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

Download \(\LaTeX\)

\begin{question}A coffee with temperature $156^\circ$ is left in a room with temperature $61^\circ$. After 7 minutes the temperature of the coffee is $128^\circ$, how long until the coffee is $105^\circ$?
    \soln{9cm}{Using $T = T_0+(T_1-T_0)e^{kt}$ gives $T = 61+(156-61)e^{kt}= 61+95e^{kt}$. Using the point $(7, 128)$ gives $128= 61+95e^{k(7)}$. Isolating the exponential gives $\frac{67}{95}=e^{7k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{67}{95} \right)}}{7}$.  Substuting $k$ back into the equation gives $T = 61+95e^{\frac{\ln{\left(\frac{67}{95} \right)}}{7}t}$ and simplifying gives $T = 95 \left(\frac{67}{95}\right)^{\frac{t}{7}} + 61$. Using $T$ gives the equation $105=95 \left(\frac{67}{95}\right)^{\frac{t}{7}} + 61$.  Isolating the exponential gives $\frac{44}{95}=\left(\frac{67}{95}\right)^{\frac{t}{7}}$. Taking the natural logarithm of both sides and solving for $t$ gives $t = \frac{7 \ln{\left(\frac{44}{95} \right)}}{\ln{\left(\frac{67}{95} \right)}}\approx 15$ minutes. }

\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.}

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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 61^\circ " src="/equation_images/%20%5Cdisplaystyle%2061%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 61^\circ " data-equation-content=" \displaystyle 61^\circ " /> . After 7 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 128^\circ " src="/equation_images/%20%5Cdisplaystyle%20128%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 128^\circ " data-equation-content=" \displaystyle 128^\circ " /> , how long until the coffee is  <img class="equation_image" title=" \displaystyle 105^\circ " src="/equation_images/%20%5Cdisplaystyle%20105%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 105^\circ " data-equation-content=" \displaystyle 105^\circ " /> ?</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 = 61+(156-61)e^{kt}= 61+95e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2061%2B%28156-61%29e%5E%7Bkt%7D%3D%2061%2B95e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = 61+(156-61)e^{kt}= 61+95e^{kt} " data-equation-content=" \displaystyle T = 61+(156-61)e^{kt}= 61+95e^{kt} " /> . Using the point  <img class="equation_image" title=" \displaystyle (7, 128) " src="/equation_images/%20%5Cdisplaystyle%20%287%2C%20128%29%20" alt="LaTeX:  \displaystyle (7, 128) " data-equation-content=" \displaystyle (7, 128) " />  gives  <img class="equation_image" title=" \displaystyle 128= 61+95e^{k(7)} " src="/equation_images/%20%5Cdisplaystyle%20128%3D%2061%2B95e%5E%7Bk%287%29%7D%20" alt="LaTeX:  \displaystyle 128= 61+95e^{k(7)} " data-equation-content=" \displaystyle 128= 61+95e^{k(7)} " /> . Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{67}{95}=e^{7k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B67%7D%7B95%7D%3De%5E%7B7k%7D%20" alt="LaTeX:  \displaystyle \frac{67}{95}=e^{7k} " data-equation-content=" \displaystyle \frac{67}{95}=e^{7k} " /> . 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{67}{95} \right)}}{7} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B67%7D%7B95%7D%20%5Cright%29%7D%7D%7B7%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{67}{95} \right)}}{7} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{67}{95} \right)}}{7} " /> .  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 = 61+95e^{\frac{\ln{\left(\frac{67}{95} \right)}}{7}t} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2061%2B95e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B67%7D%7B95%7D%20%5Cright%29%7D%7D%7B7%7Dt%7D%20" alt="LaTeX:  \displaystyle T = 61+95e^{\frac{\ln{\left(\frac{67}{95} \right)}}{7}t} " data-equation-content=" \displaystyle T = 61+95e^{\frac{\ln{\left(\frac{67}{95} \right)}}{7}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle T = 95 \left(\frac{67}{95}\right)^{\frac{t}{7}} + 61 " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2095%20%5Cleft%28%5Cfrac%7B67%7D%7B95%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B7%7D%7D%20%2B%2061%20" alt="LaTeX:  \displaystyle T = 95 \left(\frac{67}{95}\right)^{\frac{t}{7}} + 61 " data-equation-content=" \displaystyle T = 95 \left(\frac{67}{95}\right)^{\frac{t}{7}} + 61 " /> . Using  <img class="equation_image" title=" \displaystyle T " src="/equation_images/%20%5Cdisplaystyle%20T%20" alt="LaTeX:  \displaystyle T " data-equation-content=" \displaystyle T " />  gives the equation  <img class="equation_image" title=" \displaystyle 105=95 \left(\frac{67}{95}\right)^{\frac{t}{7}} + 61 " src="/equation_images/%20%5Cdisplaystyle%20105%3D95%20%5Cleft%28%5Cfrac%7B67%7D%7B95%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B7%7D%7D%20%2B%2061%20" alt="LaTeX:  \displaystyle 105=95 \left(\frac{67}{95}\right)^{\frac{t}{7}} + 61 " data-equation-content=" \displaystyle 105=95 \left(\frac{67}{95}\right)^{\frac{t}{7}} + 61 " /> .  Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{44}{95}=\left(\frac{67}{95}\right)^{\frac{t}{7}} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B44%7D%7B95%7D%3D%5Cleft%28%5Cfrac%7B67%7D%7B95%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B7%7D%7D%20" alt="LaTeX:  \displaystyle \frac{44}{95}=\left(\frac{67}{95}\right)^{\frac{t}{7}} " data-equation-content=" \displaystyle \frac{44}{95}=\left(\frac{67}{95}\right)^{\frac{t}{7}} " /> . Taking the natural logarithm of both sides and solving for  <img class="equation_image" title=" \displaystyle t " src="/equation_images/%20%5Cdisplaystyle%20t%20" alt="LaTeX:  \displaystyle t " data-equation-content=" \displaystyle t " />  gives  <img class="equation_image" title=" \displaystyle t = \frac{7 \ln{\left(\frac{44}{95} \right)}}{\ln{\left(\frac{67}{95} \right)}}\approx 15 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%20%5Cfrac%7B7%20%5Cln%7B%5Cleft%28%5Cfrac%7B44%7D%7B95%7D%20%5Cright%29%7D%7D%7B%5Cln%7B%5Cleft%28%5Cfrac%7B67%7D%7B95%7D%20%5Cright%29%7D%7D%5Capprox%2015%20" alt="LaTeX:  \displaystyle t = \frac{7 \ln{\left(\frac{44}{95} \right)}}{\ln{\left(\frac{67}{95} \right)}}\approx 15 " data-equation-content=" \displaystyle t = \frac{7 \ln{\left(\frac{44}{95} \right)}}{\ln{\left(\frac{67}{95} \right)}}\approx 15 " />  minutes. </p> </p>