## 17Calculus Precalculus - Half-Life

##### 17Calculus

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 a shortcut 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.

### 17calculus youtube playlist - precalculus half-life

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?

$$t \approx 17.51$$ 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

$$t \approx 17.51$$ 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?

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

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

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.

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?

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 $$t$$.

$$\begin{array}{rcl} 1250 & = & 2500 e^{0.1t\ln(24/25)} \\ 0.5 & = & e^{0.1t\ln(24/25)} \\ \ln(0.5) & = & 0.1t\ln(24/25) \\ t & = & \displaystyle{\frac{\ln(0.5)}{0.1\ln(24/25)}} \\ & \approx & 169.8 ~ \text{years} \end{array}$$

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 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

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 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

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 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

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 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

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 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

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 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

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 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

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 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

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 solution

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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 solution

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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 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

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 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

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 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

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 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

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 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

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 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

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 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

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 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

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 solution

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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?

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

### PatrickJMT - 672 video solution

video by PatrickJMT

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.

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

### Krista King Math - 678 video solution

video by Krista King Math

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?

a) 236g
b) 1104 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

 Carbon-14 has a half-life of 5730 years. This first line of the problem statement gives us enough information to determine the decay rate, $$k$$, in the equation $$A(t) = A_o e^{kt}$$ Half-life means the time it takes for half of the initial amount to decay. The initial amount is $$A_o$$, so the amount $$A(t) = A_o/2$$ occurs when $$t=5730$$. Plugging these into the equation gives us $$A_o/2 = A_o e^{5730k}$$ The initial amount $$A_o$$ cancels out, so we don't need to know the initial amount. Now we solve for the decay rate $$k$$. $$1/2 = e^{5730k}$$ To get $$k$$ out of the exponent, we take the natural logarithm of both sides. We chose the natural logarithm because we have $$e$$ as the base of the term $$e^{5730k}$$. $$\ln(1/2) = \ln( e^{5730k} )$$ $$-\ln 2 = 5730k$$ $$k = \dfrac{-\ln 2}{5730}$$ We will leave $$k$$ in this form for now to make our calculations more precise in the rest of the problem. a) If the initial amount is 300g, how much is left after 2000 years? Now for part a) we are given the initial amount of 300g. So $$A_o = 300$$. We know this because the word initial tells us that $$t=0$$ and so $$A(0) = A_o e^0 \to A(0) = A_o = 300$$ We are asked to find $$A(2000)$$. Plugging in what we know gives us $$A(2000) = 300e^{2000(-\ln2)/5730}$$ So our answer is $$A(2000) \approx 236$$g b) If the initial amount is 400g, when will there be 350g left? Since the initial amount is given as 400g, we know that $$A_o = 400$$ Now we are asked to find the time $$t$$ when $$A(t) = 350$$ Our partial equation is $$350 = 400 e^{kt}$$ We know $$k$$ but we will solve for $$t$$ first and then plug in our value for $$k$$. $$350/400 = e^{kt} \to \ln (350/400) = \ln(e^{kt})$$ $$\ln(350/400) = kt \to t = \ln(350/400)/k$$ Now we can plug in our exact value of $$k$$. $$t = \dfrac{\ln(350/400)}{(-\ln 2)/5730}$$ We can plug these values into our calculator at this point but it might help to simplify this a bit. $$t = \dfrac{-5730 \ln(350/400)}{\ln 2}$$ $$t = \dfrac{-5730 (\ln 350 - \ln 400)}{\ln 2}$$ $$t \approx 1104$$ years

The reason we kept $$k$$ in exact terms until the very end is that, if we found a decimal approximation too early, our final answer would have been off. This is a problem especially if you are expected to enter your answer into an online learning system.

a) 236g
b) 1104 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)

$$t = 13.86$$ years

Problem Statement

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

Solution

 The decay rate is the magnitude of $$k$$ in the equation $$A(t) = A_o e^{kt}$$. Since the problem states it the compound decays, the value of $$k$$ is negative. So $$k = -0.05$$. Since we are talking about half-life, we know that $$1/2 = e^{-0.05t}$$. Now we need to solve for $$t$$. $$\ln( 1/2 ) = \ln( e^{-0.05t} )$$ $$-\ln( 2 ) = -0.05t$$ $$\ln( 2 )/0.05 = t$$ Plugging these into a calculator, we get $$t=13.86$$ years.

$$t = 13.86$$ years

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