\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Precalculus - Half-Life

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Half-life is closely related to exponential decay.

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

The half-life 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 half-life 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 half-life 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 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

\( t \approx 17.51 \) 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

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

Solution

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

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

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

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

Problem Statement

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

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

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

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

The half-life 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 half-life?

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 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

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

3059 video

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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

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

3060 video

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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

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

3061 video

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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

The half-life of Oxygen-15 is 2 minutes. What fraction of a sample of O-15 will remain after 5 half-lives?

Solution

3062 video

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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

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

3063 video

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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

The half-life of uranium-232 is 68.9 years. How much of a 100-gram sample is present after 250 years?

Solution

3064 video

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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

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

3065 video

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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

A 30-kg sample of plutonium-239 will decay by one kg in 1180 years. What is the half-life of plutonium-239?

Solution

3066 video

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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

The half-life of Radium-226 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.

3067 video

3067 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

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

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

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

3069 video

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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

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

3070 video

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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

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

3071 video

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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

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

3072 video

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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

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

3073 video

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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

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

3074 video

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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

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

Solution

3075 video

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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

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

3076 video

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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

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

Solution

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

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

approximately 16.98 years

Problem Statement

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

672 video

video by PatrickJMT

Final Answer

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

What is the half-life of Radium-226 if its decay rate is 0.000436? Assume time is in years.

Final Answer

t = 1590 yrs

Problem Statement

What is the half-life of Radium-226 if its decay rate is 0.000436? Assume time is in years.

Solution

678 video

video by Krista King Math

Final Answer

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

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

a) 236g;
b) 1112 years

Problem Statement

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

679 video

video by Khan Academy

Final Answer

a) 236g;
b) 1112 years

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Find the half-life of a compound where the decay rate is 0.05. (time is in years)

Problem Statement

Find the half-life of a compound where the decay rate is 0.05. (time is in years)

Final Answer

t = 13.86 years

Problem Statement

Find the half-life of a compound where the decay rate is 0.05. (time is in years)

Solution

680 video

video by Khan Academy

Final Answer

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

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 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle 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

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

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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