A coffee with temperature $LaTeX: \displaystyle 156^\circ$ is left in a room with temperature $LaTeX: \displaystyle 61^\circ$ . After 7 minutes the temperature of the coffee is $LaTeX: \displaystyle 128^\circ$ , how long until the coffee is $LaTeX: \displaystyle 105^\circ$ ?

Using $LaTeX: \displaystyle T = T_0+(T_1-T_0)e^{kt}$ gives $LaTeX: \displaystyle T = 61+(156-61)e^{kt}= 61+95e^{kt}$ . Using the point $LaTeX: \displaystyle (7, 128)$ gives $LaTeX: \displaystyle 128= 61+95e^{k(7)}$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{67}{95}=e^{7k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{67}{95} \right)}}{7}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle T = 61+95e^{\frac{\ln{\left(\frac{67}{95} \right)}}{7}t}$ and simplifying gives $LaTeX: \displaystyle T = 95 \left(\frac{67}{95}\right)^{\frac{t}{7}} + 61$ . Using $LaTeX: \displaystyle T$ gives the equation $LaTeX: \displaystyle 105=95 \left(\frac{67}{95}\right)^{\frac{t}{7}} + 61$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{44}{95}=\left(\frac{67}{95}\right)^{\frac{t}{7}}$ . Taking the natural logarithm of both sides and solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = \frac{7 \ln{\left(\frac{44}{95} \right)}}{\ln{\left(\frac{67}{95} \right)}}\approx 15$ minutes.