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Using \(\displaystyle T = T_0+(T_1-T_0)e^{kt}\) gives \(\displaystyle T = 66+(178-66)e^{kt}= 66+112e^{kt}\). Using the point \(\displaystyle (8, 153)\) gives \(\displaystyle 153= 66+112e^{k(8)}\). Isolating the exponential gives \(\displaystyle \frac{87}{112}=e^{8k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{87}{112} \right)}}{8}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle T = 66+112e^{\frac{\ln{\left(\frac{87}{112} \right)}}{8}t}\) and simplifying gives \(\displaystyle T = 112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66\). Using \(\displaystyle T\) gives the equation \(\displaystyle 133=112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66\). Isolating the exponential gives \(\displaystyle \frac{67}{112}=\left(\frac{87}{112}\right)^{\frac{t}{8}}\). Taking the natural logarithm of both sides and solving for \(\displaystyle t\) gives \(\displaystyle t = \frac{8 \ln{\left(\frac{67}{112} \right)}}{\ln{\left(\frac{87}{112} \right)}}\approx 16\) minutes.

Download \(\LaTeX\)

\begin{question}A coffee with temperature $178^\circ$ is left in a room with temperature $66^\circ$. After 8 minutes the temperature of the coffee is $153^\circ$, how long until the coffee is $133^\circ$?
    \soln{9cm}{Using $T = T_0+(T_1-T_0)e^{kt}$ gives $T = 66+(178-66)e^{kt}= 66+112e^{kt}$. Using the point $(8, 153)$ gives $153= 66+112e^{k(8)}$. Isolating the exponential gives $\frac{87}{112}=e^{8k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{87}{112} \right)}}{8}$.  Substuting $k$ back into the equation gives $T = 66+112e^{\frac{\ln{\left(\frac{87}{112} \right)}}{8}t}$ and simplifying gives $T = 112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66$. Using $T$ gives the equation $133=112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66$.  Isolating the exponential gives $\frac{67}{112}=\left(\frac{87}{112}\right)^{\frac{t}{8}}$. Taking the natural logarithm of both sides and solving for $t$ gives $t = \frac{8 \ln{\left(\frac{67}{112} \right)}}{\ln{\left(\frac{87}{112} \right)}}\approx 16$ minutes. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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HTML for Canvas

<p> <p>A coffee with temperature  <img class="equation_image" title=" \displaystyle 178^\circ " src="/equation_images/%20%5Cdisplaystyle%20178%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 178^\circ " data-equation-content=" \displaystyle 178^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 66^\circ " src="/equation_images/%20%5Cdisplaystyle%2066%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 66^\circ " data-equation-content=" \displaystyle 66^\circ " /> . After 8 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 153^\circ " src="/equation_images/%20%5Cdisplaystyle%20153%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 153^\circ " data-equation-content=" \displaystyle 153^\circ " /> , how long until the coffee is  <img class="equation_image" title=" \displaystyle 133^\circ " src="/equation_images/%20%5Cdisplaystyle%20133%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 133^\circ " data-equation-content=" \displaystyle 133^\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 = 66+(178-66)e^{kt}= 66+112e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2066%2B%28178-66%29e%5E%7Bkt%7D%3D%2066%2B112e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle T = 66+(178-66)e^{kt}= 66+112e^{kt} " data-equation-content=" \displaystyle T = 66+(178-66)e^{kt}= 66+112e^{kt} " /> . Using the point  <img class="equation_image" title=" \displaystyle (8, 153) " src="/equation_images/%20%5Cdisplaystyle%20%288%2C%20153%29%20" alt="LaTeX:  \displaystyle (8, 153) " data-equation-content=" \displaystyle (8, 153) " />  gives  <img class="equation_image" title=" \displaystyle 153= 66+112e^{k(8)} " src="/equation_images/%20%5Cdisplaystyle%20153%3D%2066%2B112e%5E%7Bk%288%29%7D%20" alt="LaTeX:  \displaystyle 153= 66+112e^{k(8)} " data-equation-content=" \displaystyle 153= 66+112e^{k(8)} " /> . Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{87}{112}=e^{8k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B87%7D%7B112%7D%3De%5E%7B8k%7D%20" alt="LaTeX:  \displaystyle \frac{87}{112}=e^{8k} " data-equation-content=" \displaystyle \frac{87}{112}=e^{8k} " /> . 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{87}{112} \right)}}{8} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B87%7D%7B112%7D%20%5Cright%29%7D%7D%7B8%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{87}{112} \right)}}{8} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{87}{112} \right)}}{8} " /> .  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 = 66+112e^{\frac{\ln{\left(\frac{87}{112} \right)}}{8}t} " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%2066%2B112e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B87%7D%7B112%7D%20%5Cright%29%7D%7D%7B8%7Dt%7D%20" alt="LaTeX:  \displaystyle T = 66+112e^{\frac{\ln{\left(\frac{87}{112} \right)}}{8}t} " data-equation-content=" \displaystyle T = 66+112e^{\frac{\ln{\left(\frac{87}{112} \right)}}{8}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle T = 112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66 " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D%20112%20%5Cleft%28%5Cfrac%7B87%7D%7B112%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B8%7D%7D%20%2B%2066%20" alt="LaTeX:  \displaystyle T = 112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66 " data-equation-content=" \displaystyle T = 112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66 " /> . 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 133=112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66 " src="/equation_images/%20%5Cdisplaystyle%20133%3D112%20%5Cleft%28%5Cfrac%7B87%7D%7B112%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B8%7D%7D%20%2B%2066%20" alt="LaTeX:  \displaystyle 133=112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66 " data-equation-content=" \displaystyle 133=112 \left(\frac{87}{112}\right)^{\frac{t}{8}} + 66 " /> .  Isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{67}{112}=\left(\frac{87}{112}\right)^{\frac{t}{8}} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B67%7D%7B112%7D%3D%5Cleft%28%5Cfrac%7B87%7D%7B112%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B8%7D%7D%20" alt="LaTeX:  \displaystyle \frac{67}{112}=\left(\frac{87}{112}\right)^{\frac{t}{8}} " data-equation-content=" \displaystyle \frac{67}{112}=\left(\frac{87}{112}\right)^{\frac{t}{8}} " /> . 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{8 \ln{\left(\frac{67}{112} \right)}}{\ln{\left(\frac{87}{112} \right)}}\approx 16 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%20%5Cfrac%7B8%20%5Cln%7B%5Cleft%28%5Cfrac%7B67%7D%7B112%7D%20%5Cright%29%7D%7D%7B%5Cln%7B%5Cleft%28%5Cfrac%7B87%7D%7B112%7D%20%5Cright%29%7D%7D%5Capprox%2016%20" alt="LaTeX:  \displaystyle t = \frac{8 \ln{\left(\frac{67}{112} \right)}}{\ln{\left(\frac{87}{112} \right)}}\approx 16 " data-equation-content=" \displaystyle t = \frac{8 \ln{\left(\frac{67}{112} \right)}}{\ln{\left(\frac{87}{112} \right)}}\approx 16 " />  minutes. </p> </p>