A coffee with temperature $LaTeX: \displaystyle 168^\circ$ is left in a room with temperature $LaTeX: \displaystyle 67^\circ$ . After 6 minutes the temperature of the coffee is $LaTeX: \displaystyle 163^\circ$ , how long until the coffee is $LaTeX: \displaystyle 144^\circ$ ?

Using $LaTeX: \displaystyle T = T_0+(T_1-T_0)e^{kt}$ gives $LaTeX: \displaystyle T = 67+(168-67)e^{kt}= 67+101e^{kt}$ . Using the point $LaTeX: \displaystyle (6, 163)$ gives $LaTeX: \displaystyle 163= 67+101e^{k(6)}$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{96}{101}=e^{6k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{96}{101} \right)}}{6}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle T = 67+101e^{\frac{\ln{\left(\frac{96}{101} \right)}}{6}t}$ and simplifying gives $LaTeX: \displaystyle T = 101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67$ . Using $LaTeX: \displaystyle T$ gives the equation $LaTeX: \displaystyle 144=101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{77}{101}=\left(\frac{96}{101}\right)^{\frac{t}{6}}$ . Taking the natural logarithm of both sides and solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = \frac{6 \ln{\left(\frac{77}{101} \right)}}{\ln{\left(\frac{96}{101} \right)}}\approx 32$ minutes.