A coffee with temperature $LaTeX: \displaystyle 178^\circ$ is left in a room with temperature $LaTeX: \displaystyle 66^\circ$ . After 8 minutes the temperature of the coffee is $LaTeX: \displaystyle 153^\circ$ , how long until the coffee is $LaTeX: \displaystyle 133^\circ$ ?

Using $LaTeX: \displaystyle T = T_0+(T_1-T_0)e^{kt}$ gives $LaTeX: \displaystyle T = 66+(178-66)e^{kt}= 66+112e^{kt}$ . Using the point $LaTeX: \displaystyle (8, 153)$ gives $LaTeX: \displaystyle 153= 66+112e^{k(8)}$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{87}{112}=e^{8k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{87}{112} \right)}}{8}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle T = 66+112e^{\frac{\ln{\left(\frac{87}{112} \right)}}{8}t}$ and simplifying gives $LaTeX: \displaystyle T = 112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66$ . Using $LaTeX: \displaystyle T$ gives the equation $LaTeX: \displaystyle 133=112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{67}{112}=\left(\frac{87}{112}\right)^{\frac{t}{8}}$ . Taking the natural logarithm of both sides and solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = \frac{8 \ln{\left(\frac{67}{112} \right)}}{\ln{\left(\frac{87}{112} \right)}}\approx 16$ minutes.