17Calculus Precalculus - Half-Life
Half-life is closely related to exponential decay.
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Half-life is the time it takes for half the substance to decay and, therefore, is related only to exponential decay, not growth. The idea is to take the equation \(y = Ae^{kt}\), set the left side to \(A/2\) and solve for \(t\). Notice that you don't have to know the initial amount \(A\) since in the equation \(A/2 = Ae^{kt}\), the \(A\) cancels leaving \(1/2 = e^{kt}\). You can then use basic logarithms to solve for \(t\).
Notes and Warning About Practice Problems
Before you start working the practice problems, we would like to give you a heads-up on what to expect.
1. Warning!
If you search the internet a little bit or watch a few videos on YouTube, you will find that there is actually an equation that can be used to find the value of \(k\). We do not recommend that you use that equation. Using the equations that we give here as a starting point for your problems will help you hone your skills with exponentials and logarithms. Those skills will be vital when you reach calculus when you will not usually have a set of equations to choose from to solve problems. Making things easy now will only cause problems later.
2. Some of the practice problem videos do not give the units as part of the answer. Make sure you always give units with your answers and, as usual, check with your instructor to see what they expect.
Whew! While searching YouTube for practice problems, we found several ways that instructors showed to solve half-life problems. We also saw various types of notation and several tricks that were not helpful. We have, therefore, built a YouTube playlist that we believe shows the most common problems with the best solutions. Now, that doesn't mean your teacher is going to expect you to work problems the same way they do in these videos, so you need to check with them to see what they expect. However, IN OUR OPINION, the videos in our 17Calculus YouTube channel under Precalculus - Half-Life are the best and follow the most standard procedure.
Okay, keeping all that in mind, you are ready to work the practice problems.
Practice
The half-life of Cobalt 60 is 5.27 years. How long will it take for 90% of the material to decay?
Problem Statement |
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The half-life of Cobalt 60 is 5.27 years. How long will it take for 90% of the material to decay?
Hint |
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If 90% of the material decays, how much is left?
Problem Statement |
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The half-life of Cobalt 60 is 5.27 years. How long will it take for 90% of the material to decay?
Final Answer |
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\( t \approx 17.51 \) years
Problem Statement |
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The half-life of Cobalt 60 is 5.27 years. How long will it take for 90% of the material to decay?
Hint |
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If 90% of the material decays, how much is left?
Solution |
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For half-life, we use the equation \(A(t) = A_o e^{kt}\). Half-life means that half of the material is gone in 5.27 years. They didn't give us the initial amount but we don't need it to determine k.
To find k, let \(A(t)=A_o/2\), \(t=5.27\) and solve for k.
\(\begin{array}{rcl} \displaystyle{ \frac{A_o}{2} } & = & A_o e^{k(5.27)} \\ 1/2 & = & e^{5.27k} \\ \ln(1/2) & = & 5.27k \\ \displaystyle{ \frac{-\ln 2}{5.27} } & = & k \end{array} \)
So our general equation is \(\displaystyle{ A(t) = A_o e^{-t(\ln 2)/5.27} }\).
When 90% of the material is gone, 10% remains.
\( 0.10 = e^{-t(\ln 2)/5.27} \) |
\( \ln(0.10) = -t(\ln 2)/5.27 \) |
\( 5.27\ln(0.10) = -t(\ln 2) \) |
\( 5.27\ln(0.10)/(\ln 2) = -t \) |
\( -5.27\ln(0.10)/(\ln 2) = t \) |
\( t \approx 17.51 \) years |
Final Answer |
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\( t \approx 17.51 \) years |
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The half-life of radium is 1690 years. If 10 grams are present now, how much will be present in 50 years?
Problem Statement |
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The half-life of radium is 1690 years. If 10 grams are present now, how much will be present in 50 years?
Final Answer |
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\(\displaystyle{ A(50) = 10 e^{-50(\ln 2)/1690} \approx 9.797 }\) years
Problem Statement |
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The half-life of radium is 1690 years. If 10 grams are present now, how much will be present in 50 years?
Solution |
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For half-life, we use the equation \(A(t) = A_o e^{kt}\). Half-life means that half of the material is gone in 1690 years. They give us the initial amount but we don't need it to determine k. We will need it later to determine the final amount in 50 years, though.
To find k, let \(A(t)=A_o/2\), \(t=1690\) and solve for k.
\(\begin{array}{rcl} \displaystyle{ \frac{A_o}{2} } & = & A_o e^{k(1690)} \\ 1/2 & = & e^{1690k} \\ \ln(1/2) & = & 1690k \\ \displaystyle{ \frac{-\ln 2}{1690} } & = & k \end{array} \)
So our general equation is \(\displaystyle{ A(t) = A_o e^{-t(\ln 2)/1690} }\). To find the quantity present in 50 years, we let \(t=50\) in this equation to get
\(\displaystyle{ A(50) = 10 e^{-50(\ln 2)/1690} \approx 9.797 }\) years.
Final Answer |
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\(\displaystyle{ A(50) = 10 e^{-50(\ln 2)/1690} \approx 9.797 }\) years |
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Suppose that the half-life of a certain substance is 20 days and there are initially 10 grams of the substance. How much remains after 75 days?
Problem Statement |
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Suppose that the half-life of a certain substance is 20 days and there are initially 10 grams of the substance. How much remains after 75 days?
Final Answer |
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The amount of the substance after 75 days is approximately \(0.743\) grams.
Problem Statement |
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Suppose that the half-life of a certain substance is 20 days and there are initially 10 grams of the substance. How much remains after 75 days?
Solution |
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The equation we use is \( A(t) = A_0 e^{kt} \) where
\(A(t)\) is the amount of the substance at time t in grams
\(A_0\) is the initial amount in grams
k is the decay rate constant
time t is in days
From the problem statement, we know that \(A_0 = 10g\). In order to answer the question about how much remains after 75 days, we use the half-life information to determine the constant k.
The statement that the half-life of the substance is 20 days tells us that in 20 days, half of the initial amount remains. That would be \(10/2=5\) grams at time \(t=20\) days. We can substitute both of those values into the original \(A(t)\) equation and see if it helps us.
\( 5 = 10 e^{20k} \)
Notice that we are left with just one unknown here, k, so we can solve for it.
\(\displaystyle{\begin{array}{rcl} 5 & = & 10 e^{20k} \\ 1/2 & = & e^{20k} \\ \ln(1/2) & = & \ln(e^{20k}) \\ \ln(1) - \ln(2) & = & (20k) \ln(e) \\ -\ln(2) & = & 20k \\ -\ln(2)/20 & = & k \end{array} }\)
Now that we know the value of k, our equation is \(\displaystyle{ A(t) = 10 e^{-t\ln(2)/20} }\)
So, we have an equation that tells us the amount of the substance at every time t. To determine the amount of the substance after 75 days, we just let \(t=75\) in this last equation. This gives us
\(\displaystyle{ A(75) = 10 e^{-75\ln(2)/20} \approx 0.743 }\) grams.
See the logarithms page for a review of the logarithm rules we used above.
Final Answer |
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The amount of the substance after 75 days is approximately \(0.743\) grams. |
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Suppose a material decays at a rate proportional to the quantity of the material and there were 2500 grams 10 years ago. If there are 2400 grams now, what is the half-life?
Problem Statement |
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Suppose a material decays at a rate proportional to the quantity of the material and there were 2500 grams 10 years ago. If there are 2400 grams now, what is the half-life?
Final Answer |
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The half-life is approximately 169.8 years.
Problem Statement |
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Suppose a material decays at a rate proportional to the quantity of the material and there were 2500 grams 10 years ago. If there are 2400 grams now, what is the half-life?
Solution |
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Since the material decays proportional to the quantity of the material, the equation we need is \(A(t) = A_0e^{kt}\). With the given information we need to determine the decay rate, k. Then use that to help us determine the time \(t\) when the quantity is \((1/2)A_0\) (since we need to know the HALF life, i.e. the time when half the material remains).
Given - - at \(t=0\), \(A=2500g\) so \(A_0 = 2500\)
Also given - - \(t=10\), \(A(10) = 2400g\)
Use this to determine k.
\(\begin{array}{rcl} 2400 & = & 2500 e^{k(10)} \\ \displaystyle{\frac{24}{25}} & = & e^{10k} \\ \ln(24/25) & = & 10k \\ 0.1\ln(24/25) & = & k \end{array} \)
Half of the initial amount is \(2500/2 = 1250\), so we have \(\displaystyle{ 1250 = 2500 e^{0.1t\ln(24/25)} }\)
and we need to solve for \(t\).
\(\begin{array}{rcl} 1250 & = & 2500 e^{0.1t\ln(24/25)} \\ 0.5 & = & e^{0.1t\ln(24/25)} \\ \ln(0.5) & = & 0.1t\ln(24/25) \\ t & = & \displaystyle{\frac{\ln(0.5)}{0.1\ln(24/25)}} \\ & \approx & 169.8 ~ \text{years} \end{array} \)
Final Answer |
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The half-life is approximately 169.8 years. |
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The half-life of element X is 8 days. How long will it take for 100g of element X to decay such that only 12.5g of element X remains?
Problem Statement |
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The half-life of element X is 8 days. How long will it take for 100g of element X to decay such that only 12.5g of element X remains?
Solution |
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Iodine-131 has a half-life of 8 days. If there are 200 grams of this sample, how much of I-131 will remain after 32 days?
Problem Statement |
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Iodine-131 has a half-life of 8 days. If there are 200 grams of this sample, how much of I-131 will remain after 32 days?
Solution |
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Sodium-24 has a half-life of 15 hours. If there are 800g of Na-24 initially, how long ill it take for 750g of Na-24 to decay?
Problem Statement |
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Sodium-24 has a half-life of 15 hours. If there are 800g of Na-24 initially, how long ill it take for 750g of Na-24 to decay?
Solution |
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The half-life of Oxygen-15 is 2 minutes. What fraction of a sample of O-15 will remain after 5 half-lives?
Problem Statement |
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The half-life of Oxygen-15 is 2 minutes. What fraction of a sample of O-15 will remain after 5 half-lives?
Solution |
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It takes 35 days for a 512 gram sample of element X to decay to a final amount of 4 games. What is the half-life of element X?
Problem Statement |
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It takes 35 days for a 512 gram sample of element X to decay to a final amount of 4 games. What is the half-life of element X?
Solution |
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The half-life of uranium-232 is 68.9 years. How much of a 100-gram sample is present after 250 years?
Problem Statement |
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The half-life of uranium-232 is 68.9 years. How much of a 100-gram sample is present after 250 years?
Solution |
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The half-life of cobalt-60 is 5.3 years. If an old sample of 10 grams has now decayed to 1 gram, how much time has passed?
Problem Statement |
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The half-life of cobalt-60 is 5.3 years. If an old sample of 10 grams has now decayed to 1 gram, how much time has passed?
Solution |
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A 30-kg sample of plutonium-239 will decay by one kg in 1180 years. What is the half-life of plutonium-239?
Problem Statement |
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A 30-kg sample of plutonium-239 will decay by one kg in 1180 years. What is the half-life of plutonium-239?
Solution |
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The half-life of Radium-226 is 1590 years. If a sample contains 100mg, how many mg will remain after 2500 years?
Problem Statement |
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The half-life of Radium-226 is 1590 years. If a sample contains 100mg, how many mg will remain after 2500 years?
Solution |
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This problem is solved twice, once in each video.
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The half-life of Palladium-100 is 4 days. After 14 days a sample of Palladium-100 has been reduced to a mass of 8mg.
a. What was the initial mass (in mg) of the sample?
b. What is the mass 6 weeks after the start?
Problem Statement |
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The half-life of Palladium-100 is 4 days. After 14 days a sample of Palladium-100 has been reduced to a mass of 8mg.
a. What was the initial mass (in mg) of the sample?
b. What is the mass 6 weeks after the start?
Solution |
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3068 video
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The half-life of Palladium-100 is 4 days. After 12 days a sample of Palladium-100 has been reduced to a mass go 6mg.
a. What was the initial pass (in mg) of the sample?
b. What is the mass (in mg) 7 weeks after the start?
Problem Statement |
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The half-life of Palladium-100 is 4 days. After 12 days a sample of Palladium-100 has been reduced to a mass go 6mg.
a. What was the initial pass (in mg) of the sample?
b. What is the mass (in mg) 7 weeks after the start?
Solution |
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A wooden artifact from an ancient tomb contains 60% of the carbon-14 that is present in living trees. How long ago, to the nearest year, was the artifact made? The half-life of carbon-14 is 5730 years.
Problem Statement |
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A wooden artifact from an ancient tomb contains 60% of the carbon-14 that is present in living trees. How long ago, to the nearest year, was the artifact made? The half-life of carbon-14 is 5730 years.
Solution |
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You have 100g of a substance that has a half-life of 20 days. How long will it take for you to have 10g remaining?
Problem Statement |
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You have 100g of a substance that has a half-life of 20 days. How long will it take for you to have 10g remaining?
Solution |
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A radioactive substance has a half-life of 50 days. After 20 days, only 30g were left. Assuming that the radioactive substance decays exponentially, find the initial amount of the substance.
Problem Statement |
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A radioactive substance has a half-life of 50 days. After 20 days, only 30g were left. Assuming that the radioactive substance decays exponentially, find the initial amount of the substance.
Solution |
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At the beginning of an experiment, a scientist has 200 grams of a radioactive substance. After 40 minutes, her sample has decayed to 15.2 grams.
a. What is the half-life of the substance in minutes?
b. Find a formula for the amount of the substance remaining at time t.
c. How many grams will remain after 71 minutes?
Problem Statement |
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At the beginning of an experiment, a scientist has 200 grams of a radioactive substance. After 40 minutes, her sample has decayed to 15.2 grams.
a. What is the half-life of the substance in minutes?
b. Find a formula for the amount of the substance remaining at time t.
c. How many grams will remain after 71 minutes?
Solution |
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The radioactive isotope Uranium-237 has a half-life of 6.75 days. If you start with 1kg of U-237, how much will have not decayed after a year?
Problem Statement |
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The radioactive isotope Uranium-237 has a half-life of 6.75 days. If you start with 1kg of U-237, how much will have not decayed after a year?
Solution |
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A population decays exponentially from 100 to 20 in 1 week. What is its half-life? Find the formula in terms of days.
Problem Statement |
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A population decays exponentially from 100 to 20 in 1 week. What is its half-life? Find the formula in terms of days.
Solution |
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The half-life of iodine 123 is 13 hours. How long will it take for 1000 mg to decay to 100mg? How long will it take to decay to 10% of its initial amount?
Problem Statement |
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The half-life of iodine 123 is 13 hours. How long will it take for 1000 mg to decay to 100mg? How long will it take to decay to 10% of its initial amount?
Solution |
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A population decays exponentially from 100 to 20 in 1 week. What is its half-life? Find the formula in terms of days.
Problem Statement |
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A population decays exponentially from 100 to 20 in 1 week. What is its half-life? Find the formula in terms of days.
Solution |
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3077 video
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Suppose that radioactive ion-X decays at a constant annual rate of 4%. What is the half-life of the substance when the initial amount is 100g?
Problem Statement |
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Suppose that radioactive ion-X decays at a constant annual rate of 4%. What is the half-life of the substance when the initial amount is 100g?
Final Answer |
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approximately 16.98 years
Problem Statement |
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Suppose that radioactive ion-X decays at a constant annual rate of 4%. What is the half-life of the substance when the initial amount is 100g?
Solution |
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672 video
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Final Answer |
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approximately 16.98 years |
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What is the half-life of Radium-226 if its decay rate is 0.000436? Assume time is in years.
Problem Statement |
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What is the half-life of Radium-226 if its decay rate is 0.000436? Assume time is in years.
Final Answer |
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t = 1590 yrs
Problem Statement |
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What is the half-life of Radium-226 if its decay rate is 0.000436? Assume time is in years.
Solution |
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678 video
video by Krista King Math |
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Final Answer |
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t = 1590 yrs |
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Carbon-14 has a half-life of 5730 years.
a) If the initial amount is 300g, how much is left after 2000 years?
b) If the initial amount is 400g, when will there be 350g left?
Problem Statement |
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Carbon-14 has a half-life of 5730 years.
a) If the initial amount is 300g, how much is left after 2000 years?
b) If the initial amount is 400g, when will there be 350g left?
Final Answer |
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a) 236g;
b) 1112 years
Problem Statement |
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Carbon-14 has a half-life of 5730 years.
a) If the initial amount is 300g, how much is left after 2000 years?
b) If the initial amount is 400g, when will there be 350g left?
Solution |
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679 video
video by Khan Academy |
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Final Answer |
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a) 236g; |
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Find the half-life of a compound where the decay rate is 0.05. (time is in years)
Problem Statement |
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Find the half-life of a compound where the decay rate is 0.05. (time is in years)
Final Answer |
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t = 13.86 years
Problem Statement |
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Find the half-life of a compound where the decay rate is 0.05. (time is in years)
Solution |
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680 video
video by Khan Academy |
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Final Answer |
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t = 13.86 years |
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Really UNDERSTAND Precalculus
Topics You Need To Understand For This Page
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
|---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
|---|---|
\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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