## Precalculus (6th Edition) Blitzer

a. $T=65+120e^{-0.1438t}$ b. after about 8 minutes.
a. We are given $T_0=185^{\circ}F, C=65^{\circ}F, T(2)=155^{\circ}F$ We have $155=65+(185-65)e^{2k}$ which gives $k=\frac{ln(90/120)}{2}\approx-0.1438$ Thus the model equation is $T=65+120e^{-0.1438t}$ b. Letting $T=105$, we have $65+120e^{-0.1438t}=105$ which gives $t=-\frac{ln(40/120)}{0.1438}\approx8min$ that is, after about 8 minutes.