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Using \(\displaystyle T = T_0+(T_1-T_0)e^{kt}\) gives \(\displaystyle T = 70+(167-70)e^{kt}= 70+97e^{kt}\). Using the point \(\displaystyle (10, 161)\) gives \(\displaystyle 161= 70+97e^{k(10)}\). Isolating the exponential gives \(\displaystyle \frac{91}{97}=e^{10k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{91}{97} \right)}}{10}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle T = 70+97e^{\frac{\ln{\left(\frac{91}{97} \right)}}{10}t}\) and simplifying gives \(\displaystyle T = 97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70\). Using \(\displaystyle T\) gives the equation \(\displaystyle 156=97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70\). Isolating the exponential gives \(\displaystyle \frac{86}{97}=\left(\frac{91}{97}\right)^{\frac{t}{10}}\). Taking the natural logarithm of both sides and solving for \(\displaystyle t\) gives \(\displaystyle t = \frac{10 \ln{\left(\frac{86}{97} \right)}}{\ln{\left(\frac{91}{97} \right)}}\approx 19\) minutes.

Download \(\LaTeX\)

\begin{question}A coffee with temperature $167^\circ$ is left in a room with temperature $70^\circ$. After 10 minutes the temperature of the coffee is $161^\circ$, how long until the coffee is $156^\circ$?
    \soln{9cm}{Using $T = T_0+(T_1-T_0)e^{kt}$ gives $T = 70+(167-70)e^{kt}= 70+97e^{kt}$. Using the point $(10, 161)$ gives $161= 70+97e^{k(10)}$. Isolating the exponential gives $\frac{91}{97}=e^{10k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{91}{97} \right)}}{10}$.  Substuting $k$ back into the equation gives $T = 70+97e^{\frac{\ln{\left(\frac{91}{97} \right)}}{10}t}$ and simplifying gives $T = 97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70$. Using $T$ gives the equation $156=97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70$.  Isolating the exponential gives $\frac{86}{97}=\left(\frac{91}{97}\right)^{\frac{t}{10}}$. Taking the natural logarithm of both sides and solving for $t$ gives $t = \frac{10 \ln{\left(\frac{86}{97} \right)}}{\ln{\left(\frac{91}{97} \right)}}\approx 19$ 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 167^\circ " src="/equation_images/%20%5Cdisplaystyle%20167%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 167^\circ " data-equation-content=" \displaystyle 167^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 70^\circ " src="/equation_images/%20%5Cdisplaystyle%2070%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 70^\circ " data-equation-content=" \displaystyle 70^\circ " /> . After 10 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 161^\circ " src="/equation_images/%20%5Cdisplaystyle%20161%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 161^\circ " data-equation-content=" \displaystyle 161^\circ " /> , how long until the coffee is  <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 " /> ?</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 = 70+(167-70)e^{kt}= 70+97e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2070%2B%28167-70%29e%5E%7Bkt%7D%3D%2070%2B97e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = 70+(167-70)e^{kt}= 70+97e^{kt} " data-equation-content=" \displaystyle T = 70+(167-70)e^{kt}= 70+97e^{kt} " /> . Using the point  <img class="equation_image" title=" \displaystyle (10, 161) " src="/equation_images/%20%5Cdisplaystyle%20%2810%2C%20161%29%20" alt="LaTeX:  \displaystyle (10, 161) " data-equation-content=" \displaystyle (10, 161) " />  gives  <img class="equation_image" title=" \displaystyle 161= 70+97e^{k(10)} " src="/equation_images/%20%5Cdisplaystyle%20161%3D%2070%2B97e%5E%7Bk%2810%29%7D%20" alt="LaTeX:  \displaystyle 161= 70+97e^{k(10)} " data-equation-content=" \displaystyle 161= 70+97e^{k(10)} " /> . Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{91}{97}=e^{10k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B91%7D%7B97%7D%3De%5E%7B10k%7D%20" alt="LaTeX:  \displaystyle \frac{91}{97}=e^{10k} " data-equation-content=" \displaystyle \frac{91}{97}=e^{10k} " /> . 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{91}{97} \right)}}{10} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B91%7D%7B97%7D%20%5Cright%29%7D%7D%7B10%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{91}{97} \right)}}{10} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{91}{97} \right)}}{10} " /> .  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 = 70+97e^{\frac{\ln{\left(\frac{91}{97} \right)}}{10}t} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2070%2B97e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B91%7D%7B97%7D%20%5Cright%29%7D%7D%7B10%7Dt%7D%20" alt="LaTeX:  \displaystyle T = 70+97e^{\frac{\ln{\left(\frac{91}{97} \right)}}{10}t} " data-equation-content=" \displaystyle T = 70+97e^{\frac{\ln{\left(\frac{91}{97} \right)}}{10}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle T = 97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70 " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2097%20%5Cleft%28%5Cfrac%7B91%7D%7B97%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B10%7D%7D%20%2B%2070%20" alt="LaTeX:  \displaystyle T = 97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70 " data-equation-content=" \displaystyle T = 97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70 " /> . 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 156=97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70 " src="/equation_images/%20%5Cdisplaystyle%20156%3D97%20%5Cleft%28%5Cfrac%7B91%7D%7B97%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B10%7D%7D%20%2B%2070%20" alt="LaTeX:  \displaystyle 156=97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70 " data-equation-content=" \displaystyle 156=97 \left(\frac{91}{97}\right)^{\frac{t}{10}} + 70 " /> .  Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{86}{97}=\left(\frac{91}{97}\right)^{\frac{t}{10}} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B86%7D%7B97%7D%3D%5Cleft%28%5Cfrac%7B91%7D%7B97%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B10%7D%7D%20" alt="LaTeX:  \displaystyle \frac{86}{97}=\left(\frac{91}{97}\right)^{\frac{t}{10}} " data-equation-content=" \displaystyle \frac{86}{97}=\left(\frac{91}{97}\right)^{\frac{t}{10}} " /> . 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{10 \ln{\left(\frac{86}{97} \right)}}{\ln{\left(\frac{91}{97} \right)}}\approx 19 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%20%5Cfrac%7B10%20%5Cln%7B%5Cleft%28%5Cfrac%7B86%7D%7B97%7D%20%5Cright%29%7D%7D%7B%5Cln%7B%5Cleft%28%5Cfrac%7B91%7D%7B97%7D%20%5Cright%29%7D%7D%5Capprox%2019%20" alt="LaTeX:  \displaystyle t = \frac{10 \ln{\left(\frac{86}{97} \right)}}{\ln{\left(\frac{91}{97} \right)}}\approx 19 " data-equation-content=" \displaystyle t = \frac{10 \ln{\left(\frac{86}{97} \right)}}{\ln{\left(\frac{91}{97} \right)}}\approx 19 " />  minutes. </p> </p>