Halflife is closely related to exponential decay.
Halflife 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 headsup 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 halflife 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  HalfLife 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 halflife of Cobalt 60 is 5.27 years. How long will it take for 90% of the material to decay?
Problem Statement 

The halflife of Cobalt 60 is 5.27 years. How long will it take for 90% of the material to decay?
Hint 

If 90% of the material decays, how much is left?
Problem Statement 

The halflife of Cobalt 60 is 5.27 years. How long will it take for 90% of the material to decay?
Final Answer 

\( t \approx 17.51 \) years
Problem Statement 

The halflife of Cobalt 60 is 5.27 years. How long will it take for 90% of the material to decay?
Hint 

If 90% of the material decays, how much is left?
Solution 

For halflife, we use the equation \(A(t) = A_o e^{kt}\). Halflife 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 

\( t \approx 17.51 \) years 
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The halflife of radium is 1690 years. If 10 grams are present now, how much will be present in 50 years?
Problem Statement 

The halflife of radium is 1690 years. If 10 grams are present now, how much will be present in 50 years?
Final Answer 

\(\displaystyle{ A(50) = 10 e^{50(\ln 2)/1690} \approx 9.797 }\) years
Problem Statement 

The halflife of radium is 1690 years. If 10 grams are present now, how much will be present in 50 years?
Solution 

For halflife, we use the equation \(A(t) = A_o e^{kt}\). Halflife 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 

\(\displaystyle{ A(50) = 10 e^{50(\ln 2)/1690} \approx 9.797 }\) years 
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Suppose that the halflife of a certain substance is 20 days and there are initially 10 grams of the substance. How much remains after 75 days?
Problem Statement 

Suppose that the halflife of a certain substance is 20 days and there are initially 10 grams of the substance. How much remains after 75 days?
Final Answer 

The amount of the substance after 75 days is approximately \(0.743\) grams.
Problem Statement 

Suppose that the halflife of a certain substance is 20 days and there are initially 10 grams of the substance. How much remains after 75 days?
Solution 

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 halflife information to determine the constant k.
The statement that the halflife 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 

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 halflife?
Problem Statement 

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 halflife?
Final Answer 

The halflife is approximately 169.8 years.
Problem Statement 

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 halflife?
Solution 

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 

The halflife is approximately 169.8 years. 
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The halflife 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 

The halflife 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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Iodine131 has a halflife of 8 days. If there are 200 grams of this sample, how much of I131 will remain after 32 days?
Problem Statement 

Iodine131 has a halflife of 8 days. If there are 200 grams of this sample, how much of I131 will remain after 32 days?
Solution 

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Sodium24 has a halflife of 15 hours. If there are 800g of Na24 initially, how long ill it take for 750g of Na24 to decay?
Problem Statement 

Sodium24 has a halflife of 15 hours. If there are 800g of Na24 initially, how long ill it take for 750g of Na24 to decay?
Solution 

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The halflife of Oxygen15 is 2 minutes. What fraction of a sample of O15 will remain after 5 halflives?
Problem Statement 

The halflife of Oxygen15 is 2 minutes. What fraction of a sample of O15 will remain after 5 halflives?
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 halflife of element X?
Problem Statement 

It takes 35 days for a 512 gram sample of element X to decay to a final amount of 4 games. What is the halflife of element X?
Solution 

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The halflife of uranium232 is 68.9 years. How much of a 100gram sample is present after 250 years?
Problem Statement 

The halflife of uranium232 is 68.9 years. How much of a 100gram sample is present after 250 years?
Solution 

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The halflife of cobalt60 is 5.3 years. If an old sample of 10 grams has now decayed to 1 gram, how much time has passed?
Problem Statement 

The halflife of cobalt60 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 30kg sample of plutonium239 will decay by one kg in 1180 years. What is the halflife of plutonium239?
Problem Statement 

A 30kg sample of plutonium239 will decay by one kg in 1180 years. What is the halflife of plutonium239?
Solution 

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The halflife of Radium226 is 1590 years. If a sample contains 100mg, how many mg will remain after 2500 years?
Problem Statement 

The halflife of Radium226 is 1590 years. If a sample contains 100mg, how many mg will remain after 2500 years?
Solution 

This problem is solved twice, once in each video.
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The halflife of Palladium100 is 4 days. After 14 days a sample of Palladium100 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 

The halflife of Palladium100 is 4 days. After 14 days a sample of Palladium100 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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The halflife of Palladium100 is 4 days. After 12 days a sample of Palladium100 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 

The halflife of Palladium100 is 4 days. After 12 days a sample of Palladium100 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 carbon14 that is present in living trees. How long ago, to the nearest year, was the artifact made? The halflife of carbon14 is 5730 years.
Problem Statement 

A wooden artifact from an ancient tomb contains 60% of the carbon14 that is present in living trees. How long ago, to the nearest year, was the artifact made? The halflife of carbon14 is 5730 years.
Solution 

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You have 100g of a substance that has a halflife of 20 days. How long will it take for you to have 10g remaining?
Problem Statement 

You have 100g of a substance that has a halflife of 20 days. How long will it take for you to have 10g remaining?
Solution 

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A radioactive substance has a halflife 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 

A radioactive substance has a halflife 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 halflife 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 

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 halflife 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 Uranium237 has a halflife of 6.75 days. If you start with 1kg of U237, how much will have not decayed after a year?
Problem Statement 

The radioactive isotope Uranium237 has a halflife of 6.75 days. If you start with 1kg of U237, 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 halflife? Find the formula in terms of days.
Problem Statement 

A population decays exponentially from 100 to 20 in 1 week. What is its halflife? Find the formula in terms of days.
Solution 

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The halflife 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 

The halflife 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 halflife? Find the formula in terms of days.
Problem Statement 

A population decays exponentially from 100 to 20 in 1 week. What is its halflife? Find the formula in terms of days.
Solution 

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Suppose that radioactive ionX decays at a constant annual rate of 4%. What is the halflife of the substance when the initial amount is 100g?
Problem Statement 

Suppose that radioactive ionX decays at a constant annual rate of 4%. What is the halflife of the substance when the initial amount is 100g?
Final Answer 

approximately 16.98 years
Problem Statement 

Suppose that radioactive ionX decays at a constant annual rate of 4%. What is the halflife of the substance when the initial amount is 100g?
Solution 

video by PatrickJMT 

Final Answer 

approximately 16.98 years 
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What is the halflife of Radium226 if its decay rate is 0.000436? Assume time is in years.
Problem Statement 

What is the halflife of Radium226 if its decay rate is 0.000436? Assume time is in years.
Final Answer 

t = 1590 yrs
Problem Statement 

What is the halflife of Radium226 if its decay rate is 0.000436? Assume time is in years.
Solution 

video by Krista King Math 

Final Answer 

t = 1590 yrs 
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Carbon14 has a halflife 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 

Carbon14 has a halflife 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 

a) 236g;
b) 1112 years
Problem Statement 

Carbon14 has a halflife 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 

video by Khan Academy 

Final Answer 

a) 236g; 
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Find the halflife of a compound where the decay rate is 0.05. (time is in years)
Problem Statement 

Find the halflife of a compound where the decay rate is 0.05. (time is in years)
Final Answer 

t = 13.86 years
Problem Statement 

Find the halflife of a compound where the decay rate is 0.05. (time is in years)
Solution 

video by Khan Academy 

Final Answer 

t = 13.86 years 
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Really UNDERSTAND Precalculus
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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