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Mathematics LibreTexts

3.5: Applications of the Negative Exponential Function

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    40912
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    At the beginning of Chapter \(3,\) we worked with application problems and solved them using the graphing calculator. In this section, we will revisit some of these application problems and use the solution methods discussed in the previous sections to solve these problems algebraically.
    Radioactive Decay
    The decay of a radioactive element into its non-radioactive form occurs following a time line dictated by the "half-life" of the element. The half-life is the amount of time that it takes for half of the existing radioactive material to decay to its non-radioactive form.
    Consider the equation:
    \[
    A(t)=A_{0} e^{-k t}
    \]
    where \(A(t)\) is the amount of material left at time \(t, A_{0}\) is the amount present at \(t=0,\) and \(k\) is a constant that can be determined based on the half-life of the material.

    If we know that after one half-life, there will \(50 \%\) of the radioactive material remaining, then we can say that:
    \[
    0.5 A_{0}=A_{0} e^{-k t_{h}}
    \]

    where \(t_{h}\) is the half-life. To solve this equation for \(k,\) we would first divide on both sides by \(A_{0}:\)
    \[
    \begin{array}{c}
    \frac{0.5 A_{0}}{A_{0}}=\frac{A_{0} e^{-k t_{h}}}{A_{0}} \\
    0.5=e^{-k t_{h}}
    \end{array}
    \]

    Then take the natural logarithm of both sides and bring the exponent down in front of the expression as a coefficient:
    \[
    \begin{aligned}
    0.5 &=e^{-k t_{h}} \\
    \ln (0.5) &=\ln \left(e^{-k t_{h}}\right) \\
    \ln (0.5) &=-k t_{h} * \ln (e) \\
    \ln (0.5) &=-k t_{h} * 1 \\
    -\frac{\ln (0.5)}{t_{h}} &=k
    \end{aligned}
    \]
    The value of \(k\) can then be used in the equation \(A(t)=A_{0} e^{-k t}\) to determine the amount of material left after any time \(t\)

    Example
    The isotope Gold-198 \(\left(^{198} \mathrm{Au}\right)\) is a type of gold sometimes used in medical applications and has a half-life of 2.7 days. How much of a 65 gram sample of \(^{198} \mathrm{Au}\) will be left after 6 days? How long would it take for there to be 10 grams left?
    If we know the half-life, we can calculate the value of the constant \(k\)
    \[
    \begin{array}{l}
    k=-\frac{\ln (0.5)}{t_{h}} \\
    k=-\frac{\ln (0.5)}{2.7} \\
    k \approx 0.2567
    \end{array}
    \]

    Now that we know the value of \(k\), we can directly calculate the amount of \(^{198} \mathrm{Au}\) left after 6 days:
    \[
    \begin{aligned}
    A(t) &=A_{0} e^{-k t} \\
    &=65 e^{-0.2567 * 6} \\
    & \approx 13.932
    \end{aligned}
    \]
    So, approximately 13.932 grams of \(^{198}\) Au would be left after 6 days.
    In order to calculate how long it takes for 10 grams of \(^{198}\) Au to be left, we'll need to solve for \(t\) in the equation \(A(t)=A_{0} e^{-k t}\) with \(A(t)=10:\)
    \[
    \begin{aligned}
    A(t) &=A_{0} e^{-k t} \\
    10 &=65 e^{-0.2567 t}
    \end{aligned}
    \]
    First, we'll divide on both sides by 65
    \[
    \begin{array}{l}
    \frac{10}{65}=\frac{65 e^{-0.2567 t}}{65} \\
    \frac{10}{65}=e^{-0.2567 t}
    \end{array}
    \]

    Then, take the natural logarithm on both sides:
    \[
    \ln \left(\frac{10}{65}\right)=\ln \left(e^{-0.2567 t}\right)
    \]

    We'll calculate an approximate value for \(\ln \left(\frac{10}{65}\right)\) and restate the right hand side of the equation using the properties of logarithms:
    \[
    \begin{aligned}
    -1.872 & \approx-0.2567 t * \ln e \\
    -1.872 & \approx-0.2567 t * 1 \\
    -1.872 & \approx-0.2567 t \\
    7.3 & \approx t
    \end{aligned}
    \]
    So, it would take about 7.3 days for the 65 grams of \(^{198}\) Au to decay to 10 grams.

    Newton's Law of Cooling
    Newton's Law of Cooling states that the temperature of an object can be determined using the equation:
    \[
    T=T_{a}+C e^{-k t}
    \]
    where \(T_{a}\) is the ambient temperature of the surrounding environment. The values of the constants \(C\) and \(k\) can often be calculated from given information.

    Example
    A bottle of soda at room temperature \(\left(72^{\circ} \mathrm{F}\right)\) is placed in a refrigerator where the temperature is \(44^{\circ} \mathrm{F}\)

    After half an hour, the soda has cooled to \(61^{\circ} \mathrm{F}\). What is the temperature of the soda after another half hour?

    First, since the soda is placed in the refrigerator where the ambient temperature is \(44^{\circ} \mathrm{F},\) then \(T_{a}=44 .\) We also know that at \(t=0, T=72,\) which is the temperature of the soda can when it is first put in the refrigerator. This will allow us to calculate the constant \(C\).

    \[
    \begin{aligned}
    T &=T_{a}+C e^{-k t} \\
    72 &=44+C e^{-k * 0} \\
    72 &=44+C e^{0}=44+C * 1 \\
    72 &=44+C \\
    28 &=C
    \end{aligned}
    \]
    Now we know that \(T_{a}=44\) and \(C=28 .\) We can use the other piece of information from the problem to calculate the value of \(k .\) The problem states that after 30 minutes the soda has cooled off to \(61^{\circ} \mathrm{F}\). That means that when \(t=30\) (or \(t=0.5\) depending on which units you choose) the temperature \(T\) will be \(61^{\circ} \mathrm{F}\). We can then set up the equation to reflect this and calculate the vlaue of \(k:\)
    \[
    \begin{array}{l}
    T=44+28 e^{-k t} \\
    61=44+28 e^{-k * 30}
    \end{array}
    \]

    First, we'll subtract 44 on both sides and divide by 28 :
    \[
    \begin{array}{l}
    61=44+28 e^{-k * 30} \\
    17=28 e^{-30 k} \\
    \frac{17}{28}=e^{-30 k}
    \end{array}
    \]

    Now, we'll take the natural logarithm on both sides to bring the \(-30 k\) down from the exponent:
    \[
    \begin{array}{l}
    \ln \left(\frac{17}{28}\right)=\ln \left(e^{-30 k}\right) \\
    \ln \left(\frac{17}{28}\right)=-30 k * \ln e \\
    -0.499 \approx-30 k \\
    0.01663 \approx k
    \end{array}
    \]
    Now, we have the full formula for calculating temperature in this scenario:
    \[
    T=44+28 e^{-0.01663 t}
    \]

    To find out what happens after 60 minutes, we can simply plug in 60 for \(t\)
    \[
    \begin{array}{l}
    T=44+28 e^{-0.01663 t} \\
    T=44+28 e^{-0.01663 \cdot 60} \\
    T=44+28 e^{-0.9978} \\
    T \approx 44+28 * 0.3687 \\
    T \approx 44+10.3236 \approx 54.3^{\circ} \mathrm{F}
    \end{array}
    \]

    Exercises 3.5
    1) If the half-life of radioactive cesium- 137 is 30 years, find the value of \(k\) in the equation \(A(t)=A_{0} e^{-k t}\)
    a) Given a 10 gram sample of cesium-137, how much will remain after 80 years?
    b) How long will it take for only 2 grams of cesium- 137 to remain?
    2) The half-life for radioactive thorium- 234 is about 25 days. Use this to find the value of \(k\) in the equation \(A(t)=A_{0} e^{-k t}\)
    a) How much of a 40 gram sample will remain after 60 days?
    b) After how long will only 10 grams of thorium- 234 remain?
    3) Given a sample of strontium- 90 , it is known that after 18 years there are \(32 \mathrm{mg}\) remaining and after 65 years there are \(10 \mathrm{mg}\) remaining. Use this information to find out how much strontium- 90 was in the sample to begin with, and also determine the half-life of strontium- 90 .
    4) \(\quad\) A 12 mg sample of radioactive polonium decays to \(7.26 \mathrm{mg}\) in 100 days.
    a) What is the half-life of polonium?
    b) How much of the 12 mg sample remains after 180 days?

    5) A hot bowl of soup is served at a dinner party. It starts to cool according to Newton's Law of Cooling so that its temperature at time \(t\) is given by:
    \[
    y=65+145 e^{-0.05 t}
    \]
    where \(t\) is measured in minutes and \(y\) is measured in degrees Fahrenheit.
    a) What is the initial temperature of the soup?
    b) What is the temperature after 10 minutes?
    c) After how long will the temperature be \(100^{\circ} ?\)
    6) Newton's Law of Cooling is used in homicide investigations to determine the time of death. The normal body temperature is \(98.6^{\circ} \mathrm{F}\). Immediately following death, the body begins to cool. This process uses Newton's Law of Cooling:
    \[
    y=T_{a}+C e^{-k t}
    \]
    If the ambient temperature is \(60^{\circ},\) and the body has cooled to \(72^{\circ} \mathrm{F}\) after 6 hours, use this information to determine the value of \(k\) in the equation.

    7) The police discover the body of a murder victim. Critical to solving the crime is determining when the murder was committed. The coroner arrives at the murder scene at 12 Noon. She immediately takes the temperature of the body and finds it to be \(94.6^{\circ} \mathrm{F}\). She then takes the temperature 1.5 hours later and finds it to be \(93.4^{\circ} \mathrm{F}\). If the temperature of the room is \(70^{\circ} \mathrm{F}\), when was the murder committed?

    8) A roasted turkey is taken from the oven when its temperature has reached
    \(185^{\circ} \mathrm{F}\) and is placed on a table in a room where the temperature is \(75^{\circ} \mathrm{F}\)
    a) If the temperature of the turkey is \(150^{\circ} \mathrm{F}\) after 30 minutes, what is its temperature after 45 minutes?
    b) When will the turkey cool off to \(100^{\circ} \mathrm{F}\) ?

    9) A kettle full of water is brought to a boil in a room with an ambient temperature of \(20^{\circ} \mathrm{C}\). After 15 minutes, the temperature of the water has decreased from \(100^{\circ} \mathrm{C}\) to \(75^{\circ} \mathrm{C} .\) Find an equation using Newton's Law of Cooling to represent to temperature at time \(t .\) Find the temperature of the water after 25 minutes.

    10) \(\quad\) A cup of coffee with a temperature of \(105^{\circ} \mathrm{F}\) is placed in a freezer with a temperature of \(0^{\circ} \mathrm{F}\). After 5 minutes, the temperature of the coffee is \(70^{\circ} \mathrm{F}\). Find an equation using Newton's Law of Cooling to represent to temperature at time \(t\) What will the temperature be in 10 minutes?


    3.5: Applications of the Negative Exponential Function is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Richard W. Beveridge.

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