Using the ito integral:
$$x_t=\int_0^t b(s,x(s)) \, ds+\int_0^t \sigma (s,x(s)) \, dB(s)+x_0$$
I want to solve this:
I have $$g(s,n(s))=\log(n(s))$$
Using ito's lemma taking the derivative I get this:
$$\frac{dn(s)}{n(s)}=d\log (n(s))+\frac{\alpha ^2 dt}{2}$$
Given: $$dn(s)=\alpha n(s) dB(s)+r n(s) dt$$
When I substitute I get:
$$d\log (n(s))\to \text{$\alpha $}dB(s)+(r -\frac{1}{2} \alpha ^2 )dt$$
Integrating
$$\text{log}\left(\frac{N(s)}{N(0)}\right)=\text{$\alpha $}B(t)+(r -\frac{1}{2} \alpha ^2)t $$
My issue is not to understand what the log rules are doing my issue is why is log applied to $x_0$ and not to everything on the right hand side of the ito integral defined above.
What am I missing?
Note that the differential equation
$$dX_t = b(t) \, dt + \sigma(t) \, dB_t \tag{1}$$
is (by definition!) to be understood as an integral equation, i.e. $(1)$ is equivalent to saying that
$$X_t - X_0 = \int_0^t b(s) \, ds + \int_0^t \sigma(s) \, dB_s.$$
Now in your setting, this menas that
$$d \log(n(s)) = \alpha \, dB(s) + \left(r- \frac{1}{2} \alpha^2 \right) \, ds$$
is equivalent to
$$\log(n(t))-\log(n(0)) = \int_0^t \alpha \, dB_s + \int_0^t \left(r- \frac{1}{2} \alpha^2 \right) \, ds$$
which, in turn, is obviously equivalent to
$$\log \left( \frac{n(t)}{n(0)} \right) = \alpha B_t + \left(r- \frac{1}{2} \alpha^2 \right) t.$$