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Using \(\displaystyle T = T_0+(T_1-T_0)e^{kt}\) gives \(\displaystyle T = 72+(170-72)e^{kt}= 72+98e^{kt}\). Using the point \(\displaystyle (5, 164)\) gives \(\displaystyle 164= 72+98e^{k(5)}\). Isolating the exponential gives \(\displaystyle \frac{46}{49}=e^{5k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{46}{49} \right)}}{5}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle T = 72+98e^{\frac{\ln{\left(\frac{46}{49} \right)}}{5}t}\) and simplifying gives \(\displaystyle T = 98 \left(\frac{46}{49}\right)^{\frac{t}{5}} + 72\). Using \(\displaystyle t = 14\) gives \(\displaystyle T =98 \left(\frac{46}{49}\right)^{\frac{14}{5}} + 72\approx 154^\circ\)

Download \(\LaTeX\)

\begin{question}A coffee with temperature $170^\circ$ is left in a room with temperature $72^\circ$. After 5 minutes the temperature of the coffee is $164^\circ$, what is the temperature of the coffee after 14 minutes?
    \soln{9cm}{Using $T = T_0+(T_1-T_0)e^{kt}$ gives $T = 72+(170-72)e^{kt}= 72+98e^{kt}$. Using the point $(5, 164)$ gives $164= 72+98e^{k(5)}$. Isolating the exponential gives $\frac{46}{49}=e^{5k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{46}{49} \right)}}{5}$.  Substuting $k$ back into the equation gives $T = 72+98e^{\frac{\ln{\left(\frac{46}{49} \right)}}{5}t}$ and simplifying gives $T = 98 \left(\frac{46}{49}\right)^{\frac{t}{5}} + 72$. Using $t = 14$ gives $T =98 \left(\frac{46}{49}\right)^{\frac{14}{5}} + 72\approx 154^\circ$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>A coffee with temperature  <img class="equation_image" title=" \displaystyle 170^\circ " src="/equation_images/%20%5Cdisplaystyle%20170%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 170^\circ " data-equation-content=" \displaystyle 170^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 72^\circ " src="/equation_images/%20%5Cdisplaystyle%2072%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 72^\circ " data-equation-content=" \displaystyle 72^\circ " /> . After 5 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 164^\circ " src="/equation_images/%20%5Cdisplaystyle%20164%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 164^\circ " data-equation-content=" \displaystyle 164^\circ " /> , what is the temperature of the coffee after 14 minutes?</p> </p>

HTML for Canvas

<p> <p>Using  <img class="equation_image" title=" \displaystyle T = T_0+(T_1-T_0)e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%20T_0%2B%28T_1-T_0%29e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = T_0+(T_1-T_0)e^{kt} " data-equation-content=" \displaystyle T = T_0+(T_1-T_0)e^{kt} " />  gives  <img class="equation_image" title=" \displaystyle T = 72+(170-72)e^{kt}= 72+98e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2072%2B%28170-72%29e%5E%7Bkt%7D%3D%2072%2B98e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = 72+(170-72)e^{kt}= 72+98e^{kt} " data-equation-content=" \displaystyle T = 72+(170-72)e^{kt}= 72+98e^{kt} " /> . Using the point  <img class="equation_image" title=" \displaystyle (5, 164) " src="/equation_images/%20%5Cdisplaystyle%20%285%2C%20164%29%20" alt="LaTeX:  \displaystyle (5, 164) " data-equation-content=" \displaystyle (5, 164) " />  gives  <img class="equation_image" title=" \displaystyle 164= 72+98e^{k(5)} " src="/equation_images/%20%5Cdisplaystyle%20164%3D%2072%2B98e%5E%7Bk%285%29%7D%20" alt="LaTeX:  \displaystyle 164= 72+98e^{k(5)} " data-equation-content=" \displaystyle 164= 72+98e^{k(5)} " /> . Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{46}{49}=e^{5k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B46%7D%7B49%7D%3De%5E%7B5k%7D%20" alt="LaTeX:  \displaystyle \frac{46}{49}=e^{5k} " data-equation-content=" \displaystyle \frac{46}{49}=e^{5k} " /> . Solving for  <img class="equation_image" title=" \displaystyle k " src="/equation_images/%20%5Cdisplaystyle%20k%20" alt="LaTeX:  \displaystyle k " data-equation-content=" \displaystyle k " />  gives  <img class="equation_image" title=" \displaystyle k=\frac{\ln{\left(\frac{46}{49} \right)}}{5} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B46%7D%7B49%7D%20%5Cright%29%7D%7D%7B5%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{46}{49} \right)}}{5} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{46}{49} \right)}}{5} " /> .  Substuting  <img class="equation_image" title=" \displaystyle k " src="/equation_images/%20%5Cdisplaystyle%20k%20" alt="LaTeX:  \displaystyle k " data-equation-content=" \displaystyle k " />  back into the equation gives  <img class="equation_image" title=" \displaystyle T = 72+98e^{\frac{\ln{\left(\frac{46}{49} \right)}}{5}t} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2072%2B98e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B46%7D%7B49%7D%20%5Cright%29%7D%7D%7B5%7Dt%7D%20" alt="LaTeX:  \displaystyle T = 72+98e^{\frac{\ln{\left(\frac{46}{49} \right)}}{5}t} " data-equation-content=" \displaystyle T = 72+98e^{\frac{\ln{\left(\frac{46}{49} \right)}}{5}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle T = 98 \left(\frac{46}{49}\right)^{\frac{t}{5}} + 72 " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2098%20%5Cleft%28%5Cfrac%7B46%7D%7B49%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B5%7D%7D%20%2B%2072%20" alt="LaTeX:  \displaystyle T = 98 \left(\frac{46}{49}\right)^{\frac{t}{5}} + 72 " data-equation-content=" \displaystyle T = 98 \left(\frac{46}{49}\right)^{\frac{t}{5}} + 72 " /> . Using  <img class="equation_image" title=" \displaystyle t = 14 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%2014%20" alt="LaTeX:  \displaystyle t = 14 " data-equation-content=" \displaystyle t = 14 " />  gives  <img class="equation_image" title=" \displaystyle T =98 \left(\frac{46}{49}\right)^{\frac{14}{5}} + 72\approx 154^\circ " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D98%20%5Cleft%28%5Cfrac%7B46%7D%7B49%7D%5Cright%29%5E%7B%5Cfrac%7B14%7D%7B5%7D%7D%20%2B%2072%5Capprox%20154%5E%5Ccirc%20" alt="LaTeX:  \displaystyle T =98 \left(\frac{46}{49}\right)^{\frac{14}{5}} + 72\approx 154^\circ " data-equation-content=" \displaystyle T =98 \left(\frac{46}{49}\right)^{\frac{14}{5}} + 72\approx 154^\circ " /> </p> </p>