$$P=A{(1+i)^N-1\over i(1+i)^N}$$ To solve for $N$, divide by $A$, multiply by $i$: $${Pi\over A}={(1+i)^N-1\over(1+i)^N}=1-(1+i)^{-N}$$ Rewrtie as $$(1+i)^{-N}=1-{Pi\over A}$$ Take logs: $$-N\log(1+i)=\log\left(1-{Pi\over A}\right)$$ Now divide both sides by $-\log(1+i)$.