A coffee with temperature $LaTeX: \displaystyle 162^\circ$ is left in a room with temperature $LaTeX: \displaystyle 82^\circ$ . After 10 minutes the temperature of the coffee is $LaTeX: \displaystyle 138^\circ$ , what is the temperature of the coffee after 13 minutes?

Using $LaTeX: \displaystyle T = T_0+(T_1-T_0)e^{kt}$ gives $LaTeX: \displaystyle T = 82+(162-82)e^{kt}= 82+80e^{kt}$ . Using the point $LaTeX: \displaystyle (10, 138)$ gives $LaTeX: \displaystyle 138= 82+80e^{k(10)}$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{7}{10}=e^{10k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{7}{10} \right)}}{10}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle T = 82+80e^{\frac{\ln{\left(\frac{7}{10} \right)}}{10}t}$ and simplifying gives $LaTeX: \displaystyle T = 80 \left(\frac{7}{10}\right)^{\frac{t}{10}} + 82$ . Using $LaTeX: \displaystyle t = 13$ gives $LaTeX: \displaystyle T =80 \left(\frac{7}{10}\right)^{\frac{13}{10}} + 82\approx 132^\circ$