Calculus: Early Transcendentals 8th Edition

Let the initial temperature be $32.5^{\circ}C$ at 1:30 pm We can find $k$: $T(t) = 20+12.5~e^{kt}$ $T(1) = 20+12.5~e^{k} = 30.3$ $12.5~e^{k} = 10.3$ $e^{k} = \frac{10.3}{12.5}$ $k = ln(\frac{10.3}{12.5})$ $k = -0.193584749$ Now we can let the initial temperature be $37.0^{\circ}C$. Then: $T(t) = 20+17~e^{-0.193584749~t}$ We can find the time $t$ when the temperature is $32.5^{\circ}C$: $T(t) = 20+17~e^{-0.193584749~t}$ $20+17~e^{-0.193584749~t} = 32.5$ $e^{-0.193584749~t} = \frac{12.5}{17}$ $-0.193584749~t = ln(\frac{12.5}{17})$ $t = \frac{ln(\frac{12.5}{17})}{-0.193584749}$ $t = 1.6$ The murder took place 1.6 hours (i.e. 1 hour and 36 minutes) before 1:30 pm. Therefore, the murder took place at 11:54 am.