A coffee with temperature $LaTeX: \displaystyle 156^\circ$ is left in a room with temperature $LaTeX: \displaystyle 60^\circ$ . After 15 minutes the temperature of the coffee is $LaTeX: \displaystyle 124^\circ$ , what is the temperature of the coffee after 24 minutes?

Using $LaTeX: \displaystyle T = T_0+(T_1-T_0)e^{kt}$ gives $LaTeX: \displaystyle T = 60+(156-60)e^{kt}= 60+96e^{kt}$ . Using the point $LaTeX: \displaystyle (15, 124)$ gives $LaTeX: \displaystyle 124= 60+96e^{k(15)}$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{2}{3}=e^{15k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{2}{3} \right)}}{15}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle T = 60+96e^{\frac{\ln{\left(\frac{2}{3} \right)}}{15}t}$ and simplifying gives $LaTeX: \displaystyle T = 96 \left(\frac{2}{3}\right)^{\frac{t}{15}} + 60$ . Using $LaTeX: \displaystyle t = 24$ gives $LaTeX: \displaystyle T =96 \left(\frac{2}{3}\right)^{\frac{24}{15}} + 60\approx 110^\circ$