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The hotel Bora-Bora is having a pig roast. At noon, the chef put the pig in a large earthen oven. The pig’s original temperature was 75°F. At 2:00 pm the chef checked the pig’s temperature and was upset because it had reached only 100°F. If the oven’s temperature remains a constant 325°F

The temperature u of a heated object at a given time t is modeled by the function

$u\left(t\right)=T+\left(u_0-T\right)e^{kt}$.

At noon, the chef put the pig in a large earthen oven. The pig’s original temperature was 75°F. At 2:00 pm the chef checked the pig’s temperature and was upset because it had reached only 100°F

Substitute $u\left(t\right)=100,T=325,u_0=75,t=2$in the function $u\left(t\right)=T+\left(u_0-T\right)e^{kt}$

$$\begin{gathered} u\left(t\right)=T+\left(u_0-T\right)e^{kt} \\ 100=325+\left(75-325\right)e^{2k} \\ -225=-250e^{2k} \\ \frac{225}{250}=e^{2k} \\ \ln \left(0.9\right)=\ln \left(e^{2k}\right) \\ \ln \left(0.9\right)=2k \\ k=\frac{\ln \left(0.9\right)}{2} \\ k=-0.052 \end{gathered}$$

The time at which the hotel can serve its guests can be calculated by substituting

$u\left(t\right)=17,T=325,u_0=75,k=-0.052$in the function $u\left(t\right)=T+\left(u_0-T\right)e^{kt}$

$$\begin{gathered} u\left(t\right)=T+\left(u_0-T\right)e^{kt} \\ 175=325-250e^{-0.052t} \\ -150=-250e^{-0.052t} \\ 0.6=e^{-0.052t} \\ \ln \left(0.6\right)=\ln \left(e^{-0.052t}\right) \\ \ln \left(0.6\right)=-0.052t \\ t=\frac{\ln \left(0.6\right)}{-0.052} \\ t=9.82hours \end{gathered}$$