Exponential Decay

The concept of exponential decay is one that arises frequently in science, economics, and medicine. Therefore, knowing how to solve problems involving exponential decay is a valuable skill that we can apply to a variety of academic fields. As an example, in nuclear physics, we may be given a problem where we have an initial amount of a radioactive substance. If we know the half-life of the substance, we can create an equation that predicts how much of the substance will remain after a certain number of years. We can create similar models when dealing with the economic concept of depreciation, drug concentration in the bloodstream, and much more. Before we tackle exponential decay problems, we might want to have a look at some related concepts such as exponential growth, exponential regression, and graphing functions of the form x^n. Having a firm grasp of these topics will be useful when solving exponential decay problems.

How to solve exponential decay problems

Radioactive decay is a topic where an understanding of exponential decay will be crucial. Let's look at an example problem.

We have 4000 atoms of uranium-234, an isotope of uranium with a half-life of roughly 245,500 years. That is to say, it decays at such a rate that it will degrade by half once 245,500 years have passed. We can model radioactive decay by the equation

$N = N_{0} 2^{- (\frac{t}{τ})}$

$N_{0}$ is the initial quantity of the element. The variable t represents time in years. The Greek letter tau, or τ, represents the half-life. N represents the remaining quantity once the time has elapsed.

How much will be left after 491000 years?

We're given the initial quantity, the half-life, and the amount of time elapsed.

$N_{0} = 4000 atoms$

$τ = 245500 years$

$t = 491000 years$

We can insert each of these values into our equation to calculate our answer.

$N = 4000 \times (2^{- \frac{491000}{245500}})$

$N = 4000 \times (2^{- 2})$

$N = 4000 \times 0.25$

$N = 1000$

Therefore, after 49,000 years, we'll have 1000 atoms of uranium-234 left.

We can apply the same concept to depreciation. Depreciation can be modeled by the equation

$y = C \times {(1 - \frac{x}{100})}^{t}$

C represents the initial value, x represents the depreciation rate in percent, and t stands for time. Now we have the tools we need to solve a depreciation problem.

If a company buys an industrial supercomputer for $3.4 million and the computer depreciates by 21% every year, calculate its value in 3 years.

$c = 34000$

$x = 21$

x = 21

$t = 3$

Now we insert everything into our formula.

$y = 3400000 \times {(1 - \frac{21}{100})}^{3}$

$y = 3400000 \times {(1 - 0.21)}^{3}$

$y = 3400000 \times {0.79}^{3}$

$y = 3400000 \times 0.493039$

$y = 3400000 \times 1676332.6$

After three years, our computer will depreciate to this value.

Let's look at one final example. Doctors may need information about the amount of a certain drug in a patient's bloodstream. The concentration decreases over time, which can be expressed by the equation

$y = A \times p^{t}$

A represents the initial amount of the drug. p represents the amount of a drug left in the bloodstream after a certain amount of time. t represents the amount of time that passes and y is the concentration remaining after that time has elapsed.

If a doctor gives 240 milligrams of a drug to a patient, how much will be left in the bloodstream after two hours? Assume a p-value of .40.

$y = (240) {(0.40)}^{2}$

$y = (240) (0.16)$

$= 38.4$

Therefore, there will be 38.4 milligrams of the drug left in the patient's bloodstream after two hours.

Exponential decay practice problems

1. If we purchase a car for $50,000 and it depreciates at a rate of 20% per year, how much will it be worth after 4 years?

Recall that the formula for depreciation is

$y = C {(1 - \frac{x}{100})}^{t} .$

Our givens are the original value of the car (C), the depreciation rate (x), and the amount of time that passes (t).

$C = $ 50000$

$x = 20 %$

$t = 4 years$

We insert each value into the formula like so:

$y = 50000 \times {(1 - \frac{20}{100})}^{4}$

$y = 50000 \times {(1 - 0.20)}^{4}$

$y = 50000 \times {(0.80)}^{4}$

$y = 50000 \times (0.4096)$

$= 20480$

After four years, the car will be worth $20,480.

2. If a doctor administers 10 milligrams of a painkiller to a patient, how much will be left in the patient's bloodstream after 3 hours if the p-value is .90?

Recall that the formula for drug concentration in the bloodstream is

$y = A p^{t}$

We have the original dosage (A), the p-value, and the time that passes (t).

$A = 10 milligrams$

$p = 0.90$

$t = 3 hours$

We insert each given value in the formula.

$y = (10) \times {(0.90)}^{3}$

$y = (10) \times (0.729)$

$= 7.29$

Therefore, there will be 7.29 milligrams of the painkiller in the patient's bloodstream after 3 hours.

3. If there are five million atoms of potassium-40, how many atoms will be left after 6.4 billion years? The half-life of potassium-40 is about 1.28 billion years.

Remember that the formula for radioactive decay is

$N = N_{0} \times 2^{(- \frac{t}{τ})}$

We know the starting amount (N_0), the time that passes (t), and the half-life (tau).

$N_{0} = 5000000 atoms$

$t = 6400000000 years$

$tau = 1280000000 years$

Now we plug it all in.

$N = (5000000) \times 2^{(- \frac{6400000000}{1280000000})}$

$= (5000000) \times 2^{(- 5)}$

$= (5000000) \times 0.03125$

$= 156250$

There will be 156,250 atoms of potassium-40 remaining after 6.4 billion years.

4. If we own a car that depreciates at a rate of 10% per year and it's currently worth $15,000, how much was it worth if we bought it five years ago? Round to two decimal places.

Remember the formula for depreciation is

$y = C \times {(1 - \frac{x}{100})}^{t}$

We know the car's current value (y), the depreciation rate (x), and the amount of time that elapsed since the purchase (t). We want to know the original value (C). We can solve this equation by plugging everything in, simplifying, and then rearranging it to isolate C.

$15000 = C \times {(1 - \frac{10}{100})}^{t}$

$15000 = C \times {(1 - 0.10)}^{t}$

$15000 = C \times {(0.9)}^{t}$

15,000 = C * (0.9)^5

$15000 = C \times 0.59049$

Now we divide 15,000 by .59049 to get our answer.

$C = \frac{15000}{0,.59049}$

$\approx 25402.63$

Therefore, $25,402.63 is the approximate value of the car when we bought it 5 years ago.

5. If we have a clump of 5,000 atoms of plutonium-239 that has been undergoing radioactive decay for the past 264,000 years, how many atoms were initially present? This isotope of plutonium has a half-life of approximately 24,000 years.

We start with our original formula for radioactive decay.

$N = N_{0} \times 2^{- \frac{t}{τ}}$

We have the current quantity (N), the time that has passed (t), and the half-life (τ). We need to calculate the initial quantity (N0). Let's plug everything in and work from there.

$5000 = N_{0} \times 2^{- \frac{264000}{24000}}$

$5000 = N_{0} \times 2^{(- 11)}$

Now we divide both sides by $2^{(- 11)}$ to get our answer.

$N_{0} = \frac{5000}{2^{(- 11)}}$

$N_{0} = 10240000$

This is our original quantity of plutonium-239 atoms.

6. If a patient's bloodstream contains 50 milligrams of a drug that was administered 5 hours ago, what was the initial dosage that the doctor administered? The p-value is .80. Round to two decimal places.

Our formula is

$y = A p^{t}$ .

We have the current concentration in the bloodstream (y), the p-value, and the time (t). We need the initial dosage (A).

$50 = A \times {(0.80)}^{5}$

$50 = A \times (0.32768)$

Divide each side by .32768.

$A = \frac{50}{0.32768}$

$\approx 152.59$

152.59 milligrams was the initial approximate dose.

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Graphing Functions of the Form x^n

Exponential Regression

Exponential Growth

Flashcards covering the Exponential Decay

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College Algebra Flashcards

Practice tests covering the Exponential Decay

Algebra 1 Diagnostic Tests

College Algebra Diagnostic Tests

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