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Using \(\displaystyle T = T_0+(T_1-T_0)e^{kt}\) gives \(\displaystyle T = 60+(156-60)e^{kt}= 60+96e^{kt}\). Using the point \(\displaystyle (15, 124)\) gives \(\displaystyle 124= 60+96e^{k(15)}\). Isolating the exponential gives \(\displaystyle \frac{2}{3}=e^{15k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{2}{3} \right)}}{15}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle T = 60+96e^{\frac{\ln{\left(\frac{2}{3} \right)}}{15}t}\) and simplifying gives \(\displaystyle T = 96 \left(\frac{2}{3}\right)^{\frac{t}{15}} + 60\). Using \(\displaystyle t = 24\) gives \(\displaystyle T =96 \left(\frac{2}{3}\right)^{\frac{24}{15}} + 60\approx 110^\circ\)

Download \(\LaTeX\)

\begin{question}A coffee with temperature $156^\circ$ is left in a room with temperature $60^\circ$. After 15 minutes the temperature of the coffee is $124^\circ$, what is the temperature of the coffee after 24 minutes?
    \soln{9cm}{Using $T = T_0+(T_1-T_0)e^{kt}$ gives $T = 60+(156-60)e^{kt}= 60+96e^{kt}$. Using the point $(15, 124)$ gives $124= 60+96e^{k(15)}$. Isolating the exponential gives $\frac{2}{3}=e^{15k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{2}{3} \right)}}{15}$.  Substuting $k$ back into the equation gives $T = 60+96e^{\frac{\ln{\left(\frac{2}{3} \right)}}{15}t}$ and simplifying gives $T = 96 \left(\frac{2}{3}\right)^{\frac{t}{15}} + 60$. Using $t = 24$ gives $T =96 \left(\frac{2}{3}\right)^{\frac{24}{15}} + 60\approx 110^\circ$}

\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 156^\circ " src="/equation_images/%20%5Cdisplaystyle%20156%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 156^\circ " data-equation-content=" \displaystyle 156^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 60^\circ " src="/equation_images/%20%5Cdisplaystyle%2060%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 60^\circ " data-equation-content=" \displaystyle 60^\circ " /> . After 15 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 124^\circ " src="/equation_images/%20%5Cdisplaystyle%20124%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 124^\circ " data-equation-content=" \displaystyle 124^\circ " /> , what is the temperature of the coffee after 24 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 = 60+(156-60)e^{kt}= 60+96e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2060%2B%28156-60%29e%5E%7Bkt%7D%3D%2060%2B96e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = 60+(156-60)e^{kt}= 60+96e^{kt} " data-equation-content=" \displaystyle T = 60+(156-60)e^{kt}= 60+96e^{kt} " /> . Using the point  <img class="equation_image" title=" \displaystyle (15, 124) " src="/equation_images/%20%5Cdisplaystyle%20%2815%2C%20124%29%20" alt="LaTeX:  \displaystyle (15, 124) " data-equation-content=" \displaystyle (15, 124) " />  gives  <img class="equation_image" title=" \displaystyle 124= 60+96e^{k(15)} " src="/equation_images/%20%5Cdisplaystyle%20124%3D%2060%2B96e%5E%7Bk%2815%29%7D%20" alt="LaTeX:  \displaystyle 124= 60+96e^{k(15)} " data-equation-content=" \displaystyle 124= 60+96e^{k(15)} " /> . Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{2}{3}=e^{15k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B2%7D%7B3%7D%3De%5E%7B15k%7D%20" alt="LaTeX:  \displaystyle \frac{2}{3}=e^{15k} " data-equation-content=" \displaystyle \frac{2}{3}=e^{15k} " /> . 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{2}{3} \right)}}{15} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B2%7D%7B3%7D%20%5Cright%29%7D%7D%7B15%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{2}{3} \right)}}{15} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{2}{3} \right)}}{15} " /> .  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 = 60+96e^{\frac{\ln{\left(\frac{2}{3} \right)}}{15}t} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2060%2B96e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B2%7D%7B3%7D%20%5Cright%29%7D%7D%7B15%7Dt%7D%20" alt="LaTeX:  \displaystyle T = 60+96e^{\frac{\ln{\left(\frac{2}{3} \right)}}{15}t} " data-equation-content=" \displaystyle T = 60+96e^{\frac{\ln{\left(\frac{2}{3} \right)}}{15}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle T = 96 \left(\frac{2}{3}\right)^{\frac{t}{15}} + 60 " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2096%20%5Cleft%28%5Cfrac%7B2%7D%7B3%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B15%7D%7D%20%2B%2060%20" alt="LaTeX:  \displaystyle T = 96 \left(\frac{2}{3}\right)^{\frac{t}{15}} + 60 " data-equation-content=" \displaystyle T = 96 \left(\frac{2}{3}\right)^{\frac{t}{15}} + 60 " /> . Using  <img class="equation_image" title=" \displaystyle t = 24 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%2024%20" alt="LaTeX:  \displaystyle t = 24 " data-equation-content=" \displaystyle t = 24 " />  gives  <img class="equation_image" title=" \displaystyle T =96 \left(\frac{2}{3}\right)^{\frac{24}{15}} + 60\approx 110^\circ " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D96%20%5Cleft%28%5Cfrac%7B2%7D%7B3%7D%5Cright%29%5E%7B%5Cfrac%7B24%7D%7B15%7D%7D%20%2B%2060%5Capprox%20110%5E%5Ccirc%20" alt="LaTeX:  \displaystyle T =96 \left(\frac{2}{3}\right)^{\frac{24}{15}} + 60\approx 110^\circ " data-equation-content=" \displaystyle T =96 \left(\frac{2}{3}\right)^{\frac{24}{15}} + 60\approx 110^\circ " /> </p> </p>