Answer
8:01am.
Work Step by Step
Step 1. Identify the given conditions:
$C=70^\circ F, T_0=98.6^\circ F, T(x+30)=85.6^\circ F,T(x+60)=82.7^\circ F$
where the time intervals $x+30, x+60$ are in minutes with the assumption that the person died $x$ minutes before 9am.
Step 2. Using Newton's Law of Cooling with the above conditions, we have
$85.6=70+(98.6-70)e^{k(x+30)}$ and $82.7=70+(98.6-70)e^{k(x+60)}$
Step 3. Rewrite the above two equations as
$\begin{cases} e^{k(x+30)}\approx0.54545 \\ e^{k(x+60)}\approx0.44406 \end{cases}$ or $\begin{cases} k(x+30)=ln(0.54545) \\ k(x+60)=ln(0.44406) \end{cases}$
Step 4. Cancel the parameter $k$, to get
$ln(0.54545)(x+60)=ln(0.44406) (x+30)$ or $x+60=1.337x+1.337(30)$
which gives $x\approx59min$
Step 5. We conclude that the person died around 8:01am.
