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Using \(\displaystyle T = T_0+(T_1-T_0)e^{kt}\) gives \(\displaystyle T = 67+(168-67)e^{kt}= 67+101e^{kt}\). Using the point \(\displaystyle (6, 163)\) gives \(\displaystyle 163= 67+101e^{k(6)}\). Isolating the exponential gives \(\displaystyle \frac{96}{101}=e^{6k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{96}{101} \right)}}{6}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle T = 67+101e^{\frac{\ln{\left(\frac{96}{101} \right)}}{6}t}\) and simplifying gives \(\displaystyle T = 101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67\). Using \(\displaystyle T\) gives the equation \(\displaystyle 144=101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67\). Isolating the exponential gives \(\displaystyle \frac{77}{101}=\left(\frac{96}{101}\right)^{\frac{t}{6}}\). Taking the natural logarithm of both sides and solving for \(\displaystyle t\) gives \(\displaystyle t = \frac{6 \ln{\left(\frac{77}{101} \right)}}{\ln{\left(\frac{96}{101} \right)}}\approx 32\) minutes.

Download \(\LaTeX\)

\begin{question}A coffee with temperature $168^\circ$ is left in a room with temperature $67^\circ$. After 6 minutes the temperature of the coffee is $163^\circ$, how long until the coffee is $144^\circ$?
    \soln{9cm}{Using $T = T_0+(T_1-T_0)e^{kt}$ gives $T = 67+(168-67)e^{kt}= 67+101e^{kt}$. Using the point $(6, 163)$ gives $163= 67+101e^{k(6)}$. Isolating the exponential gives $\frac{96}{101}=e^{6k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{96}{101} \right)}}{6}$.  Substuting $k$ back into the equation gives $T = 67+101e^{\frac{\ln{\left(\frac{96}{101} \right)}}{6}t}$ and simplifying gives $T = 101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67$. Using $T$ gives the equation $144=101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67$.  Isolating the exponential gives $\frac{77}{101}=\left(\frac{96}{101}\right)^{\frac{t}{6}}$. Taking the natural logarithm of both sides and solving for $t$ gives $t = \frac{6 \ln{\left(\frac{77}{101} \right)}}{\ln{\left(\frac{96}{101} \right)}}\approx 32$ minutes. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\begin{document}\begin{question}(10pts) The question goes here!
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HTML for Canvas

<p> <p>A coffee with temperature  <img class="equation_image" title=" \displaystyle 168^\circ " src="/equation_images/%20%5Cdisplaystyle%20168%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 168^\circ " data-equation-content=" \displaystyle 168^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 67^\circ " src="/equation_images/%20%5Cdisplaystyle%2067%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 67^\circ " data-equation-content=" \displaystyle 67^\circ " /> . After 6 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 163^\circ " src="/equation_images/%20%5Cdisplaystyle%20163%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 163^\circ " data-equation-content=" \displaystyle 163^\circ " /> , how long until the coffee is  <img class="equation_image" title=" \displaystyle 144^\circ " src="/equation_images/%20%5Cdisplaystyle%20144%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 144^\circ " data-equation-content=" \displaystyle 144^\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 = 67+(168-67)e^{kt}= 67+101e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2067%2B%28168-67%29e%5E%7Bkt%7D%3D%2067%2B101e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = 67+(168-67)e^{kt}= 67+101e^{kt} " data-equation-content=" \displaystyle T = 67+(168-67)e^{kt}= 67+101e^{kt} " /> . Using the point  <img class="equation_image" title=" \displaystyle (6, 163) " src="/equation_images/%20%5Cdisplaystyle%20%286%2C%20163%29%20" alt="LaTeX:  \displaystyle (6, 163) " data-equation-content=" \displaystyle (6, 163) " />  gives  <img class="equation_image" title=" \displaystyle 163= 67+101e^{k(6)} " src="/equation_images/%20%5Cdisplaystyle%20163%3D%2067%2B101e%5E%7Bk%286%29%7D%20" alt="LaTeX:  \displaystyle 163= 67+101e^{k(6)} " data-equation-content=" \displaystyle 163= 67+101e^{k(6)} " /> . Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{96}{101}=e^{6k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B96%7D%7B101%7D%3De%5E%7B6k%7D%20" alt="LaTeX:  \displaystyle \frac{96}{101}=e^{6k} " data-equation-content=" \displaystyle \frac{96}{101}=e^{6k} " /> . 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{96}{101} \right)}}{6} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B96%7D%7B101%7D%20%5Cright%29%7D%7D%7B6%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{96}{101} \right)}}{6} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{96}{101} \right)}}{6} " /> .  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 = 67+101e^{\frac{\ln{\left(\frac{96}{101} \right)}}{6}t} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2067%2B101e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B96%7D%7B101%7D%20%5Cright%29%7D%7D%7B6%7Dt%7D%20" alt="LaTeX:  \displaystyle T = 67+101e^{\frac{\ln{\left(\frac{96}{101} \right)}}{6}t} " data-equation-content=" \displaystyle T = 67+101e^{\frac{\ln{\left(\frac{96}{101} \right)}}{6}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle T = 101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67 " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%20101%20%5Cleft%28%5Cfrac%7B96%7D%7B101%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B6%7D%7D%20%2B%2067%20" alt="LaTeX:  \displaystyle T = 101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67 " data-equation-content=" \displaystyle T = 101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67 " /> . 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 144=101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67 " src="/equation_images/%20%5Cdisplaystyle%20144%3D101%20%5Cleft%28%5Cfrac%7B96%7D%7B101%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B6%7D%7D%20%2B%2067%20" alt="LaTeX:  \displaystyle 144=101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67 " data-equation-content=" \displaystyle 144=101 \left(\frac{96}{101}\right)^{\frac{t}{6}} + 67 " /> .  Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{77}{101}=\left(\frac{96}{101}\right)^{\frac{t}{6}} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B77%7D%7B101%7D%3D%5Cleft%28%5Cfrac%7B96%7D%7B101%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B6%7D%7D%20" alt="LaTeX:  \displaystyle \frac{77}{101}=\left(\frac{96}{101}\right)^{\frac{t}{6}} " data-equation-content=" \displaystyle \frac{77}{101}=\left(\frac{96}{101}\right)^{\frac{t}{6}} " /> . 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{6 \ln{\left(\frac{77}{101} \right)}}{\ln{\left(\frac{96}{101} \right)}}\approx 32 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%20%5Cfrac%7B6%20%5Cln%7B%5Cleft%28%5Cfrac%7B77%7D%7B101%7D%20%5Cright%29%7D%7D%7B%5Cln%7B%5Cleft%28%5Cfrac%7B96%7D%7B101%7D%20%5Cright%29%7D%7D%5Capprox%2032%20" alt="LaTeX:  \displaystyle t = \frac{6 \ln{\left(\frac{77}{101} \right)}}{\ln{\left(\frac{96}{101} \right)}}\approx 32 " data-equation-content=" \displaystyle t = \frac{6 \ln{\left(\frac{77}{101} \right)}}{\ln{\left(\frac{96}{101} \right)}}\approx 32 " />  minutes. </p> </p>