= Solution
First justify that the logarithm is defined at all times from one onward. For planar <Brownian motion> started at $z\ne0$, stop on exiting the annulus $\varepsilon<|x|<R$, where $\varepsilon<|z|<R$. This exit time $\rho$ is finite almost surely: the stopped identity $|B_t|^2-2t$ gives $\mathbb E(t\wedge\rho)\le R^2/2$. The function $f(x)=\log|x|$ is harmonic on the annulus, so its <stopped process> is a bounded <martingale>. Optional stopping gives
$$
\mathbb P_z(\text{hit }\varepsilon\text{ before }R)=\frac{\log R-\log|z|}{\log R-\log\varepsilon}\longrightarrow0.
$$
Hitting zero before the outer circle would require hitting every inner circle first. It therefore has probability zero. Taking a countable sequence of outer radii tending to infinity proves that <planar Brownian motion avoids a fixed point>. Since $B_1$ has a density and is nonzero almost surely, condition on $\mathcal F_1$ to apply this result to the future path.
For $x\ne0$, $\nabla f(x)=x/|x|^2$ and $\Delta f(x)=0$ in two dimensions. The <Itô formula> on annuli therefore gives
$$
\boxed{X_t=X_1+\int_1^t\frac{B_s}{|B_s|^2}\cdot dB_s\qquad(t\ge1).}
$$
To localize explicitly, let $\tau_n=\inf\{t\ge1:|B_t|\notin(1/n,n)\}$. These <stopping times> increase to infinity almost surely: on each compact time interval the continuous path has a finite maximum and, by point avoidance, a positive minimum radius. Up to $\tau_n$, the stochastic integrand has norm at most $n$. Its stopped integral is square-integrable on every finite horizon. The initial value $\log|B_1|$ is integrable, as follows from the Rayleigh-density calculation in part d. Hence $X^{\tau_n}$ is a true <martingale> from time one and \b[$X$ is a <local martingale>]. This also gives its <quadratic variation> from time one as $\int_1^t|B_s|^{-2}\,ds$.
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