If the decrease in body temp is exponential then it will be in the form of y= ab^x where y is body temp at any given time x after 5Am , a is body temp at 5Am when police arrive (starting value) and b is decay ratio
We have 2 data points t=88.2 at time x=0 and t=86.7 at t=1 hour later
we have the equation 88.2= ab^0 since b^0 =1 a=88.2
and 86.7=ab^1
86.7= 88.2b. Solve for b=86.7/88.2= .983
At time of death body temp would be 98.6 so
98.6=88.2(.983)^x. Now solve for x
(.983)^x= 98.6/88.2
(.983)^x =1.118 now take the log of both sides I will use natural log‘ can use logbase10 also
ln(.983)^x =ln1.118 By property of logs
xln(.983) = ln1.118
x= ln1.118/ln.983
x=-6.5 hours so person died 6 1/2 hours prior to 5AM which would be 10:30 PM