A coffee with temperature $LaTeX: \displaystyle 167^\circ$ is left in a room with temperature $LaTeX: \displaystyle 70^\circ$ . After 10 minutes the temperature of the coffee is $LaTeX: \displaystyle 161^\circ$ , how long until the coffee is $LaTeX: \displaystyle 156^\circ$ ?

Using $LaTeX: \displaystyle T = T_0+(T_1-T_0)e^{kt}$ gives $LaTeX: \displaystyle T = 70+(167-70)e^{kt}= 70+97e^{kt}$ . Using the point $LaTeX: \displaystyle (10, 161)$ gives $LaTeX: \displaystyle 161= 70+97e^{k(10)}$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{91}{97}=e^{10k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{91}{97} \right)}}{10}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle T = 70+97e^{\frac{\ln{\left(\frac{91}{97} \right)}}{10}t}$ and simplifying gives $LaTeX: \displaystyle T = 97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70$ . Using $LaTeX: \displaystyle T$ gives the equation $LaTeX: \displaystyle 156=97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{86}{97}=\left(\frac{91}{97}\right)^{\frac{t}{10}}$ . Taking the natural logarithm of both sides and solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = \frac{10 \ln{\left(\frac{86}{97} \right)}}{\ln{\left(\frac{91}{97} \right)}}\approx 19$ minutes.