A coffee with temperature $LaTeX: \displaystyle 154^\circ$ is left in a room with temperature $LaTeX: \displaystyle 66^\circ$ . After 15 minutes the temperature of the coffee is $LaTeX: \displaystyle 147^\circ$ , how long until the coffee is $LaTeX: \displaystyle 139^\circ$ ?

Using $LaTeX: \displaystyle T = T_0+(T_1-T_0)e^{kt}$ gives $LaTeX: \displaystyle T = 66+(154-66)e^{kt}= 66+88e^{kt}$ . Using the point $LaTeX: \displaystyle (15, 147)$ gives $LaTeX: \displaystyle 147= 66+88e^{k(15)}$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{81}{88}=e^{15k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{81}{88} \right)}}{15}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle T = 66+88e^{\frac{\ln{\left(\frac{81}{88} \right)}}{15}t}$ and simplifying gives $LaTeX: \displaystyle T = 88 \left(\frac{81}{88}\right)^{\frac{t}{15}} + 66$ . Using $LaTeX: \displaystyle T$ gives the equation $LaTeX: \displaystyle 139=88 \left(\frac{81}{88}\right)^{\frac{t}{15}} + 66$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{73}{88}=\left(\frac{81}{88}\right)^{\frac{t}{15}}$ . Taking the natural logarithm of both sides and solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = \frac{15 \ln{\left(\frac{73}{88} \right)}}{\ln{\left(\frac{81}{88} \right)}}\approx 34$ minutes.