why this holds. I'd be really grateful if you could be of an assistance.